First write the two numbers in base‑10.

\[
17_{b}=1\cdot b+7=b+7,\qquad 
97_{b}=9\cdot b+7=9b+7 .
\]

We need  

\[
b+7 \mid 9b+7 .
\]

Use the Euclidean algorithm:

\[
9b+7 = 9(b+7)-56 .
\]

Hence  

\[
b+7 \mid 9b+7 \iff b+7 \mid 56 .
\]

So \(b+7\) must be a positive divisor of \(56\).

Since the base must satisfy \(b>9\), we have \(b+7>16\).  
The divisors of \(56\) exceeding \(16\) are \(28\) and \(56\).

\[
\begin{cases}
b+7=28 \ \Rightarrow\ b=21,\\[2pt]
b+7=56 \ \Rightarrow\ b=49.
\end{cases}
\]

Both bases indeed work:

\[
\frac{97_{21}}{17_{21}}=\frac{196}{28}=7,\qquad 
\frac{97_{49}}{17_{49}}=\frac{448}{56}=8 .
\]

Therefore the required bases are \(21\) and \(49\), and their sum is  

\[
21+49 = 70.
\]

\[
\boxed{70}
\]

Let  

\[
c=\#\text{ chocolate},\qquad v=\#\text{ vanilla},\qquad s=\#\text{ strawberry}
\]

with  

\[
c+v+s =9,\qquad c>v>s\ge1 .
\]

---

### 1. Find all possible triples \((c,v,s)\).

Since \(c>v>s\ge 1\),

\[
c\ge s+2,\; v\ge s+1 \Longrightarrow c+v+s\ge 3s+3\le9\Longrightarrow s\le2 .
\]

- **If \(s=1\)**: \(v\ge2,\ c\ge3\) and \(c+v=8\).  
  Possibilities: \((c,v)=(6,2),(5,3)\).

- **If \(s=2\)**: \(v\ge3,\ c\ge4\) and \(c+v=7\).  
  Possibility: \((c,v)=(4,3)\).

Thus the only triples are  

\[
(6,2,1),\quad (5,3,1),\quad (4,3,2).
\]

---

### 2. Count assignments for each triple.

For a given \((c,v,s)\) the number of assignments of flavors to the nine distinct players is the multinomial coefficient  

\[
\frac{9!}{c!\,v!\,s!}.
\]

\[
\begin{aligned}
(6,2,1):\quad &\frac{9!}{6!\,2!\,1!}= \binom{9}{6}\binom{3}{2}=84\cdot3=252,\\[4pt]
(5,3,1):\quad &\frac{9!}{5!\,3!\,1!}= \binom{9}{5}\binom{4}{3}=126\cdot4=504,\\[4pt]
(4,3,2):\quad &\frac{9!}{4!\,3!\,2!}= \binom{9}{4}\binom{5}{3}=126\cdot10=1260.
\end{aligned}
\]

---

### 3. Total number of assignments

\[
N = 252+504+1260 = 2016.
\]

---

### 4. Remainder modulo \(1000\)

\[
N \bmod 1000 = 2016 \bmod 1000 = 16.
\]

\[
\boxed{16}
\]

Let the trapezoid have parallel sides \(r\) and \(s\) \((r\neq s)\) and equal legs \(L\).

---

### 1.  Semiperimeter from the incircle

A convex quadrilateral that contains an incircle (a tangential quadrilateral) satisfies  

\[
\text{Area}= (\text{inradius})\times (\text{semiperimeter}).
\]

The inradius is \(3\) and the area is \(72\); therefore  

\[
p=\frac{\text{Area}}{r_{\text{in}}}= \frac{72}{3}=24,
\]

where \(p\) is the semiperimeter.

---

### 2.  Relation among the sides

For a tangential quadrilateral, Pitot’s theorem gives  

\[
\text{(sum of one pair of opposite sides)} = \text{(sum of the other pair)} .
\]

In an isosceles trapezoid the opposite sides are the two bases and the two equal legs, so  

\[
r+s = L+L =2L \quad\Longrightarrow\quad L=\frac{r+s}{2}.
\]

The semiperimeter is  

\[
p = \frac{r+s+2L}{2}= \frac{r+s+r+s}{2}=r+s .
\]

Since \(p=24\), we obtain  

\[
\boxed{r+s=24}.
\]

Consequently  

\[
L=\frac{r+s}{2}=12 .
\]

---

### 3.  Height of the trapezoid

The incircle is tangent to both parallel sides, so the distance between those sides (the height \(h\)) equals twice the radius:

\[
h = 2\cdot 3 = 6 .
\]

---

### 4.  Using the leg length

Place the trapezoid with its bases horizontal. The horizontal offset between the ends of the bases is \(\dfrac{|r-s|}{2}\). Hence, by the Pythagorean theorem,

\[
L^{2}= \left(\frac{r-s}{2}\right)^{2}+h^{2}.
\]

With \(L=12\) and \(h=6\),

\[
12^{2}= \left(\frac{r-s}{2}\right)^{2}+6^{2}
\quad\Longrightarrow\quad
144 = \frac{(r-s)^{2}}{4}+36
\quad\Longrightarrow\quad
(r-s)^{2}=432.
\]

---

### 5.  Compute \(r^{2}+s^{2}\)

\[
\begin{aligned}
(r+s)^{2} &= r^{2}+2rs+s^{2}=24^{2}=576,\\
(r-s)^{2} &= r^{2}-2rs+s^{2}=432.
\end{aligned}
\]

Adding these equations eliminates the \(rs\) term:

\[
(r+s)^{2}+(r-s)^{2}=2(r^{2}+s^{2})\Longrightarrow
576+432=2(r^{2}+s^{2}).
\]

Thus

\[
r^{2}+s^{2}= \frac{1008}{2}=504.
\]

---

\[
\boxed{504}
\]

The first equation  

\[
|25+20i-z|=5
\]

is a circle with centre  

\[
C=(25,20)
\]

and radius \(r=5\).

The second equation  

\[
|z-4-k| = |z-3i-k|
\]

means that \(z\) is equidistant from the points  

\[
A=4+k \qquad\text{and}\qquad B=3i+k .
\]

Thus \(A=(k+4,0),\; B=(k,3)\).  
The set of points equidistant from \(A\) and \(B\) is the perpendicular bisector of \(\overline{AB}\).

---

### Equation of the perpendicular bisector

Midpoint of \(AB\):

\[
M=\Bigl(\frac{k+4+k}{2},\frac{0+3}{2}\Bigr)=(k+2,\frac32).
\]

The vector \(\overrightarrow{AB}=(-4,3)\).  
A direction vector perpendicular to this is \((3,4)\) (since \((-4,3)\cdot(3,4)=0\)).  

Hence the line has slope \(\frac43\) and equation  

\[
y-\frac32 = \frac43\bigl(x-(k+2)\bigr)
\]
\[
\Longrightarrow\; y = \frac43 x - \frac43 k - \frac76 .
\]

In standard form:

\[
8x-6y = 8k+7 .
\]

---

### Tangency condition

The system will have exactly one solution iff this line is tangent to the circle.  
The distance from the centre \(C=(25,20)\) to the line \(8x-6y-(8k+7)=0\) must equal the radius \(5\):

\[
\frac{|8\cdot25-6\cdot20-(8k+7)|}{\sqrt{8^{2}+(-6)^{2}}}=5.
\]

Compute the numerator:

\[
8\cdot25-6\cdot20-(8k+7)=200-120-8k-7=73-8k.
\]

Since \(\sqrt{8^{2}+6^{2}}=10\),

\[
\frac{|73-8k|}{10}=5 \quad\Longrightarrow\quad |73-8k|=50.
\]

Thus  

\[
73-8k = 50 \quad\text{or}\quad 73-8k = -50.
\]

\[
\begin{cases}
8k = 23 \;\;\Rightarrow\; k = \dfrac{23}{8},\\[4pt]
8k = 123\;\Rightarrow\; k = \dfrac{123}{8}.
\end{cases}
\]

Both values give a tangent line, so each yields exactly one solution \(z\).

---

### Sum of all possible \(k\)

\[
k_1+k_2 = \frac{23}{8}+\frac{123}{8}= \frac{146}{8}= \frac{73}{4}.
\]

Write \(\frac{73}{4}\) as \(\frac{m}{n}\) with \(\gcd(m,n)=1\): \(m=73,\ n=4\).

\[
m+n = 73+4 = 77.
\]

\[
\boxed{77}
\]

The equation  

\[
12x^{2}-xy-6y^{2}=0
\]

is a homogeneous quadratic form.  Factor it over the rationals.

\[
12x^{2}-xy-6y^{2}= -6\Bigl(y-\frac43x\Bigr)\Bigl(y+\frac32x\Bigr)
\]

Hence  

\[
12x^{2}-xy-6y^{2}=0\iff 
\Bigl(y-\frac43x\Bigr)\Bigl(y+\frac32x\Bigr)=0 .
\]

So any integer solution must satisfy one of the two linear equations  

\[
y=\frac{4}{3}x\quad\text{or}\quad y=-\frac32 x .
\]

---

### 1.  Solutions of \(y=\frac43x\)

Write \(3y=4x\).  
Since \(\gcd(3,4)=1\), the integer solutions are  

\[
x=3k,\qquad y=4k\qquad(k\in\mathbb Z).
\]

Both coordinates must lie in \([-100,100]\):

\[
-100\le 3k\le100,\qquad -100\le4k\le100
\]

The second inequality is stricter, giving  

\[
-25\le k\le 25 .
\]

Thus there are  

\[
25-(-25)+1 = 51
\]

integer points on this line.

---

### 2.  Solutions of \(y=-\frac32x\)

Write \(2y=-3x\).  
Since \(\gcd(2,3)=1\), the integer solutions are  

\[
x=2k,\qquad y=-3k\qquad(k\in\mathbb Z).
\]

Both coordinates must satisfy  

\[
-100\le2k\le100,\qquad -100\le-3k\le100 .
\]

These give  

\[
-50\le k\le50,\qquad -\frac{100}{3}\le k\le\frac{100}{3}\Longrightarrow -33\le k\le33 .
\]

Hence there are  

\[
33-(-33)+1 = 67
\]

integer points on this line.

---

### 3.  Remove double‑counted point

The two lines intersect only at \((0,0)\) (when \(k=0\) in both families).  
Thus the total number of distinct ordered pairs is  

\[
51+67-1 = 117 .
\]

\[
\boxed{117}
\]

**Step 1.  Divisibility conditions**

For an integer to be divisible by \(22\) it must be divisible by \(2\) and by \(11\).

*Divisible by 2* – the last digit (the units digit) must be even, i.e. one of \(\{2,4,6,8\}\).

*Divisible by 11* – for an 8‑digit number \(a_1a_2\ldots a_8\) (with \(a_1\) the leftmost digit) the rule is  

\[
(a_8+a_6+a_4+a_2)-(a_7+a_5+a_3+a_1)\equiv0\pmod{11}.
\]

Thus the sum of the digits in the even positions (2,4,6,8) equals the sum of the digits in the odd positions (1,3,5,7) modulo \(11\).

The total sum of the digits \(1+2+\dots+8=36\).  
If the two sums are equal modulo \(11\), then

\[
2\cdot(\text{odd–position sum})\equiv36\equiv3\pmod{11}
\Longrightarrow\text{odd–position sum}\equiv7\pmod{11}.
\]

The possible values are \(7,18,29,\ldots\).  
Since four distinct digits from \(\{1,\dots ,8\}\) can sum only between \(10\) and \(26\), the only feasible value is  

\[
\boxed{\text{odd–position sum}=18}.
\]

Consequently the even–position sum is also \(18\).

**Step 2.  Choose the digits for the odd positions**

We need all 4‑element subsets of \(\{1,2,3,4,5,6,7,8\}\) whose sum is \(18\).

Listing them (or by noting the complementary pairs) gives  

\[
\begin{aligned}
&\{1,2,7,8\},\; \{1,3,6,8\},\; \{1,4,5,8\},\; \{1,4,6,7\},\\
&\{2,3,5,8\},\; \{2,3,6,7\},\; \{2,4,5,7\},\; \{3,4,5,6\}.
\end{aligned}
\]

There are  

\[
\boxed{M=8}
\]

such subsets.  
Each of them contains exactly two even digits (because a sum of 18 cannot be obtained with a different parity distribution), so for every subset \(A\) (odd‑position digits) we have  

\[
e_A=2,\qquad E_B=4-e_A=2,
\]

where \(E_B\) is the number of even digits among the complementary set \(B\) (the even‑position digits).

**Step 3.  Arrange the digits**

*Odd positions* (1,3,5,7): the four digits of \(A\) can be placed in \(4! = 24\) ways.

*Even positions* (2,4,6,8): the four digits of \(B\) must be placed so that the digit in position 8 (the units digit) is even.  
Since \(B\) contains exactly \(E_B=2\) even digits, we have  

\[
\text{choices for the units digit}=2,
\qquad
\text{ways to arrange the remaining three digits}=3! = 6.
\]

Thus the number of admissible placements for \(B\) is  

\[
f(B)=2\cdot 6=12.
\]

**Step 4.  Count all numbers**

For each of the \(M=8\) valid subsets \(A\),

\[
\text{numbers contributed}=24\times12=288.
\]

Hence  

\[
N=8\times288=2304.
\]

**Step 5.  Compute the required difference**

\[
N-2025=2304-2025=279.
\]

\[
\boxed{279}
\]

The original parabola is  

\[
y=x^{2}-4 .
\]

---

### 1.  Equation of the rotated parabola  

Rotate the plane \(60^{\circ}\) counter‑clockwise about the origin.
If \((X,Y)\) is a point of the rotated parabola, then rotating it back
\(60^{\circ}\) clockwise gives a point \((x',y')\) on the original parabola.

\[
\begin{pmatrix}x'\\y'\end{pmatrix}
=
\begin{pmatrix}\cos60^\circ&\sin60^\circ\\
-\sin60^\circ&\cos60^\circ\end{pmatrix}
\begin{pmatrix}X\\Y\end{pmatrix}
=
\begin{pmatrix}\frac12X+\frac{\sqrt3}{2}Y\\[2pt]
-\frac{\sqrt3}{2}X+\frac12Y\end{pmatrix}.
\]

Thus  

\[
y' = x'^2-4\Longrightarrow 
\frac{-\sqrt3X+Y}{2} =\Bigl(\frac{X+\sqrt3Y}{2}\Bigr)^{\!2}-4.
\]

Multiplying by \(2\) and simplifying gives the equation of the rotated
parabola:

\[
(x+\sqrt3\,y)^{2}+2\sqrt3\,x-2y-16=0. \tag{1}
\]

---

### 2.  Intersection with the original parabola  

Substituting \(y=x^{2}-4\) from the original parabola into (1) yields

\[
\bigl(x+\sqrt3(x^{2}-4)\bigr)^{2}+2\sqrt3\,x-2(x^{2}-4)-16=0,
\]

which simplifies to

\[
3x^{4}+2\sqrt3\,x^{3}-25x^{2}-6\sqrt3\,x+40=0. \tag{2}
\]

---

### 3.  Factoring the quartic  

Look for a factorisation of the form  
\((ax^{2}+b\sqrt3\,x+c)(dx^{2}+e\sqrt3\,x+f)\).
Matching coefficients gives the system  

\[
\begin{cases}
ad=3\\
ae+bd=2\\
af+3be+cd=-25\\
bf+ce=-6\\
cf=40
\end{cases}
\]

which is solved by  

\[
a=1,\;b=1,\;c=-4,\qquad d=3,\;e=-1,\;f=-10 .
\]

Hence

\[
3x^{4}+2\sqrt3\,x^{3}-25x^{2}-6\sqrt3\,x+40
   =(x^{2}+\sqrt3\,x-4)(3x^{2}-\sqrt3\,x-10). \tag{3}
\]

---

### 4.  Solving for the intersection points  

From (3) we obtain four possible \(x\)-coordinates.

1. \(x^{2}+\sqrt3\,x-4=0\)

\[
x=\frac{-\sqrt3\pm\sqrt{19}}{2}.
\]

2. \(3x^{2}-\sqrt3\,x-10=0\)

\[
x=\frac{\sqrt3\pm\sqrt{123}}{6}.
\]

For each \(x\) we compute \(y=x^{2}-4\).

| \(x\) | sign of \(x\) | \(y=x^{2}-4\) | quadrant |
|---|---|---|---|
| \(\displaystyle\frac{-\sqrt3+\sqrt{19}}{2}\) | \(>0\) | \(\displaystyle\frac{3-\sqrt{57}}{2}<0\) | **IV** |
| \(\displaystyle\frac{-\sqrt3-\sqrt{19}}{2}\) | \(<0\) | \(>0\) | II |
| \(\displaystyle\frac{\sqrt3+\sqrt{123}}{6}\) | \(>0\) | \(>0\) | I |
| \(\displaystyle\frac{\sqrt3-\sqrt{123}}{6}\) | \(<0\) | \(<0\) | III |

Only the first solution lies in the fourth quadrant, as required.

---

### 5.  The required \(y\)-coordinate  

For  
\[
x=\frac{-\sqrt3+\sqrt{19}}{2},
\]
\[
y=x^{2}-4
   =\left(\frac{-\sqrt3+\sqrt{19}}{2}\right)^{2}-4
   =\frac{19+3-2\sqrt{57}}{4}-4
   =\frac{22-2\sqrt{57}}{4}-\frac{16}{4}
   =\frac{3-\sqrt{57}}{2}.
\]

Thus  

\[
y=\frac{a-\sqrt{b}}{c}\quad\text{with}\quad a=3,\;b=57,\;c=2,
\]
and \(\gcd(a,c)=1\).

Finally  

\[
a+b+c=3+57+2=62.
\]

\[
\boxed{62}
\]

Let  

\[
\vec{AB}= \mathbf b ,\qquad \vec{AC}= \mathbf c ,\qquad 
|\mathbf b| =28,\; |\mathbf c| =91 .
\]

Write every point on the sides as a linear combination of \(\mathbf b,\mathbf c\).

\[
\begin{aligned}
D&=\frac{4}{28}\mathbf b =\frac17\mathbf b ,&
E&=\frac{20}{28}\mathbf b =\frac57\mathbf b ,\\[2mm]
F&=\frac{13}{91}\mathbf c ,&
G&=\frac{65}{91}\mathbf c .
\end{aligned}
\]

The reflections about a point are central symmetries, thus  

\[
M=2F-D=-\frac17\mathbf b+\frac{26}{91}\mathbf c ,\qquad
N=2E-G=\frac{10}{7}\mathbf b-\frac{65}{91}\mathbf c .
\]

--------------------------------------------------------------------
### 1.  Area of \(\triangle ABC\)

The quadrilateral \(DEGF\) can be split into \(\triangle DE G\) and \(\triangle DGF\).

For any three points \(P=\alpha\mathbf b+\beta\mathbf c\),
\(Q=\gamma\mathbf b+\delta\mathbf c\) we have  

\[
\bigl[(Q-P)\times(R-P)\bigr]
    =(\gamma-\alpha)(\zeta-\beta)-(\delta-\beta)(\xi-\alpha)\;\bigl(\mathbf b\times\mathbf c\bigr),
\]

so the area of a triangle equals the absolute value of that determinant
times \(\dfrac12|\mathbf b\times\mathbf c|\).

*Triangle \(DEG\)*  

\[
\det\!\begin{pmatrix}
\frac57-\frac17 & 0\\[2mm]
0-\frac17&\frac{65}{91}
\end{pmatrix}
  =\frac{20}{49}.
\]

Thus \(\displaystyle
\text{Area}(DEG)=\frac{20}{49}\cdot\frac12|\mathbf b\times\mathbf c|.
\)

*Triangle \(DGF\)*  

\[
\det\!\begin{pmatrix}
0-\frac17&\frac{65}{91}\\[2mm]
0-\frac17&\frac{13}{91}
\end{pmatrix}
  =\frac{4}{49},
\qquad
\text{Area}(DGF)=\frac{4}{49}\cdot\frac12|\mathbf b\times\mathbf c|.
\]

Hence  

\[
\text{Area}(DEGF)=\Bigl(\frac{20}{49}+\frac{4}{49}\Bigr)
                  \frac12|\mathbf b\times\mathbf c|
                =\frac{24}{49}\,\text{Area}(\triangle ABC).
\]

Given \(\text{Area}(DEGF)=288\),

\[
\frac{24}{49}\,\text{Area}(\triangle ABC)=288
\Longrightarrow
\text{Area}(\triangle ABC)=288\cdot\frac{49}{24}=12\cdot49=588.
\]

--------------------------------------------------------------------
### 2.  Area of the heptagon \(AFNBCEM\)

Place the points in the \((\alpha,\beta)\)–plane where \((\alpha,\beta)\) are the
coefficients of \(\mathbf b,\mathbf c\):

\[
\begin{array}{c|c}
\text{Vertex}&(\alpha,\beta)\\\hline
A&(0,0)\\
F&(0,\frac{13}{91})\\
N&(\frac{10}{7},-\frac{65}{91})\\
B&(1,0)\\
C&(0,1)\\
E&(\frac57,0)\\
M&\bigl(-\frac17,\frac{26}{91}\bigr)
\end{array}
\]

Applying the shoelace formula to these seven points gives

\[
\begin{aligned}
\sum x_i y_{i+1}-y_i x_{i+1}
&=1,\\[2mm]
\text{Area}_{\text{coeff}}&=\frac12\cdot1=\frac12 .
\end{aligned}
\]

In the original plane a figure’s area is multiplied by
\(|\mathbf b\times\mathbf c|=2\,\text{Area}(\triangle ABC)\).
Thus

\[
\text{Area}(AFNBCEM)=|\mathbf b\times\mathbf c|\,
\frac12
   =\text{Area}(\triangle ABC)=588.
\]

--------------------------------------------------------------------
\[
\boxed{588}
\]

The 12 letters are  

\[
A,B,C,D,E,F,G,H,I,J,K,L .
\]

A random grouping into six unordered pairs is performed, each pair is written with its
letters in alphabetical order, and then the six two‑letter “words’’ are listed
alphabetically.  

Let the word that appears last (the lexicographically greatest word) be called the **max word**.
The max word is determined by the *first* letters of the six pairs:
its first letter is the largest among the six first letters, and if several pairs have that
first letter, the one with the larger second letter wins.

--------------------------------------------------------------------
### 1.   Total number of possible pairings
\[
\#\text{pairings}= \frac{12!}{2!^{6}\,6!}=10395 .
\]

--------------------------------------------------------------------
### 2.   When does the max word contain \(G\)?

Two mutually exclusive possibilities:

* **(A) \(G\) is the first letter of its pair.**  
  Then the pair is \((G,X)\) with \(X\in\{H,I,J,K,L\}\).
  For \((G,X)\) to be the max word, no other pair may have a first
  letter larger than \(G\).  
  Hence each of the other four letters \(\{H,I,J,K,L\}\setminus\{X\}\) must be paired
  with a letter \(\le F\) (otherwise its first letter would exceed \(G\)).
  The remaining letters are  

  * 4 “high’’ letters \(\{H,I,J,K,L\}\setminus\{X\}\);
  * 6 “low’’ letters \(\{A,B,C,D,E,F\}\).

  Choose which four low letters are matched with the four high letters
  and then match them (an injection).  
  The number of ways is  

  \[
  {}_{6}P_{4}=6\cdot5\cdot4\cdot3=360 .
  \]

  The two low letters left over are forced to pair with each other.
  Thus for a fixed \(X\) there are \(360\) pairings, and there are
  \(5\) possible choices for \(X\):

  \[
  N_A=5\cdot360=1800 .
  \]

* **(B) \(G\) is the second letter of its pair.**  
  Then the pair is \((X,G)\) with \(X<G\), and \(X\) must be the
  largest first letter of any pair.  
  If \(X\le E\) there are not enough letters \(\le X\) to pair with all
  letters \(>X\) (there would be more “high’’ letters than “low’’ ones).  
  The only possible value is \(X=F\).

  Pair \(G\) with \(F\).  The five letters \(\{H,I,J,K,L\}\) must each be paired
  with a distinct letter from \(\{A,B,C,D,E\}\).  There are  

  \[
  5! =120
  \]

  ways to do this, and no letters remain.

  Hence  

  \[
  N_B =120 .
  \]

--------------------------------------------------------------------
### 3.   Favourable pairings
\[
N_{\text{fav}}=N_A+N_B=1800+120=1920 .
\]

--------------------------------------------------------------------
### 4.   Desired probability
\[
P=\frac{N_{\text{fav}}}{\#\text{pairings}}
   =\frac{1920}{10395}
   =\frac{128}{693}\qquad(\gcd(128,693)=1).
\]

Thus \(m=128,\;n=693\) and  

\[
m+n=128+693=821 .
\]

\[
\boxed{821}
\]

Let the line be the \(x\)-axis and put  

\[
A=(0,0),\qquad B=(b,0),\qquad C=(c,0),\qquad D=(d,0),\qquad E=(e,0),\qquad F=(f,0)
\]

with the points in the order \(A\!<\!B\!<\!C\!<\!D\!<\!E\!<\!F\).

---

### 1.  Locate the collinear points

\[
\begin{aligned}
AC&=26 \Longrightarrow c-0 = 26 \; \Rightarrow\; c=26,\\[2pt]
CE&=31 \Longrightarrow e-c =31 \; \Rightarrow\; e=26+31=57,\\[2pt]
AF&=73 \Longrightarrow f=73 .
\end{aligned}
\]

From \(DF=33\) we get  

\[
f-d =33 \Longrightarrow d = f-33 = 73-33 =40 .
\]

From \(BD=22\),

\[
d-b =22 \Longrightarrow b = d-22 = 40-22 =18 .
\]

Thus  

\[
A=(0,0),\; B=(18,0),\; C=(26,0),\; D=(40,0),\; E=(57,0),\; F=(73,0).
\]

All distances given in the statement are satisfied.

---

### 2.  Position of \(G\)

Let \(G=(x,y)\) with \(y\neq0\).  
The distances to \(C\) and \(D\) give

\[
\begin{cases}
(x-26)^2+y^2 = 40^2 =1600,\\[2pt]
(x-40)^2+y^2 = 30^2 = 900 .
\end{cases}
\]

Subtracting the equations:

\[
(x-26)^2-(x-40)^2 = 700.
\]

Expanding,

\[
(-52x+26^2)+(80x-40^2)=700\;\Longrightarrow\;28x-924=700,
\]

so  

\[
28x=1624\quad\Longrightarrow\quad x=58 .
\]

Plug \(x=58\) into \((x-26)^2+y^2=1600\):

\[
(58-26)^2+y^2 = 32^2 + y^2 = 1600\;\Longrightarrow\; y^2 = 1600-1024 = 576,
\]

hence \(|y|=24\).  
Take \(G=(58,24)\) (the other choice \((58,-24)\) gives the same area).

---

### 3.  Area of \(\triangle BGE\)

Points:

\[
B=(18,0),\qquad E=(57,0),\qquad G=(58,24).
\]

The base \(BE\) lies on the \(x\)-axis, so its length is  

\[
BE = 57-18 = 39.
\]

The altitude from \(G\) to the line \(BE\) is simply \(|y| = 24\).

Thus  

\[
\text{Area} = \frac12 \times (\text{base}) \times (\text{height})
            = \frac12 \times 39 \times 24
            = 39 \times 12
            = 468 .
\]

(Equivalently, the shoelace formula gives \(\frac12|1368-432|=468\).)

---

\[
\boxed{468}
\]

Let  

\[
d=n+2\qquad (d\ge 1,\; n=d-2).
\]

Then  

\[
n+3=d+1 ,\qquad n^{2}+9=(d-2)^{2}+9=d^{2}-4d+13 .
\]

Hence  

\[
3(n+3)(n^{2}+9)=3(d+1)(d^{2}-4d+13).
\]

The condition \(d\mid 3(n+3)(n^{2}+9)\) becomes  

\[
d \mid 3(d+1)(d^{2}-4d+13).
\]

Take this congruence modulo \(d\).  
Since \(d\equiv0\pmod d\),

\[
d+1\equiv1,\qquad d^{2}-4d+13\equiv13\pmod d,
\]

so  

\[
3(d+1)(d^{2}-4d+13)\equiv 3\cdot1\cdot13=39\pmod d.
\]

Thus \(d\) must divide \(39\):

\[
d\in\{1,3,13,39\}.
\]

Recall \(d=n+2\) and we need \(n>0\).  
Hence

\[
n=d-2\in\{-1,1,11,37\}.
\]

Discard the non‑positive value \(-1\). The admissible positive integers are  

\[
n=1,\;11,\;37 .
\]

Check:

* \(n=1\): \(n+2=3\) and \(3(1+3)(1^{2}+9)=120\) is divisible by \(3\).
* \(n=11\): \(n+2=13\) and \(3(14)(130)=5460\) is divisible by \(13\).
* \(n=37\): \(n+2=39\) and \(3(40)(1378)=165360\) is divisible by \(39\).

All satisfy the condition.  

The required sum is  

\[
1+11+37 = 49.
\]

\[
\boxed{49}
\]

For the periodic saw‑tooth  

\[
f(x)=\begin{cases}
x-4k,&x\in[-1+4k,\,1+4k)\\[2pt]
4k+2-x,&x\in[1+4k,\,3+4k)
\end{cases}\qquad(k\in\mathbb Z)
\]

the parabola is  

\[
x = 34y^{2}\qquad (y\in[-1,1]).
\]

Hence an intersection point \((x,y)\) must satisfy  

\[
y=f(x),\qquad x=34y^{2}.
\]

---

### 1.  Intersections on an “up” segment  

On an up segment \(x= y+4k\).  
Substituting \(x=34y^{2}\) gives  

\[
34y^{2}=y+4k\;\Longrightarrow\;34y^{2}-y-4k=0 .
\]

For each integer \(k\) with an up segment inside \([0,34]\) (i.e. \(k=0,1,\dots ,8\))
the two roots are  

\[
y_{k}^{\pm}=\frac{1\pm\sqrt{1+544k}}{68}.
\]

Both satisfy the interval \([-1,1]\) and give \(x\in[-1+4k,\,1+4k)\);
hence each \(k\) contributes **two** intersection points.

The sum of the two roots for a fixed \(k\) is, by Vieta,

\[
y_{k}^{+}+y_{k}^{-}= \frac{1}{34}.
\]

Thus the total contribution of all up‑segments is  

\[
\sum_{k=0}^{8}(y_{k}^{+}+y_{k}^{-})=9\cdot\frac1{34}=\frac9{34}.
\]

---

### 2.  Intersections on a “down’’ segment  

On a down segment \(x=4k+2-y\).  
Using \(x=34y^{2}\),

\[
34y^{2}=4k+2-y\;\Longrightarrow\;34y^{2}+y-(4k+2)=0 .
\]

For each \(k=0,1,\dots ,8\) the roots are  

\[
z_{k}^{\pm}= \frac{-1\pm\sqrt{273+544k}}{68}.
\]

The admissible \(y\)-values must satisfy \(-1<y\le 1\).

* For \(k=0,1,\dots ,7\) we have \(\sqrt{273+544k}<67\); both roots lie in
\((-1,1]\) and give valid points.
* For \(k=8\) the negative root satisfies \(y<-1\) and is rejected; only the
positive root remains.

Hence we obtain two points for each \(k=0,\dots ,7\) and one point for
\(k=8\) – a total of \(17\) points.

The sum of the two roots of the quadratic for any \(k\) is  

\[
z_{k}^{+}+z_{k}^{-}= -\frac1{34}.
\]

Consequently  

\[
\sum_{k=0}^{7}(z_{k}^{+}+z_{k}^{-})=8\!\left(-\frac1{34}\right)=-\frac8{34},
\]
and we must add the remaining positive root for \(k=8\),

\[
z_{8}^{+}= \frac{-1+\sqrt{4625}}{68}= \frac{-1+5\sqrt{185}}{68}.
\]

Therefore the total contribution of the down‑segments is  

\[
-\frac8{34}+z_{8}^{+}= -\frac{16}{68}+ \frac{-1+5\sqrt{185}}{68}
= \frac{-17+5\sqrt{185}}{68}.
\]

---

### 3.  Sum of all \(y\)-coordinates  

\[
\begin{aligned}
S&=\frac9{34}+ \left(-\frac8{34}+z_{8}^{+}\right)\\[2pt]
  &=\frac1{34}+z_{8}^{+}
   =\frac{2}{68}+ \frac{-1+5\sqrt{185}}{68}
   =\frac{1+5\sqrt{185}}{68}.
\end{aligned}
\]

Thus the sum can be written as  

\[
S=\frac{a+b\sqrt{c}}{d}\quad\text{with}\quad 
a=1,\;b=5,\;c=185,\;d=68.
\]

All conditions are satisfied (\(\gcd(a,b,d)=1\), \(c\) square‑free).

Finally  

\[
a+b+c+d = 1+5+185+68 = \boxed{259}.
\]

The grid has three rows and three vertical \(3\times3\) blocks.  
Let the rows be \(R_1,R_2,R_3\) and the blocks be \(B_1,B_2,B_3\).

---

### 1.  How the numbers are placed in the rows and blocks  

For a given number \(k\in\{1,\ldots ,9\}\) it must appear  

* once in each row (three appearances), and  
* once in each block (three appearances).

Hence the three positions of \(k\) correspond to a **bijection** between the set of rows and the set of blocks.  
Thus for each \(k\) we may choose a permutation \(\sigma_k\in S_3\) :

\[
\sigma_k(r)=\text{the block in which }k\text{ occurs in row }r .
\]

Conversely, a choice of a permutation \(\sigma_k\) for each \(k\) determines in which block every entry of the grid lies.

---

### 2.  Counting the possible collections of permutations  

Write a column of the \(3\times9\) “block‑matrix’’ \(M\) as a permutation of \((1,2,3)\) – the three symbols denote the block in which the number of that column appears in rows \(1,2,3\).

Thus we need a sequence of nine permutations \((p_1,\dots ,p_9)\) such that each ordered pair \((\text{row},\text{block})\) occurs exactly three times; equivalently each number appears three times in each row and three times in each block.

Let  

\[
x_1,\dots ,x_6
\]

be the numbers of the six possible permutations of \(S_3\) in the sequence.
The conditions give the linear system  

\[
\begin{aligned}
x_1+x_2&=3\\
x_3+x_4&=3\\
x_5+x_6&=3\\
x_3+x_5&=3\\
x_1+x_6&=3\\
x_2+x_4&=3\\
x_4+x_6&=3\\
x_2+x_5&=3\\
x_1+x_3&=3
\end{aligned}
\]

which has the solutions  

\[
\begin{array}{c|cccccc}
 &x_1&x_2&x_3&x_4&x_5&x_6\\\hline
t=0&0&3&3&0&0&3\\
t=1&1&2&2&1&1&2\\
t=2&2&1&1&2&2&1\\
t=3&3&0&0&3&3&0
\end{array}
\qquad (t=0,1,2,3).
\]

For a fixed solution the columns can be ordered in

\[
\frac{9!}{x_1!x_2!x_3!x_4!x_5!x_6!}
\]

ways.  Summing over the four solutions gives  

\[
\begin{aligned}
N_M &= \frac{9!}{3!^3}+2\cdot\frac{9!}{1!\,2!^2\,1!\,1!\,2!}
      +\frac{9!}{3!^3}\\
    &=1680+2\cdot45360+1680=94\,080 .
\end{aligned}
\]

So there are \(94\,080\) possible matrices \(M\) describing in which block each number lies.

---

### 3.  Filling the inner columns of each block  

Inside a block \(B\) each row already has three distinct numbers (the three numbers assigned to that row in the block).  
These three numbers can be placed in the three inner columns of the block in any order, i.e. in \((3!)^3=6^3\) ways.

Since the three blocks are independent, the total number of ways to complete the grid is  

\[
94\,080\cdot (6^3)^3 = 94\,080\cdot 6^9 .
\]

Because \(6^9 = (2\cdot3)^9 =2^{9}\,3^{9}\),

\[
\begin{aligned}
94\,080\cdot6^9 &= \bigl(2^7\cdot3\cdot5\cdot7^2\bigr)\cdot\bigl(2^9\cdot3^9\bigr)\\
                &=2^{16}\cdot3^{10}\cdot5^1\cdot7^2 .
\end{aligned}
\]

Thus the total number of fillings is  

\[
2^{16}\cdot 3^{10}\cdot 5^{1}\cdot 7^{2}.
\]

---

### 4.  Computing the required sum  

Take \(p=2,\ a=16;\; q=3,\ b=10;\; r=5,\ c=1;\; s=7,\ d=2\).

\[
p\,a+q\,b+r\,c+s\,d = 2\cdot16 + 3\cdot10 + 5\cdot1 + 7\cdot2
                     = 32 + 30 + 5 + 14 = 81 .
\]

\[
\boxed{81}
\]

First rewrite the chain of inequalities  

\[
x-yz<y-zx<z-xy
\]

as two separate inequalities.

\[
\begin{aligned}
x-yz &< y-zx \\[2mm]
y-zx &< z-xy
\end{aligned}
\]

\[
\begin{aligned}
x-yz-y+zx&<0\qquad\Longrightarrow\qquad (x-y)(1+z)<0,\\[2mm]
y-zx-z+xy&<0\qquad\Longrightarrow\qquad (y-z)(1+x)<0 .
\end{aligned}
\]

The third inequality \( (x-z)(1+y)<0\) follows by adding the two, so the region is
described by  

\[
\boxed{(x-y)(1+z)<0,\qquad (y-z)(1+x)<0}
\]

together with the plane \(x+y+z=75\).

---

### 1.   Sign patterns

Let  

\[
S_x=\operatorname{sgn}(1+x),\quad S_y=\operatorname{sgn}(1+y),\quad S_z=\operatorname{sgn}(1+z).
\]

From the two product‑inequalities we obtain  

\[
\operatorname{sgn}(x-y)=-S_z,\qquad 
\operatorname{sgn}(y-z)=-S_x .
\]

Thus  

\[
\begin{array}{c|c|c|c}
S_x & S_y & S_z &\text{order of }x,y,z\\ \hline
+ & + & + & x<y<z\\
+ & - & - & y<z<x\\
- & + & + & x<z<y\\
- & - & - & \text{impossible (sum }\;75) 
\end{array}
\]

Hence the points satisfying the chain split into three convex regions, corresponding
to the three admissible sign patterns above.

---

### 2.   Which region is bounded?

*Region  \(x<y<z\)  (all three numbers \(> -1\)).*  
Using the plane equation \(z=75-x-y\) the conditions become  

\[
-1<x<y,\qquad y<\frac{75-x}{2}.
\]

In the \((x,y)\)-plane this is the triangle bounded by  

\[
x=-1,\; y=x,\; y=\frac{75-x}{2}.
\]

The three vertices are  

\[
(-1,-1),\;(-1,38),\;(25,25).
\]

Its area in the \(xy\)-plane is

\[
\frac12\,(39)(26)=507 .
\]

Since the plane \(x+y+z=75\) is tilted, an element \(dx\,dy\) on the
\(xy\)-plane corresponds to an area element  

\[
\sqrt{1+(\partial z/\partial x)^2+(\partial z/\partial y)^2}\;dx\,dy
        =\sqrt{1+1+1}\;dx\,dy=\sqrt3\,dx\,dy .
\]

Thus the area of this region in the plane is  

\[
507\sqrt3 .
\]

*The other two regions* (\(y<z<x\) and \(x<z<y\)) project onto unbounded
half‑planes, so they have infinite area.

Hence the **unique finite region** is the one with \(x<y<z\), and its area is
\(507\sqrt3\).

---

### 3.   Answer

The area is \(a\sqrt b\) with \(a=507\) and \(b=3\); therefore  

\[
a+b=507+3=510 .
\]

\[
\boxed{510}
\]

\[
\prod_{k=4}^{63}\frac{\log_k\bigl(5^{k^2-1}\bigr)}{\log_{k+1}\bigl(5^{k^2-4}\bigr)}
\]

---

### 1.  Use change‑of‑base

\[
\log_k\bigl(5^{k^2-1}\bigr)=\frac{\log_5\bigl(5^{k^2-1}\bigr)}{\log_5 k}
      =\frac{k^{2}-1}{\log_5 k},
\qquad
\log_{k+1}\bigl(5^{k^2-4}\bigr)=\frac{k^{2}-4}{\log_5 (k+1)} .
\]

Hence each factor of the product equals  

\[
\frac{k^2-1}{k^2-4}\;
      \frac{\log_5(k+1)}{\log_5k}.
\]

So

\[
P = \underbrace{\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}}_{\displaystyle P_1}
    \;
    \underbrace{\prod_{k=4}^{63}\frac{\log_5(k+1)}{\log_5k}}_{\displaystyle P_2}.
\]

---

### 2.  Simplify \(P_1\)

\[
\frac{k^2-1}{k^2-4}= \frac{(k-1)(k+1)}{(k-2)(k+2)}
                  =\frac{k-1}{k-2}\cdot\frac{k+1}{k+2}.
\]

Thus  

\[
P_1=\Bigl(\prod_{k=4}^{63}\frac{k-1}{k-2}\Bigr)
    \Bigl(\prod_{k=4}^{63}\frac{k+1}{k+2}\Bigr).
\]

Both products telescope:

\[
\prod_{k=4}^{63}\frac{k-1}{k-2}
   =\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{62}{61}
   =\frac{62}{2}=31,
\]

\[
\prod_{k=4}^{63}\frac{k+1}{k+2}
   =\frac{5}{6}\cdot\frac{6}{7}\cdots\frac{64}{65}
   =\frac{5}{65}=\frac1{13}.
\]

Therefore  

\[
P_1 =31\cdot\frac1{13} =\frac{31}{13}.
\]

---

### 3.  Simplify \(P_2\)

\[
P_2 =\prod_{k=4}^{63}\frac{\log_5(k+1)}{\log_5k}
    =\frac{\log_5 5\;\log_5 6\;\cdots\;\log_5 64}
           {\log_5 4\;\log_5 5\;\cdots\;\log_5 63}
    =\frac{\log_5 64}{\log_5 4}.
\]

Using the change‑of‑base formula,

\[
\frac{\log_5 64}{\log_5 4}= \log_4 64.
\]

Since \(4^3=64\), we have \(\log_4 64 = 3\). Hence  

\[
P_2 = 3.
\]

---

### 4.  Combine the two parts

\[
P = P_1\cdot P_2 = \frac{31}{13}\times 3 = \frac{93}{13}.
\]

Thus the product equals \(\displaystyle\frac{m}{n}\) with \(m=93,\; n=13\).

\[
m+n = 93+13 = 106.
\]

\[
\boxed{106}
\]

Factor  

\[
2025=45^{2}= (3^{2}\cdot5)^{2}=3^{4}\cdot5^{2}.
\]

Hence a divisor of \(2025\) is of the form \(3^{a}5^{b}\) with  

\(0\le a\le4\) and \(0\le b\le2\).  
Thus  

\[
|A|=(4+1)(2+1)=15 .
\]

All \(2^{15}\) subsets of \(A\) are equally likely.

---

### Divisors needed for the LCM

For a subset \(B\subseteq A\),

\[
\operatorname{lcm}(B)=3^{\max_{d\in B}a_d}\;5^{\max_{d\in B}b_d}.
\]

To obtain \(\operatorname{lcm}(B)=2025=3^{4}5^{2}\) we must have  

* at least one element whose exponent of \(3\) equals \(4\);
* at least one element whose exponent of \(5\) equals \(2\).

Define  

\[
D_3=\{d\in A:\, \text{the exponent of }3\text{ in }d\text{ is }4\},\qquad 
D_5=\{d\in A:\, \text{the exponent of }5\text{ in }d\text{ is }2\}.
\]

Counting these divisors:

* \(D_3 = \{3^{4}5^{0}, 3^{4}5^{1}, 3^{4}5^{2}\}\) ⇒ \(|D_3|=3\);
* \(D_5 = \{3^{0}5^{2}, 3^{1}5^{2}, 3^{2}5^{2}, 3^{3}5^{2}, 3^{4}5^{2}\}\) ⇒ \(|D_5|=5\);
* The intersection \(D_3\cap D_5=\{3^{4}5^{2}\}= \{2025\}\) ⇒ \(|D_3\cap D_5|=1\).

Thus a subset \(B\) has \(\operatorname{lcm}(B)=2025\) iff it contains **at least one element of \(D_3\) and at least one element of \(D_5\)** (the same element may satisfy both conditions).

---

### Counting such subsets

Let \(U\) be the set of all subsets of \(A\) (including the empty set).  
Use inclusion–exclusion.

\[
\begin{aligned}
|U| &=2^{15},\\[2mm]
\text{Subsets missing }D_3 &: 2^{15-|D_3|}=2^{12},\\[2mm]
\text{Subsets missing }D_5 &: 2^{15-|D_5|}=2^{10},\\[2mm]
\text{Subsets missing both }D_3\text{ and }D_5 &: 
2^{15-|D_3\cup D_5|}
 =2^{15-(3+5-1)}=2^{8}.
\end{aligned}
\]

Hence the number of subsets that contain at least one element of each set is  

\[
\begin{aligned}
N_{\text{good}}
&=|U|-2^{12}-2^{10}+2^{8}\\
&=32768-4096-1024+256\\
&=27904.
\end{aligned}
\]

Every such subset is non‑empty, so these are exactly the favorable subsets.

---

### Probability

\[
P=\frac{N_{\text{good}}}{2^{15}}=\frac{27904}{32768}.
\]

Reduce the fraction:

\[
27904=2^{8}\cdot 109,\qquad 32768=2^{15}=2^{8}\cdot 128
\Longrightarrow P=\frac{109}{128}.
\]

Thus \(m=109\) and \(n=128\), and  

\[
m+n = 109+128 = 237.
\]

\[
\boxed{237}
\]

Let  

* \(\omega _2\) have centre \(O\) and radius \(15\);
* \(BC\) be a diameter of \(\omega _2\);
* \(\omega _1\) have centre \(A\) and radius \(6\), internally tangent to \(\omega _2\) at \(B\).

--------------------------------------------------------------------
### 1.  Coordinates

Place \(BC\) on the \(x\)-axis with  

\[
B=(-15,0),\qquad C=(15,0),\qquad O=(0,0).
\]

Since the circles are internally tangent, the centres are collinear with the
tangency point and  

\[
OA=R_2-R_1=15-6=9 .
\]

Thus \(A\) lies on the ray \(OB\) at distance \(9\) from \(O\); consequently  

\[
A=(-9,0).
\]

--------------------------------------------------------------------
### 2.  Point \(D\)

\(AD\perp BC\), so \(AD\) is the vertical line \(x=-9\).
Intersecting this line with \(\omega _2\) (\(x^2+y^2=225\)) gives  

\[
y^2=225-(-9)^2=144\quad\Longrightarrow\quad y=\pm12 .
\]

Because the statement “\(D\) is closer to \(\overline{FG}\) than to \(\overline{EH}\)’’ puts \(D\) above the centre, we take  

\[
D=(-9,12).
\]

--------------------------------------------------------------------
### 3.  Rectangle \(EFGH\)

The rectangle is inscribed in \(\omega _1\), so its circum‑centre is the
centre of \(\omega _1\); hence the centre of the rectangle is \(A\).

Let  

* half‑width \(a\) (distance from the centre to each vertical side),
* half‑height \(b\) (distance from the centre to each horizontal side).

Because the vertices lie on \(\omega _1\),

\[
a^{2}+b^{2}=6^{2}=36. \tag{1}
\]

The condition “\(\overline{EF}\perp\overline{BC}\)” makes \(\overline{EF}\) vertical,
so the rectangle’s sides are parallel/perpendicular to \(BC\).  
Taking the usual labelling (counter‑clockwise)

\[
\begin{aligned}
E&=(-9-a,\,-b),\qquad 
F=(-9-a,\,b),\\
G&=(-9+a,\,b),\qquad 
H=(-9+a,\,-b).
\end{aligned}
\]

--------------------------------------------------------------------
### 4.  Equality of the two triangle areas

\[
\begin{aligned}
\text{Area}(\triangle DGF)&=\frac12\cdot GF\cdot(\text{vertical distance }D\text{ to }y=b)\\
&=\frac12\cdot(2a)\cdot(12-b)=a(12-b),\\[4pt]
\text{Area}(\triangle CHG)&=\frac12\cdot GH\cdot(\text{horizontal distance }C\text{ to }x=-9+a)\\
&=\frac12\cdot(2b)\cdot(24-a)=b(24-a).
\end{aligned}
\]

Given the areas are equal,

\[
a(12-b)=b(24-a)\quad\Longrightarrow\quad12a=24b\quad\Longrightarrow\quad a=2b. \tag{2}
\]

--------------------------------------------------------------------
### 5.  Solving for \(a\) and \(b\)

Substituting \(a=2b\) into (1):

\[
(2b)^{2}+b^{2}=36\;\Longrightarrow\;5b^{2}=36\;\Longrightarrow\;b^{2}=\frac{36}{5}.
\]

Hence  

\[
a^{2}=(2b)^{2}=4b^{2}= \frac{144}{5}.
\]

--------------------------------------------------------------------
### 6.  Area of the rectangle

\[
\text{Area}= (2a)(2b)=4ab=4(2b)b=8b^{2}=8\left(\frac{36}{5}\right)=\frac{288}{5}.
\]

Thus the area is \(\dfrac{288}{5}\).  In lowest terms \(m=288,\;n=5\), and

\[
m+n=288+5=293.
\]

--------------------------------------------------------------------
\[
\boxed{293}
\]

**Step 1.  How many regions are created by chords?**  

If a convex region (here the disk) is cut by \(N\) line‑segments whose interior intersection points are all distinct, the number of regions is  

\[
R=N+1+I,
\]

where \(I\) is the total number of interior intersection points.  
(Each new segment adds one region, plus one extra region for each interior point where it meets a previous segment.)

For our disk  

\[
N=27\qquad\text{(the two diameters + 25 random chords)} .
\]

Hence  

\[
E[R]=28+E[I].
\]

So we must find the expected number \(E[I]\) of interior intersection points.

--------------------------------------------------------------------

**Step 2.  Intersections involving the two diameters.**

The two diameters intersect at the centre; this contributes one guaranteed intersection.

For a random chord \(C\) we ask for the probability that it meets a given diameter.

*Vertical diameter.*  
The chord meets the vertical line \(x=0\) iff its endpoints lie in opposite half‑planes (one in \(\{Q_1,Q_4\}\) and the other in \(\{Q_2,Q_3\}\)).  
Among the six unordered pairs of distinct quadrants, four have this property:

\[
\{Q_1,Q_2\},\{Q_1,Q_3\},\{Q_2,Q_4\},\{Q_3,Q_4\},
\]

so  

\[
P(C\text{ meets the vertical diameter})=\frac{4}{6}=\frac23 .
\]

Exactly the same reasoning holds for the horizontal diameter.  
Thus for each random chord

\[
P(C\text{ meets a given diameter})=\frac23 .
\]

With 25 random chords we obtain  

\[
E[\text{intersections chord–diameter}] = 25\cdot 2\cdot\frac23=\frac{100}{3}.
\]

--------------------------------------------------------------------

**Step 3.  Intersections among the 25 random chords.**

Each chord is obtained by picking two points on the circle that lie in different quadrants.  
The unordered pair of quadrants a chord uses is equally likely to be any of the six possibilities

* four *adjacent* pairs: \(\{01\},\{12\},\{23\},\{30\}\);
* two *opposite* pairs: \(\{02\},\{13\}\).

Thus a chord is *adjacent* with probability \(\frac23\) and *opposite* with probability \(\frac13\).

--------------------------------------------------------------------
### 3.1  Classifying a pair of chords

Let chord 1 belong to unordered pair \(P\) and chord 2 to unordered pair \(Q\).  
There are three possible relationships between \(P\) and \(Q\):

| relationship | how many ordered \((P,Q)\) | intersection probability |
|--------------|---------------------------|--------------------------|
| same pair (\(P=Q\)) | 6 | \(\displaystyle\frac12\) |
| disjoint pairs (no common quadrant) | 6 (4 adjacent‑adjacent, 2 opposite‑opposite) | \(0\) for adjacent–adjacent, \(1\) for opposite–opposite |
| share exactly one quadrant | 24 (8 adjacent‑adjacent, 16 adjacent‑opposite) | \(\displaystyle\frac12\) |

*Why the numbers?*  

* Two chords of the **same type** intersect iff the order of the two points in the first quadrant is opposite to the order of the two points in the second quadrant – probability \(1/2\).

* Two **disjoint adjacent** chords lie in quadrants \(\{0,1\}\) and \(\{2,3\}\); all points of the first lie before those of the second, so they never intersect.

* Two **disjoint opposite** chords are \(\{0,2\}\) and \(\{1,3\}\); the four points appear in the order \(0,1,2,3\), which is alternating, so they always intersect.

* When the chords **share exactly one quadrant**, the shared quadrant contains two independent uniform points; whichever of the two lies first decides whether the endpoints are interleaved. Hence the intersection probability is \(1/2\).

--------------------------------------------------------------------
### 3.2  Expected intersection probability for a random pair

Summing the contributions of all 36 ordered possibilities:

\[
\begin{aligned}
\sum_{P,Q} p_{P,Q}&=
6\!\left(\frac12\right) 
+ 4\!\cdot\!0 
+ 2\!\cdot\!1 
+ 24\!\left(\frac12\right) \\
&= 3 + 2 + 12 = 17 .
\end{aligned}
\]

Therefore for two independently chosen random chords

\[
P(\text{they intersect}) = \frac{17}{36}.
\]

--------------------------------------------------------------------
### 3.3  Expected number of chord–chord intersections

There are \(\binom{25}{2}=300\) unordered pairs of the 25 chords, so

\[
E[\text{chord–chord intersections}]
=300\cdot\frac{17}{36}= \frac{425}{3}.
\]

--------------------------------------------------------------------

**Step 4.  Total expected number of interior intersections.**

\[
\begin{aligned}
E[I] &= \underbrace{1}_{\text{diameters}} 
      + \underbrace{\frac{100}{3}}_{\text{chord–diameter}} 
      + \underbrace{\frac{425}{3}}_{\text{chord–chord}} \\[2mm]
     &= \frac{525}{3}+1 = 175+1 = 176 .
\end{aligned}
\]

--------------------------------------------------------------------

**Step 5.  Expected number of regions.**

\[
E[R]=N+1+E[I]=27+1+176=204.
\]

--------------------------------------------------------------------

\[
\boxed{204}
\]

Label the twelve unit segments that form the sides of the four unit squares as in the picture  

\[
\begin{array}{c}
\text{top left square }A:\; a\;( \text{top}) ,\; a'(\text{left}) ,\; u (\text{vertical middle up}) ,\; x (\text{horizontal middle left})\\[2mm]
\text{top right square }B:\; b ,\; b' ,\; u ,\; y\\[2mm]
\text{bottom left square }C:\; c ,\; c' ,\; v ,\; x\\[2mm]
\text{bottom right square }D:\; d ,\; d' ,\; v ,\; y
\end{array}
\]

All variables are binary:  

\[
0=\text{blue},\qquad 1=\text{red}.
\]

---

### 1.  Express the condition for each square  

For a square the number of red sides must be exactly two.  
If we let  

\[
\begin{aligned}
s_A &=x+u, & s_B &=y+u,\\
s_C &=x+v, & s_D &=y+v,
\end{aligned}
\]

then the two **boundary** edges of a square must contain  

\[
r_A =2-s_A,\qquad r_B =2-s_B,\qquad r_C =2-s_C,\qquad r_D =2-s_D
\]

red sides.  
For each square the two boundary edges are distinct from the other squares, so the choices for the
four pairs are independent.

For a given pair of boundary edges the number of colourings is  

\[
\begin{cases}
1 &\text{if }r_i\in\{0,2\}\;(00\text{ or }11),\\[2mm]
2 &\text{if }r_i =1\;(01\text{ or }10).
\end{cases}
\]

Hence, when the interior pattern \((x,y,u,v)\) is fixed, the number of ways to colour the eight
outer edges equals  

\[
2^{\#\{\,i\in\{A,B,C,D\}\mid s_i=1\,\}},
\]

because a square contributes a factor \(2\) exactly when its interior sum \(s_i\) equals \(1\).

Thus the total number of admissible colourings is  

\[
N=\sum_{(x,y,u,v)\in\{0,1\}^4} 2^{\#\{i\mid s_i=1\}} .
\]

---

### 2.  Enumerate the \(2^4=16\) interior patterns  

\[
\begin{array}{c|c|c|c|c|c}
(x,y,u,v) & s_A & s_B & s_C & s_D &
\#\{i\mid s_i=1\}\\\hline
0000 &0&0&0&0&0\\
0001 &0&0&1&1&2\\
0010 &1&1&0&0&2\\
0011 &1&1&1&1&4\\
0100 &0&1&0&1&2\\
0101 &0&1&1&2&2\\
0110 &1&2&0&1&2\\
0111 &1&2&1&2&2\\
1000 &1&0&1&0&2\\
1001 &1&0&2&1&2\\
1010 &2&1&1&0&2\\
1011 &2&1&2&1&2\\
1100 &1&1&1&1&4\\
1101 &1&1&2&2&2\\
1110 &2&2&1&1&2\\
1111 &2&2&2&2&0
\end{array}
\]

(The entry “2’’ means the sum of the two interior edges of that square is \(2\); it forces the
corresponding pair of boundary edges to be both blue.)

The contribution of each pattern is \(2^{\#\{i\mid s_i=1\}}\).  Adding all contributions we get  

\[
\begin{aligned}
N&=1\;(0000)+4\;(0001)+4\;(0010)+16\;(0011)\\
  &\quad+4\;(0100)+4\;(0101)+4\;(0110)+4\;(0111)\\
  &\quad+4\;(1000)+4\;(1001)+4\;(1010)+4\;(1011)\\
  &\quad+16\;(1100)+4\;(1101)+4\;(1110)+1\;(1111)\\[2mm]
  &=1+16+48+16+1\\
  &=\boxed{82}.
\end{aligned}
\]

Therefore there are **82** ways to colour the twelve unit segments of a \(2\times2\) grid so that
every unit square has exactly two red and two blue sides.

\[
f(x)=\sin \bigl(7\pi\sin(5x)\bigr)
\qquad (0<x<2\pi)
\]

---

### 1.  Zeros of \(f\)

\(f(x)=0\iff \sin\bigl(7\pi\sin(5x)\bigr)=0\)

\[
7\pi\sin(5x)=k\pi\qquad(k\in\mathbb Z)
\Longrightarrow\sin(5x)=\frac{k}{7}.
\]

Since \(\sin(5x)\in[-1,1]\), we must have \(-7\le k\le7\).  
Thus \(k\) can be any integer from \(-7\) to \(7\) (15 values).

Put \(\theta =5x\). Then \(\theta\in(0,10\pi)\) and we need the number of solutions of  

\[
\sin\theta =\frac{k}{7}\qquad (k=-7,\ldots,7)
\]
in \((0,10\pi)\).

*If \(|k|=7\) (\(\sin\theta=\pm1\)):* each period \((0,2\pi)\) contributes one solution, and there are five periods in \((0,10\pi)\).  
\(\;\Rightarrow\) 5 solutions for \(k=7\) and 5 solutions for \(k=-7\).

*If \(k=0\) (\(\sin\theta=0\)):* solutions are \(\theta=n\pi\) with \(n=1,\dots ,9\).  
\(\;\Rightarrow\) 9 solutions.

*If \(|k|=1,2,\dots ,6\) (\(|\sin\theta|<1\) and \(\neq0\)):* in each \(2\pi\)‑interval there are two solutions, so across five intervals there are \(2\cdot5=10\) solutions.

Hence the total number of zeros is  

\[
n =\underbrace{5+5}_{k=\pm7}+ \underbrace{9}_{k=0}+ \underbrace{12\times10}_{k=\pm1,\dots ,\pm6}
      =10+9+120 = 139 .
\]

---

### 2.  Tangent zeros

A zero is tangent to the \(x\)-axis when \(f'(x)=0\) as well.

\[
f'(x)=35\pi\cos\!\bigl(7\pi\sin(5x)\bigr)\,\cos(5x).
\]

At any zero we have \(\cos\!\bigl(7\pi\sin(5x)\bigr)=\cos(k\pi)=(-1)^k\neq0\); therefore  

\[
f'(x)=0 \iff \cos(5x)=0.
\]

\(\cos(5x)=0\) gives \(5x=\dfrac{\pi}{2}+m\pi\;(m\in\mathbb Z)\), i.e.

\[
x=\frac{(2m+1)\pi}{10}.
\]

Within \(0<x<2\pi\) the odd integers \(2m+1\) run from \(1\) to \(19\); there are \(10\) such values.
At these points \(\sin(5x)=\pm1\), i.e. they correspond exactly to the solutions for \(k=\pm7\).

Thus  

\[
t=10.
\]

---

### 3.  Required sum

\[
n+t = 139+10 = 149.
\]

\[
\boxed{149}
\]

**Step 1 – Greedy description**

For a given amount \(N\) let  

\[
q=\Bigl\lfloor \frac{N}{25}\Bigr\rfloor ,\qquad r=N-25q\;(0\le r<25)
\]

The greedy algorithm first takes \(q\) quarters, then uses as many dimes as possible and finally pennies.  
Hence the greedy number of coins is  

\[
g(N)=q+\Bigl\lfloor\frac{r}{10}\Bigr\rfloor +(r\bmod 10)
      =q+f(r),
\]

where  

\[
f(x)=\Bigl\lfloor\frac{x}{10}\Bigr\rfloor +(x\bmod10)
\]

is the optimal number of coins when only dimes and pennies are allowed.

---

**Step 2 – Optimal solution with a given number of quarters**

If we use \(a\) quarters (\(0\le a\le q\)), we must make the remainder \(N-25a\) with dimes and pennies.  
The minimum number of coins for that remainder is \(f(N-25a)\).  
Thus the optimal number of coins for \(N\) is  

\[
\min_{0\le a\le q}\; \bigl[a+f(N-25a)\bigr].
\]

Because \(f\) is optimal for the two‑coin system \(\{1,10\}\), we only need to compare the greedy choice
\(a=q\) with the possibilities where we use fewer quarters.

---

**Step 3 – Effect of removing one quarter**

Write \(r=N-25q\) (the greedy remainder).  
If we use one fewer quarter (\(a=q-1\)), the new remainder is \(r+25\), and the coin count becomes  

\[
(q-1)+f(r+25).
\]

Greedy is optimal precisely when  

\[
q+f(r)\le (q-1)+f(r+25)\quad\Longleftrightarrow\quad f(r)+1\le f(r+25).
\]

Compute \(f(r+25)\) in terms of \(f(r)\).  
Since adding \(25\) cents is the same as adding two dimes and five pennies,  

\[
f(r+25)=r+25-9\Bigl\lfloor\frac{r+25}{10}\Bigr\rfloor .
\]

The floor changes by  

\[
\Bigl\lfloor\frac{r+25}{10}\Bigr\rfloor-\Bigl\lfloor\frac{r}{10}\Bigr\rfloor=
\begin{cases}
2,& r\bmod10\le 4,\\[2pt]
3,& r\bmod10\ge 5 .
\end{cases}
\]

Hence  

\[
f(r+25)=
\begin{cases}
f(r)+7, & r\bmod10\le4,\\[2pt]
f(r)-2, & r\bmod10\ge5 .
\end{cases}
\]

Consequently  

* if \(r\bmod10\le4\) then \(f(r+25)=f(r)+7>f(r)+1\) – greedy stays optimal;
* if \(r\bmod10\ge5\) then \(f(r+25)=f(r)-2\le f(r)\) – using one fewer quarter gives fewer coins.

Thus the greedy algorithm fails **exactly** when the greedy remainder \(r\) satisfies  

\[
r\bmod10\ge5 \;\Longleftrightarrow\; r\in\{5,6,7,8,9,15,16,17,18,19\}.
\]

---

**Step 4 – Dropping more than one quarter never helps**

For two quarters removed we add \(50\) cents to the remainder.  
A similar computation gives  

\[
f(r+50)=f(r)+5 .
\]

The change in total coin count is \(-2+5=+3\); for three quarters removed the change is \(-3+12=+9\), etc.  
Hence removing two or more quarters can never improve on the greedy solution.  
Therefore the condition found in Step 3 is both necessary and sufficient.

---

**Step 5 – Counting the successful \(N\) (1 ≤ N ≤ 1000)**  

The greedy fails for those \(N\ge25\) whose remainder modulo \(25\) lies in  

\[
R=\{5,6,7,8,9,15,16,17,18,19\}.
\]

For a fixed residue \(r\in R\),

\[
N=r+25k\qquad (k\ge1,\; N\le1000).
\]

The largest possible \(k\) is \(\bigl\lfloor\frac{1000-r}{25}\bigr\rfloor =39\)
(because \(1000-r\) ranges from \(981\) to \(995\)).  
Thus each residue contributes \(39\) failing numbers.

\[
\#\text{failures}=|R|\times 39 = 10\times39 = 390 .
\]

All other numbers (including the ten numbers \(5,6,7,8,9,15,16,17,18,19\) themselves, which are \(<25\) and therefore not failures) are successes.

\[
\#\text{successes}=1000-390 = 610 .
\]

---

\[
\boxed{610}
\]

Let  

\[
a_1a_2\dots a_{16},\qquad a_i\in\{0,1\}
\]

be a binary string where \(a_i=1\) means the \(i^{\text{th}}\) chair is occupied.
The condition *no person sits next to two other people* means that a
\(1\) may not have a \(1\) on both sides; equivalently the string contains
no three consecutive \(1\)’s.  
Hence we must count the binary strings of length \(16\) that

* have exactly eight \(1\)’s,
* contain no substring \(111\).

---

### 1.  Decompose the occupied chairs into blocks

Since a block of consecutive \(1\)’s can have length at most \(2\), the eight
\(1\)’s are split into blocks of length \(1\) or \(2\).

Let  

* \(x\) = number of blocks of length \(1\),
* \(y\) = number of blocks of length \(2\),
* \(m=x+y\) = total number of blocks.

Then  

\[
x+2y=8,\qquad x+y=m\Longrightarrow y=8-m,\;x=2m-8 .
\]

Thus \(m\) can be any integer with \(\lceil 8/2\rceil\le m\le 8\); i.e.  

\[
m=4,5,6,7,8 .
\]

For a given \(m\) the positions of the \(y=8-m\) blocks of length \(2\)
among the \(m\) blocks can be chosen in  

\[
\binom{m}{\,8-m\,}
\]

ways.

---

### 2.  Insert the separating zeros

Between two consecutive blocks there must be at least one \(0\).
With \(m\) blocks there are \(m-1\) mandatory zeros.
There are \(16-8=8\) zeros altogether, so after placing the mandatory ones we
still have  

\[
8-(m-1)=9-m
\]

zeros left.  

These remaining zeros may be distributed freely among the \(m+1\) gaps
(the two ends and the \(m-1\) interior gaps).  By stars‑and‑bars the number
of ways is

\[
\binom{(9-m)+(m+1)-1}{(m+1)-1}
   =\binom{9}{m}.
\]

---

### 3.  Count all possibilities

For each admissible \(m\),

\[
\text{choices}(m)=\binom{m}{8-m}\binom{9}{m}.
\]

Hence  

\[
\begin{aligned}
N&=\sum_{m=4}^{8}\binom{m}{8-m}\binom{9}{m}\\[2mm]
 &=\binom{4}{4}\binom{9}{4}
   +\binom{5}{3}\binom{9}{5}
   +\binom{6}{2}\binom{9}{6}
   +\binom{7}{1}\binom{9}{7}
   +\binom{8}{0}\binom{9}{8}\\[2mm]
 &=1\cdot126+10\cdot126+15\cdot84+7\cdot36+1\cdot9\\
 &=126+1260+1260+252+9\\
 &=2907 .
\end{aligned}
\]

---

### 4.  Remainder modulo \(1000\)

\[
2907\equiv 907\pmod{1000}.
\]

\[
\boxed{907}
\]

Let  

\[
\angle A=84^{\circ},\qquad \angle B=60^{\circ},\qquad \angle C=36^{\circ}.
\]

--------------------------------------------------------------------
### 1.  The nine‑point circle in complex form  

Put the circumcircle of \(\triangle ABC\) as the unit circle and let  

\[
a=1,\qquad b=e^{i\beta},\qquad c=e^{i\gamma},
\]

where  

\[
\beta =72^{\circ},\qquad \gamma =240^{\circ}
\]

because the central arcs are twice the opposite angles:
\(\widehat{BC}=2A=168^{\circ},\;\widehat{CA}=2B=120^{\circ},
\;\widehat{AB}=2C=72^{\circ}\).

The nine‑point centre is

\[
N=\frac{a+b+c}{2},
\]

and its radius is \(\frac12\) (since \(|a|=|b|=|c|=1\)).  
The three midpoints are  

\[
D=\frac{b+c}{2},\qquad 
E=\frac{c+a}{2},\qquad 
F=\frac{a+b}{2}.
\]

Notice that

\[
D-N=-\frac{a}{2},\qquad 
E-N=-\frac{b}{2},\qquad 
F-N=-\frac{c}{2} .
\tag{1}
\]

Hence the central angle \(\widehat{DE}\) equals the angle between vectors
\(-a\) and \(-b\); it is the same as the angle between \(a\) and \(b\).

\[
\widehat{DE}= \angle aOb = 2\angle ACB = 2\cdot36^{\circ}=72^{\circ}.
\tag{2}
\]

--------------------------------------------------------------------
### 2.  The other intersection points  

The nine‑point circle is the image of the circumcircle under the similarity

\[
X\longmapsto N-\frac{X}{2},
\tag{3}
\]

i.e. the homothety with centre the centroid (factor \(-\tfrac12\)).
Consequently, if a point \(Y\) of the nine‑point circle is the image of
\(X\) on the circumcircle, then  

\[
Y = N-\frac{X}{2}\qquad\Longleftrightarrow\qquad X=2(N-Y).
\tag{4}
\]

--------------------------------------------------------------------
#### (a) Point \(G\)

\(G\) lies on line \(BD\).  Since \(D\) is the image of \(A\) and
\(B\) is the image of the point \(X\) with \(X=b\), the line \(BD\) is the
image of the line through \(A\) parallel to chord \(BC\).
Thus \(G\) corresponds to the second intersection of the line through
\(A\;(=a)\) parallel to \(BC\) with the circumcircle.

For a line through a point \(e^{i\alpha}\) parallel to chord
\(e^{i\beta}e^{i\gamma}\) the second intersection is
\(e^{i(\beta+\gamma-\alpha)}\).  
Here \(\alpha=0,\;\beta=72^{\circ},\;\gamma=240^{\circ}\); therefore

\[
X_G = e^{i(\beta+\gamma)}=e^{i312^{\circ}} .
\]

From (3) the point on the nine‑point circle is  

\[
G = N-\frac{X_G}{2}.
\]

Hence the vector \(NG=-\frac{X_G}{2}\) has direction \(312^{\circ}+180^{\circ}=132^{\circ}\).

--------------------------------------------------------------------
#### (b) Point \(H\)

\(H\) lies on line \(AE\).  This line is parallel to chord \(CA\); the
starting point on the circumcircle is \(B\,(=e^{i72^{\circ}})\).  Thus

\[
X_H = e^{i(0+240^{\circ}-72^{\circ})}=e^{i168^{\circ}} ,
\qquad
H=N-\frac{X_H}{2},
\]

so the direction of \(NH\) is \(168^{\circ}+180^{\circ}=348^{\circ}\).

--------------------------------------------------------------------
#### (c) Point \(J\)

\(J\) lies on line \(AF\), which is parallel to chord \(AB\); it starts
from \(C\,(=e^{i240^{\circ}})\).  Therefore

\[
X_J = e^{i(0+72^{\circ}-240^{\circ})}=e^{i192^{\circ}},
\qquad
J=N-\frac{X_J}{2},
\]

and the direction of \(NJ\) is \(192^{\circ}+180^{\circ}=12^{\circ}\).

--------------------------------------------------------------------
### 3.  The required arcs on the nine‑point circle  

From the directions obtained:

\[
\begin{aligned}
&\text{Arc }DE = 252^{\circ}-180^{\circ}=72^{\circ},\\[2mm]
&\text{Arc }FG = 132^{\circ}-60^{\circ}=72^{\circ},\\[2mm]
&\text{Arc }HJ = (360^{\circ}-348^{\circ})+12^{\circ}=24^{\circ}.
\end{aligned}
\]

All are the minor arcs (each \(<180^{\circ}\)).

--------------------------------------------------------------------
### 4.  Computing the requested expression  

\[
\widehat{DE}+2\widehat{HJ}+3\widehat{FG}
=72^{\circ}+2\!\cdot\!24^{\circ}+3\!\cdot\!72^{\circ}
=72+48+216=336^{\circ}.
\]

--------------------------------------------------------------------
\[
\boxed{336}
\]

For a regular 24‑gon label the vertices \(0,1,\ldots ,23\) in order.
If a segment joins vertex \(i\) to vertex \(j\) then the chord length depends only on  

\[
k=\min \{\,|i-j|\;,\;24-|i-j|\,\},
\qquad 1\le k\le 12 .
\]

Thus a configuration in which all 12 segments have the same length corresponds to a
perfect matching in the graph whose edges are all chords of a fixed step \(k\).

---

### 1.  Structure of the step‑\(k\) graph  

Let \(d=\gcd(24,k)\).  
The vertices split into \(d\) disjoint cycles, each of length  

\[
L=\frac{24}{d}.
\]

Every vertex is incident with exactly the two edges of its cycle.  
A perfect matching can exist only if each cycle has even length, i.e. \(L\) must be even.
\(L\) is odd only when \(d=8\) (then \(L=3\)); this occurs for \(k=8\).  
Hence the step \(k=8\) gives **no** valid configuration.

For an even cycle of length \(L\ge4\) there are exactly two perfect matchings
(the two possible “alternating” choices).  
If \(L=2\) (the case \(k=12\)), the matching is forced and there is only one.

Consequently, for a permissible \(k\)

\[
\#\text{ matchings for step }k=
\begin{cases}
2^{d}, & L>2,\\[2pt]
1, & L=2 .
\end{cases}
\]

---

### 2.  Count for each step \(k\)

\[
\begin{array}{c|c|c|c}
k & d=\gcd(24,k) & L=24/d & \#\text{ matchings}\\ \hline
1,5,7,11 & 1 & 24 & 2\\
2,10 & 2 & 12 & 2^{2}=4\\
3,9 & 3 & 8 & 2^{3}=8\\
4 & 4 & 6 & 2^{4}=16\\
6 & 6 & 4 & 2^{6}=64\\
12 & 12& 2 & 1\\
8 & 8 & 3 & \text{none}
\end{array}
\]

Now sum over all admissible \(k\) ( \(k=1,\dots ,12\) except \(k=8\) ):

\[
\begin{aligned}
\text{Total}&=4\cdot2\;+\;2\cdot4\;+\;2\cdot8\;+\;1\cdot16\;+\;1\cdot64\;+\;1\cdot1\\
&=8+8+16+16+64+1\\
&=113 .
\end{aligned}
\]

---

\[
\boxed{113}
\]

Let  

\[
17_{b}=1\cdot b+7=b+7,\qquad 
97_{b}=9\cdot b+7=9b+7 .
\]

We need  

\[
b+7 \mid 9b+7 .
\]

Write \(9b+7\) in terms of \(b+7\):

\[
9b+7 = 9(b+7) - 56 .
\]

Hence the remainder upon division by \(b+7\) is \(-56\).  
Thus \(b+7\) must divide \(56\):

\[
b+7 \mid 56 .
\]

List the positive divisors of \(56\):

\[
1,\;2,\;4,\;7,\;8,\;14,\;28,\;56 .
\]

Since the base must satisfy \(b>9\) (so that the digit 9 is allowed), we need  

\[
b+7>9+7=16 .
\]

The only divisors of \(56\) larger than \(16\) are \(28\) and \(56\).

\[
\begin{aligned}
b+7 &= 28 \;\Longrightarrow\; b=21,\\[2pt]
b+7 &= 56 \;\Longrightarrow\; b=49.
\end{aligned}
\]

Both bases exceed \(9\) and thus are admissible.

The sum of all such bases is

\[
21+49 = 70 .
\]

\[
\boxed{70}
\]

Let  

\[
a_i=A_1A_i\qquad (i=2,\dots ,11),\qquad s_i=A_iA_{i+1}\qquad (i=2,\dots ,10).
\]

For each triangle \(A_iA_1A_{i+1}\;(2\le i\le10)\)

* \(\displaystyle \text{Area}=1\),
* \(\displaystyle\cos\angle A_iA_1A_{i+1}= \frac{12}{13}\).

--------------------------------------------------------------------
### 1.  Product of the two sides from \(A_1\)

With \(\theta=\angle A_iA_1A_{i+1}\) we have  

\[
\sin\theta=\sqrt{1-\cos ^2\theta}= \frac{5}{13}.
\]

The area of \(\triangle A_iA_1A_{i+1}\) is  

\[
\frac12 a_i a_{i+1}\sin\theta =1
\Longrightarrow a_i a_{i+1}= \frac{2}{\sin\theta}= \frac{2}{5/13}= \frac{26}{5}\equiv c .
\tag{1}
\]

Hence for all \(i\)

\[
a_i a_{i+1}=c=\frac{26}{5}.
\]

--------------------------------------------------------------------
### 2.  Length of the side \(A_iA_{i+1}\)

Apply the law of cosines in \(\triangle A_iA_1A_{i+1}\):

\[
s_i^2=a_i^{\,2}+a_{i+1}^{\,2}-2a_i a_{i+1}\cos\theta
      =a_i^{\,2}+a_{i+1}^{\,2}-2c\Bigl(\frac{12}{13}\Bigr).
\]

Because \(2c\frac{12}{13}= \frac{624}{65}= \frac{48}{5}\),

\[
s_i^{\,2}=a_i^{\,2}+a_{i+1}^{\,2}-\frac{48}{5}. \tag{2}
\]

--------------------------------------------------------------------
### 3.  The alternating pattern of the radii

From (1) we have \(a_{i+1}=c/a_i\). Consequently  

\[
a_{i+2}=c/a_{i+1}=c/(c/a_i)=a_i .
\]

Thus  

\[
a_{2}=a_{4}=a_{6}=a_{8}=a_{10}\equiv x, \qquad 
a_{3}=a_{5}=a_{7}=a_{9}=a_{11}\equiv \frac{c}{x}.
\]

All sides \(s_i\;(i=2,\dots ,10)\) are equal, because each uses the
pair \((x,\;c/x)\) in (2).  Set  

\[
S^2=x^{2}+\Bigl(\frac{c}{x}\Bigr)^2-\frac{48}{5}.
\tag{3}
\]

--------------------------------------------------------------------
### 4.  Express everything via \(t=x+\dfrac{c}{x}\)

Define  

\[
t=x+\frac{c}{x}=a_2+a_{11}.
\]

Then  

\[
x^{2}+\Bigl(\frac{c}{x}\Bigr)^2=t^{2}-2c .
\]

Insert this into (3):

\[
S^{2}=t^{2}-2c-\frac{48}{5}=t^{2}-(2c+\frac{48}{5}).
\]

Since \(c=\dfrac{26}{5}\),

\[
2c+\frac{48}{5}= \frac{52}{5}+\frac{48}{5}=20.
\]

Hence  

\[
S^{2}=t^{2}-20,\qquad S=\sqrt{t^{2}-20}. \tag{4}
\]

--------------------------------------------------------------------
### 5.  Use the perimeter

The perimeter of the 11‑gon is  

\[
a_2+a_{11}+ \sum_{i=2}^{10}s_i = t+9S =20.
\]

Substituting (4):

\[
t+9\sqrt{t^{2}-20}=20.
\]

Isolate the radical and square:

\[
9\sqrt{t^{2}-20}=20-t \quad\Longrightarrow\quad 
81(t^{2}-20)=(20-t)^{2}.
\]

Simplifying:

\[
80t^{2}+40t-2020=0
\Longrightarrow 4t^{2}+2t-101=0.
\]

Solve for \(t\):

\[
t=\frac{-2\pm\sqrt{4+1616}}{8}
   =\frac{-2\pm 18\sqrt5}{8}
   =\frac{-1\pm 9\sqrt5}{4}.
\]

Since \(t>0\), we take the plus sign:

\[
\boxed{t=\frac{9\sqrt5-1}{4}}.
\]

Thus  

\[
A_1A_2+A_1A_{11}= \frac{9\sqrt5-1}{4}
            =\frac{m\sqrt n-p}{q},
\]
with \(m=9,\;n=5,\;p=1,\;q=4\).

--------------------------------------------------------------------
### 6.  Compute \(m+n+p+q\)

\[
m+n+p+q = 9+5+1+4 = 19.
\]

\[
\boxed{19}
\]

The nine players are distinct.  Let  

\[
c=\#\text{chocolate},\qquad v=\#\text{vanilla},\qquad s=\#\text{strawberry}
\]

with  

\[
c+v+s=9,\qquad c>v>s\ge 1 .
\]

---

### 1.  Find all possible triples \((c,v,s)\)

Since \(c>v>s\),

\[
c\ge v+1,\qquad v\ge s+1.
\]

Write \(v=s+d_2\) and \(c=v+d_1\) where \(d_1,d_2\ge 1\).  
Then  

\[
c+v+s=(s+d_2+d_1)+(s+d_2)+s=3s+2d_2+d_1=9 .
\]

Because \(s\ge1\), we have \(3s\le 9\Rightarrow s=1,2,3\).

* **\(s=1\):** \(2d_2+d_1=6\).  
  \(d_2=1\Rightarrow d_1=4\) → \((c,v,s)=(6,2,1)\).  
  \(d_2=2\Rightarrow d_1=2\) → \((c,v,s)=(5,3,1)\).

* **\(s=2\):** \(2d_2+d_1=3\).  
  \(d_2=1\Rightarrow d_1=1\) → \((c,v,s)=(4,3,2)\).

* **\(s=3\):** \(2d_2+d_1=0\) – impossible with \(d_1,d_2\ge1\).

Thus the only feasible triples are  

\[
(6,2,1),\;(5,3,1),\;(4,3,2).
\]

---

### 2.  Count assignments for each triple

For a given \((c,v,s)\) the number of ways to assign the flavors to the 9 distinct players is the multinomial coefficient  

\[
\frac{9!}{c!\,v!\,s!}.
\]

\[
\begin{aligned}
(6,2,1):\;&\frac{9!}{6!\,2!\,1!}= \frac{362880}{720\cdot2}=252,\\[2mm]
(5,3,1):\;&\frac{9!}{5!\,3!\,1!}= \frac{362880}{120\cdot6}=504,\\[2mm]
(4,3,2):\;&\frac{9!}{4!\,3!\,2!}= \frac{362880}{24\cdot6\cdot2}=1260.
\end{aligned}
\]

---

### 3.  Total number of assignments

\[
N = 252+504+1260 = 2016.
\]

---

### 4.  Remainder modulo \(1000\)

\[
2016 \equiv 16 \pmod{1000}.
\]

\[
\boxed{16}
\]

The equation  

\[
12x^{2}-xy-6y^{2}=0
\]

is a homogeneous quadratic form.  
Factor it over the integers:

\[
12x^{2}-xy-6y^{2}= (3x+2y)(4x-3y).
\]

Indeed  

\[
(3x+2y)(4x-3y)=12x^{2}-9xy+8xy-6y^{2}=12x^{2}-xy-6y^{2}.
\]

Hence a pair \((x,y)\) of integers satisfies the equation iff

\[
3x+2y=0\qquad\text{or}\qquad 4x-3y=0 .
\]

---

### 1. Solutions of \(3x+2y=0\)

\[
3x+2y=0 \;\Longrightarrow\; y=-\frac{3}{2}x .
\]

For \(y\) to be an integer, \(x\) must be even: write \(x=2k\). Then  

\[
(x,y)=(2k,-3k),\qquad k\in\mathbb Z .
\]

Both coordinates must lie in \([-100,100]\):

\[
-100\le 2k\le 100\;\Longrightarrow\;-50\le k\le 50,\\
-100\le -3k\le 100\;\Longrightarrow\;-33\le k\le 33.
\]

Thus \(k=-33,-32,\dots ,33\); there are  

\[
33-(-33)+1 = 67
\]

integer solutions on this line.

---

### 2. Solutions of \(4x-3y=0\)

\[
4x-3y=0 \;\Longrightarrow\; y=\frac{4}{3}x .
\]

For \(y\) to be an integer, \(x\) must be a multiple of \(3\): put \(x=3t\). Then  

\[
(x,y)=(3t,4t),\qquad t\in\mathbb Z .
\]

Again impose the bounds:

\[
-100\le 3t\le 100\;\Longrightarrow\;-33\le t\le 33,\\
-100\le 4t\le 100\;\Longrightarrow\;-25\le t\le 25.
\]

Thus \(t=-25,-24,\dots ,25\); the number of solutions here is  

\[
25-(-25)+1 = 51.
\]

---

### 3. Intersection of the two lines

Both equations hold simultaneously only at the origin:

\[
3x+2y=0,\;4x-3y=0 \Longrightarrow (x,y)=(0,0).
\]

This point has been counted in both previous totals, so we subtract it once.

---

### 4. Total number of ordered pairs

\[
\#\text{solutions}=67+51-1=117.
\]

\[
\boxed{117}
\]

Let the trapezoid be \(ABCD\) with \(AB\parallel CD\) and  

\[
AB=s ,\qquad CD=r ,\qquad AD=BC=L \;( \text{isosceles}).
\]

The incircle has radius \(R=3\) and the area is \(72\).

---

### 1.  Semiperimeter

For any tangential quadrilateral (one with an incircle)

\[
\text{Area}=R\cdot(\text{semiperimeter}).
\]

Hence  

\[
72=3\cdot s_{\!p}\quad\Longrightarrow\quad s_{\!p}=24 .
\]

Thus the perimeter is  

\[
2s_{\!p}=48\quad\Longrightarrow\quad r+s+2L=48. \tag{1}
\]

---

### 2.  Tangential condition

In a tangential quadrilateral the sums of opposite sides are equal:

\[
AB+CD = AD+BC\quad\Longrightarrow\quad r+s = 2L. \tag{2}
\]

From (1) and (2),

\[
2(r+s)=48 \;\Longrightarrow\; r+s=24,\qquad
2L=r+s=24\;\Longrightarrow\; L=12.
\]

---

### 3.  Height

The incircle touches both parallel sides, so the distance between the bases equals twice the radius:

\[
h=2R=6.
\]

(Alternatively, \(\displaystyle 72=\tfrac12(r+s)h\) with \(r+s=24\) also gives \(h=6\).)

---

### 4.  Relating the bases

Place the trapezoid with its symmetry axis on the \(y\)-axis; the bases are horizontal.
The horizontal half‑difference of the bases is \(\dfrac{s-r}{2}\).
Hence the leg length satisfies

\[
L^{2}= \left(\frac{s-r}{2}\right)^{2}+h^{2}.
\]

Using \(L=12,\;h=6\),

\[
12^{2}= \frac{(s-r)^{2}}{4}+6^{2}
\quad\Longrightarrow\quad
\frac{(s-r)^{2}}{4}=108
\quad\Longrightarrow\quad
s-r=12\sqrt3. \tag{3}
\]

---

### 5.  Solving for \(r\) and \(s\)

\[
\begin{cases}
r+s=24,\\[2pt]
s-r=12\sqrt3,
\end{cases}
\qquad\Longrightarrow\qquad
\begin{aligned}
s&=\frac{24+12\sqrt3}{2}=12+6\sqrt3,\\
r&=\frac{24-12\sqrt3}{2}=12-6\sqrt3.
\end{aligned}
\]

---

### 6.  Required sum of squares

\[
\begin{aligned}
r^{2}+s^{2}
&=(12-6\sqrt3)^{2}+(12+6\sqrt3)^{2}\\
&= \bigl(144-144\sqrt3+108\bigr)+\bigl(144+144\sqrt3+108\bigr)\\
&= 504.
\end{aligned}
\]

\[
\boxed{504}
\]

Let  

* \(A=(0,0)\),  
* \(B=(u,0)\) on the \(x\)-axis,  
* \(C=(0,v)\) on the \(y\)-axis.  

Since \(\angle A=90^{\circ}\), \(BC=38\) gives  

\[
u^{2}+v^{2}=38^{2}=1444\tag{1}
\]

---

### 1.  The equilateral triangle \(AKL\)

The conditions  

\[
AK=AL=KL=14
\]

show that \(\triangle AKL\) is equilateral.  
Let the direction of \(\overrightarrow{AK}\) form an angle \(\theta\) with the
positive \(x\)-axis. Then

\[
K=(14\cos\theta,\;14\sin\theta),\qquad 
L=(14\cos(\theta+60^\circ),\;14\sin(\theta+60^\circ)).\tag{2}
\]

---

### 2.  Using \(BK=14\) and \(CL=14\)

\[
BK^{2}= (u-14\cos\theta)^2+(14\sin\theta)^2=14^{2}
\]

\[
\Longrightarrow\; u=28\cos\theta.\tag{3}
\]

Similarly  

\[
CL^{2}= (14\cos(\theta+60^\circ))^{2}+(v-14\sin(\theta+60^\circ))^{2}=14^{2}
\]

\[
\Longrightarrow\; v=28\sin(\theta+60^\circ).\tag{4}
\]

---

### 3.  Determining \(\theta\)

Insert (3) and (4) into (1):

\[
(28\cos\theta)^{2}+(28\sin(\theta+60^\circ))^{2}=1444
\]
\[
\Longrightarrow\;\cos^{2}\theta+\sin^{2}(\theta+60^\circ)=\frac{361}{196}.
\]

Using \(\sin(\theta+60^\circ)=\tfrac12\sin\theta+\tfrac{\sqrt3}{2}\cos\theta\),
the equation reduces to  

\[
6\cos^{2}\theta+2\sqrt3\sin\theta\cos\theta= \frac{312}{49}.
\]

Solving gives \(\cos\theta=\frac{13}{14}\) (the other root gives the same
triangle with the legs interchanged).  

Hence  

\[
u=28\cos\theta=28\cdot\frac{13}{14}=26,\qquad
v=28\sin(\theta+60^\circ)=28\cdot\frac{4\sqrt3}{7}=16\sqrt3 .
\]

Thus  

\[
AB=26,\qquad AC=16\sqrt3,\qquad BC=38.
\]

From (2) we obtain the coordinates of the interior points

\[
K=(13,\;3\sqrt3),\qquad L=(2,\;8\sqrt3).
\]

---

### 4.  Area of \(BKLC\)

Use the shoelace formula for the polygon \(B(26,0),K(13,3\sqrt3),L(2,8\sqrt3),C(0,16\sqrt3)\):

\[
\begin{aligned}
S_1&=26\cdot3\sqrt3+13\cdot8\sqrt3+2\cdot16\sqrt3+0\cdot0=214\sqrt3,\\
S_2&=0\cdot13+3\sqrt3\cdot2+8\sqrt3\cdot0+16\sqrt3\cdot26=422\sqrt3.
\end{aligned}
\]

\[
\text{Area}=\frac12|S_2-S_1|
          =\frac12(422-214)\sqrt3
          =104\sqrt3 .
\]

Therefore the area can be written as \(n\sqrt3\) with  

\[
\boxed{n=104}.
\]

For an eight‑digit number formed by a permutation of  

\[
\{1,2,3,4,5,6,7,8\}
\]

let  

\[
a_1a_2a_3a_4a_5a_6a_7a_8
\]

be its digits (left to right).  
A number is divisible by \(22\iff\) it is divisible by \(2\) and by \(11\).

---

### 1.  Divisibility by 2  

The last digit \(a_8\) must be even, so \(a_8\in\{2,4,6,8\}\).

---

### 2.  Divisibility by 11  

For an 8‑digit number the rule for 11 is  

\[
(a_1+a_3+a_5+a_7)-(a_2+a_4+a_6+a_8)\equiv 0\pmod{11}.
\]

Since the total sum of the digits is  

\[
1+2+\cdots+8 = 36,
\]

let  

\[
S_{\text{odd}}=a_1+a_3+a_5+a_7,\qquad 
S_{\text{even}}=a_2+a_4+a_6+a_8 .
\]

Then \(S_{\text{odd}}+S_{\text{even}}=36\) and the condition gives  

\[
S_{\text{odd}}-S_{\text{even}}\equiv0\pmod{11}.
\]

The only possible values for \(S_{\text{odd}}-S_{\text{even}}\) are \(-22,0,22\);   
\(-22\) would give \(S_{\text{odd}}=7\) and \(22\) would give \(S_{\text{odd}}=29\), both impossible because a sum of four distinct digits from \(\{1,\dots ,8\}\) cannot be smaller than \(10\) nor larger than \(26\).  
Hence

\[
S_{\text{odd}} = S_{\text{even}} = 18 .
\]

Thus the four digits in the odd positions must sum to \(18\); the same holds for the even positions.

---

### 3.  Choosing the four digits for the odd positions  

We need 4‑element subsets of \(\{1,\dots ,8\}\) whose sum is \(18\).  
A quick enumeration gives the eight subsets

\[
\begin{aligned}
&\{1,2,7,8\},\{1,3,6,8\},\{1,4,5,8\},\{1,4,6,7\},\\
&\{2,3,5,8\},\{2,3,6,7\},\{2,4,5,7\},\{3,4,5,6\}.
\end{aligned}
\]

Each contains exactly two even digits (the set \(\{2,4,6,8\}\)) and two odd digits.  
Hence every admissible odd‑position set has **2 even digits**.

Let a chosen subset be \(S\) (the odd‑position digits) and its complement \(S^c\) (the even‑position digits).

---

### 4.  Arranging the digits  

* The four digits of \(S\) can be placed in the odd positions \((1,3,5,7)\) in \(4! = 24\) ways.  
* The four digits of \(S^c\) must fill the even positions \((2,4,6,8)\) with the extra requirement that the last digit \(a_8\) be even.  
  Since \(S^c\) contains exactly \(2\) even digits, we have  

\[
\text{choices for }a_8 = 2,\qquad
\text{arrangements of the remaining three even‑position digits}=3! = 6 .
\]

Thus for each set \(S\)

\[
\#\text{numbers}=4!\times(2\times3!) = 24\times12 = 288.
\]

---

### 5.  Total number \(N\)

There are \(8\) possible subsets \(S\); consequently

\[
N = 8 \times 288 = 2304 .
\]

---

### 6.  Required difference

\[
N-2025 = 2304-2025 = 279 .
\]

\[
\boxed{279}
\]

Let  

\[
C=25+20i,\qquad r=5 ,
\]

so  

\[
|C-z|=5
\]

is the circle with centre \(C\) and radius \(5\).

The second condition  

\[
|z-4-k|=|z-3i-k|
\]

means that \(z\) is equally distant from the points  

\[
P_1=4+k\quad(\text{on the real axis}), \qquad   
P_2=k+3i .
\]

Hence the locus of \(z\) is the perpendicular bisector of the segment \(\overline{P_1P_2}\); this is a straight line.

---

### 1. Equation of the line

Write \(z=x+yi\).  

\[
(x-(4+k))^2+y^2=(x-k)^2+(y-3)^2 .
\]

Expanding and simplifying gives  

\[
-8(x-k)+16=-6y+9,
\]

\[
-8x+8k+6y+7=0 .
\]

Thus the line \(L(k)\) is  

\[
-8x+6y+8k+7=0 \qquad\text{or}\qquad 8x-6y=8k+7 .
\]

All such lines have slope \(\dfrac{4}{3}\); they are a family of parallel lines depending on \(k\).

---

### 2. Tangency condition

The system has exactly one solution precisely when the line \(L(k)\) is tangent to the circle \(|z-C|=5\).  

For a line \(Ax+By+C=0\) the distance from a point \((x_0,y_0)\) is  

\[
d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} .
\]

Here \(A=-8,\;B=6,\;C=8k+7\) and the centre of the circle is \((25,20)\). Hence

\[
d(k)=\frac{\bigl|-8\cdot25+6\cdot20+(8k+7)\bigr|}{\sqrt{(-8)^2+6^2}}
      =\frac{|8k-73|}{10}.
\]

Tangency requires \(d(k)=r=5\):

\[
\frac{|8k-73|}{10}=5\quad\Longrightarrow\quad|8k-73|=50 .
\]

Thus  

\[
8k-73=50\quad\text{or}\quad8k-73=-50,
\]

\[
k=\frac{123}{8},\qquad k=\frac{23}{8}.
\]

Both values make the line tangent, giving exactly one intersection point with the circle.

---

### 3. Sum of the possible \(k\)

\[
\frac{123}{8}+\frac{23}{8}= \frac{146}{8}= \frac{73}{4}.
\]

Write \(\dfrac{73}{4}\) as \(\dfrac{m}{n}\) with \(\gcd(m,n)=1\); then \(m=73,\;n=4\).

\[
m+n = 73+4 = 77.
\]

\[
\boxed{77}
\]

Let the letters be  

\[
A<B<C<D<E<F<G<H<I<J<K<L .
\]

In a random pairing each pair is written with its smaller letter first, so a
pair \(\{x,y\}\;(x<y)\) becomes the word \(xy\).
The six words are then listed alphabetically; the last word is the
lexicographically largest of the six two‑letter words.

Because the first letters of the six words are all different (each letter can
appear as the smaller element of at most one pair), the largest word is the
pair whose **smaller** element is the largest among the six smaller elements.
Hence the last word contains \(G\) precisely when the pair containing \(G\)
has the largest smaller element.

---

### 1.  Choose the partner of \(G\)

\(G\) can be paired with any of the other 11 letters, each with probability
\(\frac1{11}\).

*If \(G\) is paired with a letter larger than \(G\)*  
(let the partner be \(X\in\{H,I,J,K,L\}\)).  
Then the smaller element of the \(G\!-\!X\) pair is \(G\).

The remaining 10 letters consist of  

- 6 letters \(\{A,B,C,D,E,F\}\) smaller than \(G\) (call them “low”),  
- 4 letters among \(\{H,I,J,K,L\}\setminus\{X\}\) larger than \(G\) (call them “high”).

For the pair \(G\!-\!X\) to have the largest smaller element, no other pair
may have a smaller element \(\ge G\); i.e. no “high” letter may be the smaller
letter of its pair.  Hence each high letter must be matched with a low
letter.  

Number of matchings of the 10 remaining letters with no high–high pair:

\[
\binom{6}{4}\,4!\;(=15\cdot24)=360 .
\]

Total matchings of 10 letters: \((10-1)!!=945\).
Thus

\[
P(\text{lexicographically largest pair }|\,G\!-\!X,\;X>G)=\frac{360}{945}
=\frac{8}{21}.
\]

Since there are 5 possible partners larger than \(G\),

\[
\sum_{X>G}\frac1{11}\cdot\frac{8}{21}=5\cdot\frac{1}{11}\cdot\frac{8}{21}
=\frac{40}{231}.
\]

*If \(G\) is paired with a letter smaller than \(G\)*  
the only possible partner is \(F\); for any smaller partner there are fewer
than five letters below it, making it impossible for its pair to have the
largest smaller element.

When \(G\) is paired with \(F\), the smaller element of the pair is \(F\).
The other 10 letters are  

\[
\{A,B,C,D,E\}\ (\text{low})\quad\text{and}\quad\{H,I,J,K,L\}\ (\text{high}).
\]

All five high letters must be paired with the five low letters; otherwise a
high letter would become a smaller element.  The number of such matchings is
\(5!=120\).  Hence

\[
P(\text{largest pair }|\,G\!-\!F)=\frac{120}{945}=\frac{8}{63},
\]
and the contribution is \(\displaystyle\frac1{11}\cdot\frac{8}{63}
= \frac{8}{693}\).

No other partner of \(G\) can work, so the total probability that the last
word contains \(G\) is

\[
\frac{40}{231}+\frac{8}{693}
=\frac{120}{693}+\frac{8}{693}
=\frac{128}{693}.
\]

The fraction \(\frac{128}{693}\) is already in lowest terms.  
Thus \(m=128,\ n=693\) and

\[
m+n = 128+693 = 821.
\]

\[
\boxed{821}
\]

**Solution**

--------------------------------------------------------------------

### 1.  The ratios on the two sides  

\[
AB=AD+DE+EB=4+16+8=28,\qquad 
AC=AF+FG+GC=13+52+26=91 .
\]

\[
\frac{AD}{AB}=\frac{4}{28}=\frac17,\qquad 
\frac{AF}{AC}= \frac{13}{91}= \frac17 .
\]

Hence  

\[
\frac{AD}{AB}= \frac{AF}{AC}= \frac17\qquad\Longrightarrow\qquad 
DF\parallel BC .
\]

Similarly  

\[
\frac{AE}{AB}= \frac{20}{28}= \frac57,\qquad 
\frac{AG}{AC}= \frac{65}{91}= \frac57 ,
\]

so  

\[
\frac{AE}{AB}= \frac{AG}{AC}= \frac57\qquad\Longrightarrow\qquad 
EG\parallel BC .
\]

Let  

\[
p=\frac17,\qquad q=\frac57 .
\]

--------------------------------------------------------------------

### 2.  Area of \(\triangle ABC\)

Because \(DF\parallel EG\parallel BC\), the quadrilateral \(DEGF\) is the
region of \(\triangle ABC\) between the two lines through \(D\) and \(E\)
parallel to \(BC\).  

Triangles \(ADF\) and \(AEG\) are similar to \(\triangle ABC\) with similarity
ratios \(p\) and \(q\) respectively, so

\[
[ADF]=p^{2}[ABC],\qquad[AEG]=q^{2}[ABC].
\]

Hence  

\[
[DEGF]=[AEG]-[ADF]=(q^{2}-p^{2})[ABC]
        =\Bigl(\frac{5^{2}}{7^{2}}-\frac{1^{2}}{7^{2}}\Bigr)[ABC]
        =\frac{24}{49}[ABC].
\]

Given \([DEGF]=288\),

\[
[ABC]=288\cdot\frac{49}{24}=12\cdot 49 =\boxed{588}.
\]

--------------------------------------------------------------------

### 3.  Coordinates

Place  

\[
A=(0,0),\qquad B=(28,0),\qquad C=(c,\,42)
\]

(the height is \(42\) because \([ABC]=\frac12\cdot28\cdot42=588\)).  
Using the ratios \(p,q\),

\[
\begin{aligned}
D &=pB=(4,0),                   &E&=qB=(20,0),\\[2mm]
F &=pC=\Bigl(\frac{c}{7},6\Bigr), &
G &=qC=\Bigl(\frac{5c}{7},30\Bigr).
\end{aligned}
\]

Reflections give  

\[
M=2F-D=\Bigl(\frac{2c}{7}-4,\;12\Bigr),\qquad 
N=2E-G=\Bigl(40-\frac{5c}{7},\;-30\Bigr).
\]

--------------------------------------------------------------------

### 4.  Area of the heptagon \(AFNBCEM\)

List the vertices in order:

\[
\begin{array}{c|c}
\text{Vertex}&(x,y)\\ \hline
A&(0,0)\\
F&\bigl(\frac{c}{7},6\bigr)\\
N&\bigl(40-\frac{5c}{7},-30\bigr)\\
B&(28,0)\\
C&(c,42)\\
E&(20,0)\\
M&\bigl(\frac{2c}{7}-4,12\bigr)
\end{array}
\]

Apply the Shoelace formula.  The sum  

\[
\sum_{i=1}^{7}\bigl(x_i y_{i+1}-y_i x_{i+1}\bigr)
\]  

(with \(x_8=x_1,\;y_8=y_1\)) simplifies dramatically:

\[
\begin{aligned}
&0\\
&\;-\;240\\
&\;+\;840\\
&\;+\;1176\\
&\;-\;840\\
&\;+\;240\\
&0
\end{aligned}
\qquad =\;1176 .
\]

Thus

\[
[AFNBCEM]=\frac12\cdot1176=588 .
\]

The result does **not** depend on the unknown coordinate \(c\); the
cancellation occurs because the vectors \(DF\) and \(EG\) are parallel to
\(BC\).

--------------------------------------------------------------------

### 5.  Answer

\[
\boxed{588}
\]

Let  

\[
P:\;y=x^{2}-4
\]

and rotate \(P\) \(60^{\circ}\) counter‑clockwise about the origin.  
If \((x,y)\) lies on the rotated curve, then there is a point \((u,v)\) on
\(P\) such that  

\[
\begin{pmatrix}x\\y\end{pmatrix}=R\begin{pmatrix}u\\v\end{pmatrix},
\qquad 
R=\begin{pmatrix}\tfrac12&-\tfrac{\sqrt3}{2}\\[2pt]\tfrac{\sqrt3}{2}&\tfrac12\end{pmatrix}.
\]

Equivalently, a point \((x,y)\) is on both curves iff

\[
y=x^{2}-4 \qquad\text{and}\qquad 
R^{T}\!\begin{pmatrix}x\\y\end{pmatrix}
   =\begin{pmatrix}x'\\y'\end{pmatrix}
\text{ satisfies }y'=(x')^{2}-4,
\]

where  

\[
R^{T}= \begin{pmatrix}\tfrac12&\tfrac{\sqrt3}{2}\\[2pt]-\tfrac{\sqrt3}{2}&\tfrac12\end{pmatrix},
\quad 
x'=\frac{x}{2}+\frac{\sqrt3\,y}{2},\qquad
y'=-\frac{\sqrt3\,x}{2}+\frac{y}{2}.
\]

Using \(y=x^{2}-4\),

\[
x'=\frac{\sqrt3 x^{2}+x-4\sqrt3}{2},\qquad 
y'=\frac{x^{2}-\sqrt3 x-4}{2}.
\]

The condition \(y'=(x')^{2}-4\) yields  

\[
\frac{x^{2}-\sqrt3 x-4}{2}
   =\Bigl(\frac{\sqrt3 x^{2}+x-4\sqrt3}{2}\Bigr)^{2}-4 .
\]

Multiplying by \(2\) and simplifying gives the quartic

\[
3x^{4}+2\sqrt3\,x^{3}-25x^{2}-6\sqrt3\,x+40=0.
\]

Factor this over \(\mathbb{Q}(\sqrt3)\):

\[
\begin{aligned}
3x^{4}+2\sqrt3\,x^{3}-25x^{2}-6\sqrt3\,x+40
&=(x^{2}+\sqrt3\,x-4)(3x^{2}-\sqrt3\,x-10).
\end{aligned}
\]

Hence the possible \(x\)-coordinates are the roots of the two quadratics.

\[
\begin{cases}
x^{2}+\sqrt3\,x-4=0 &\Longrightarrow
\displaystyle x=\frac{-\sqrt3\pm\sqrt{19}}2,\\[6pt]
3x^{2}-\sqrt3\,x-10=0 &\Longrightarrow
\displaystyle x=\frac{\sqrt3\pm\sqrt{123}}6 .
\end{cases}
\]

We need a point in the fourth quadrant, i.e. \(x>0\) and \(y=x^{2}-4<0\).
Only  

\[
x=\frac{-\sqrt3+\sqrt{19}}2\;(>0,\;<2)
\]

satisfies this.  Its \(y\)-coordinate is

\[
\begin{aligned}
y&=x^{2}-4
   =\frac{(-\sqrt3+\sqrt{19})^{2}}{4}-4
   =\frac{22-2\sqrt{57}}4-4\\
  &=\frac{11-\sqrt{57}}2-4
   =\frac{3-\sqrt{57}}2 .
\end{aligned}
\]

Thus  

\[
y=\frac{a-\sqrt{b}}{c}\quad\text{with}\quad a=3,\;b=57,\;c=2,
\]

and \(a\) and \(c\) are relatively prime.  

Finally

\[
a+b+c=3+57+2=62.
\]

\[
\boxed{62}
\]

\[
\begin{aligned}
&x-yz<y-zx<z-xy \\
\Longleftrightarrow\;&x-yz<y-zx\quad\text{and}\quad y-zx<z-xy  \\
\Longleftrightarrow\;&(x-y)(1+z)<0\quad\text{and}\quad (y-z)(1+x)<0 .
\end{aligned}
\]

Hence  

\[
\begin{cases}
(x-y)(1+z)<0,\\[2pt]
(y-z)(1+x)<0 .
\end{cases}
\]

Each product being negative means the two factors have opposite signs.  
Thus we have four possible sign‑combinations:

\[
\begin{array}{c|c|c}
\text{Case}& (x-y,\;1+z) & (y-z,\;1+x)\\ \hline
\text{I}& (+,\; -) & (+,\; -)\\
\text{II}& (+,\; -) & (-,\; +)\\
\text{III}& (-,\; +) & (+,\; -)\\
\text{IV}& (-,\; +) & (-,\; +)
\end{array}
\]

Translating them into linear inequalities:

\[
\begin{aligned}
\text{I}&:\; x>y,\;z<-1,\;y>z,\;x<-1 \qquad\text{(impossible since }x+y+z=75)\\
\text{II}&:\; x>y,\;z<-1,\;y<z,\;x>-1 \\
\text{III}&:\; x<y,\;z>-1,\;y>z,\;x<-1 \\
\text{IV}&:\; x<y,\;z>-1,\;y<z,\;x>-1 .
\end{aligned}
\]

Thus three non‑empty convex regions arise from cases II, III, IV.
Cases II and III are unbounded, while case IV gives a bounded region.

---

### Bounded region (case IV)

Case IV is equivalent to  

\[
-1<x<y<z ,\qquad x+y+z=75 .
\]

The boundary of this region in the plane \(x+y+z=75\) consists of the three lines  

\[
\begin{cases}
x=-1,\\[2pt]
y=x,\\[2pt]
z=y .
\end{cases}
\]

Intersecting each pair of lines with the plane gives the vertices

\[
\begin{aligned}
A &: x=-1,\;y=-1 \;\Rightarrow\;z=77 ,&&A(-1,-1,77),\\
B &: x=-1,\;z=y \;\Rightarrow\;y=z=38 ,&&B(-1,38,38),\\
C &: y=x,\;z=y \;\Rightarrow\;x=y=z=25 ,&&C(25,25,25).
\end{aligned}
\]

Hence the finite region is the triangle \(ABC\).

---

### Area of the triangle

Vectors along two sides:

\[
\begin{aligned}
\vec{AB}&=B-A=(0,39,-39),\\
\vec{AC}&=C-A=(26,26,-52).
\end{aligned}
\]

\[
\vec{AB}\times\vec{AC}=(-1014,-1014,-1014)
\quad\Longrightarrow\quad 
\bigl\lVert\vec{AB}\times\vec{AC}\bigr\rVert
=1014\sqrt3 .
\]

Therefore

\[
\text{Area}= \frac12\bigl\lVert\vec{AB}\times\vec{AC}\bigr\rVert
           =\frac12\cdot1014\sqrt3
           =507\sqrt3 .
\]

Thus the area is \(a\sqrt b\) with \(a=507,\;b=3\), and

\[
a+b=507+3=510.
\]

\[
\boxed{510}
\]

First write the equation for the intersection points.

The parabola is  

\[
x=34y^{2},
\]

and the periodic saw‑tooth satisfies  

\[
f(x+4)=f(x),\qquad   
f(x)=\begin{cases}
x,&-1\le x<1,\\[2pt]
2-x,&1\le x<3 .
\end{cases}
\]

Hence an intersection must satisfy  

\[
y=f(34y^{2}). \tag{1}
\]

--------------------------------------------------------------------
### 1.  Express \(f\) using the modulus

For any real \(x\) let  

\[
t=(x+1)\pmod 4,\qquad 0\le t<4 .
\]

Then  

\[
f(x)=
\begin{cases}
t-1,&0\le t<2,\\[2pt]
3-t,&2\le t<4 .
\end{cases}
\]

Put \(x=34y^{2}\).  Let  

\[
s=34y^{2}+1,\qquad t=s\pmod4 .
\]

--------------------------------------------------------------------
### 2.  Two linear possibilities

*If \(0\le t<2\) (the “rising” part) then \(y=t-1\) and  
\(t=y+1\).  Hence  

\[
34y^{2}+1=4n+y+1\quad\Longrightarrow\quad 34y^{2}-y-4n=0,\tag{2}
\]

with the condition \(y\in[-1,1)\).

*If \(2\le t<4\) (the “falling” part) then \(y=3-t\) and  
\(t=3-y\).  Hence  

\[
34y^{2}+1=4n+3-y\quad\Longrightarrow\quad 34y^{2}+y-(2+4n)=0,\tag{3}
\]

with the condition \(y\in(-1,1]\).

Here \(n\) is the integer \(\displaystyle n=\Big\lfloor\frac{34y^{2}+1}{4}\Big\rfloor\).
Since \(0\le 34y^{2}+1\le 35\), we must have \(0\le n\le 8\).

--------------------------------------------------------------------
### 3.  Solve the quadratics

**From (2):**  

\[
y=\frac{1\pm\sqrt{1+544n}}{68}\qquad (n=0,1,\dots ,8).
\]

Both roots lie in \([-1,1)\); for each \(n\) their sum is  

\[
\frac{1}{34}.
\]

Thus  

\[
S_I=\sum_{n=0}^{8}\frac1{34}= \frac{9}{34}. \tag{4}
\]

**From (3):**  

\[
y=\frac{-1\pm\sqrt{273+544n}}{68}\qquad (n=0,1,\dots ,8).
\]

For \(n=0,\dots ,7\) both roots satisfy \((-1,1]\); for \(n=8\) the
negative root is \(\le -1\) and must be discarded.
Hence for \(n=0,\dots ,7\) each pair sums to \(-\dfrac1{34}\), and we add
the single remaining root for \(n=8\):

\[
S_{II}=8\!\left(-\frac1{34}\right)+\frac{-1+\sqrt{4625}}{68}
        =-\frac{8}{34}+\frac{-1+5\sqrt{185}}{68}. \tag{5}
\]

--------------------------------------------------------------------
### 4.  Total sum of \(y\)-coordinates

\[
\begin{aligned}
S &=S_I+S_{II}\\
  &=\frac{9}{34}-\frac{8}{34}+\frac{-1+5\sqrt{185}}{68}\\
  &=\frac{1}{34}+\frac{-1+5\sqrt{185}}{68}\\[2pt]
  &=\frac{2-1+5\sqrt{185}}{68}
   =\frac{1+5\sqrt{185}}{68}.
\end{aligned}
\]

Thus the sum can be written as \(\displaystyle \frac{a+b\sqrt{c}}{d}\) with  

\[
a=1,\qquad b=5,\qquad c=185,\qquad d=68,
\]

where \(a,b,d\) are coprime and \(c\) is square‑free.

Finally  

\[
a+b+c+d = 1+5+185+68 = 259.
\]

\[
\boxed{259}
\]

The condition  

\[
a^3+b^3+c^3\equiv 0\pmod {3^7}\qquad (3^7=2187)
\]

depends only on the residues of \(a,b,c\) modulo \(3^7\).  
Write each number as  

\[
a=3^{3\alpha}q,\qquad \alpha\in\{0,1,2\},\;q\in Q,
\]

where  

* \(Q\) = the set of cubic residues among units modulo \(3^7\);  
  \(|Q|=486\);
* for \(\alpha=0,1,2\) the sets  

\[
R_\alpha =\{3^{3\alpha}q\;|\;q\in Q\}
\]

have sizes  

\[
|R_0|=486,\;|R_1|=162,\;|R_2|=54 .
\]

All numbers divisible by \(27\) contribute \(0\) to the sum, so we only need the
ordered triples whose components lie in  

\[
R=R_0\cup R_1\cup R_2 .
\]

Denote  

\[
M_2 =\#\{(x,y)\in R^2 : x+y\equiv 0\},\qquad 
M_3 =\#\{(x,y,z)\in R^3 : x+y+z\equiv 0\}.
\]

Because \(-x\in R\) for every \(x\in R\), we have \(M_2=|R|=702\).

The total number of required triples of positive integers is  

\[
\begin{aligned}
N&=27^3      \;(\text{all three multiples of }27) \\
  &\quad+ 3\cdot27\cdot M_2 \;(\text{exactly two non‑multiples of }27)\\
  &\quad+ M_3 .
\end{aligned}
\]

Thus  

\[
N=19683+81\cdot702+M_3=19683+56862+M_3 .
\]

--------------------------------------------------------------------
### Computing \(M_3\)

Let  

\[
f(z)=\mathbf 1_R(z),\qquad
F(t)=\sum_{z\in\mathbf Z/3^7\mathbf Z}f(z) e^{2\pi i t z/3^7}
\]

be the Fourier transform of the indicator of \(R\).  
Then

\[
M_3=\frac1{3^7}\sum_{t=0}^{3^7-1}F(t)^3 .
\]

Writing \(R\) as \(Q\cup27Q\cup729Q\) and using that each element of
\(R_\alpha\) is obtained from \(|Q|/3^\alpha\) elements of \(Q\), we obtain

\[
F(t)=\sum_{\alpha=0}^{2}\frac1{3^{\alpha}}
        \sum_{q\in Q} e^{2\pi i t\cdot3^{3\alpha} q/3^7}.
\]

The inner sums are evaluated with the three characters of order 3 on
\((\mathbf Z/3^7\mathbf Z)^\times\).  One finds that \(F(t)\) depends only on
\(v=v_3(t)\) (the 3‑adic valuation of \(t\)):

\[
\begin{array}{c|c|c}
v &\text{number of }t &F(t)\\\hline
0 &1458 & -27\\
1 &486  & 54\\
2 &162  & 54\bigl(1+\operatorname{Re}A\bigr)\\
3 &54   & -27\\
4 &18   & 216\\
5 &6    & 216+162\,\operatorname{Re}A\\
6 &2    & -27
\end{array}
\]

where  

\[
A=3e^{2\pi i/9}=3\zeta_9,\qquad 
\operatorname{Re}A=3\cos\frac{2\pi}{9}.
\]

From the values
\[
\operatorname{Re}A=3\cos\frac{2\pi}{9},\qquad
\cos\frac{2\pi}{9}+ \cos\frac{4\pi}{9}+ \cos\frac{8\pi}{9}=0,
\]
one obtains  

\[
\sum_{i=1}^3 (1+r_i)^3=\frac{267}{8},\qquad
\sum_{i=1}^3\Bigl(\frac43+r_i\Bigr)^3=\frac{3671}{72},
\]
where \(r_1,r_2,r_3\) are the three real numbers
\(3\cos\frac{2\pi}{9},\,3\cos\frac{4\pi}{9},\,3\cos\frac{8\pi}{9}\).

Carrying out the sum

\[
\sum_{t}F(t)^3
 =702^3-1514\cdot27^3+486\cdot54^3+18\cdot216^3
   +54^4\!\left(\frac{267}{8}\right)+8\,503\,056\!\left(\frac{3671}{72}\right)
 =1\,291\,401\,630 .
\]

Since \(3^7=2187\),

\[
M_3=\frac{1\,291\,401\,630}{2187}=590\,490 .
\]

--------------------------------------------------------------------
### Final count

\[
\begin{aligned}
N &=19683+56862+590490\\
  &=667\,035 .
\end{aligned}
\]

The problem asks for \(N\bmod 1000\):

\[
\boxed{35}.
\]

**Step 1 –   What the conditions mean**

The grid has three rows and nine columns.  
The columns are grouped into three \(3\times3\) blocks  

\[
B_{1}=\{1,2,3\},\qquad B_{2}=\{4,5,6\},\qquad B_{3}=\{7,8,9\}.
\]

For each row we must use the numbers \(1,2,\dots ,9\) exactly once, and for each
block the nine cells of that block must also contain the numbers \(1,\dots ,9\) exactly once.

Hence for each digit \(d\) :

* it appears once in each of the three rows,  
* it appears once in each of the three blocks.

Consequently the three positions of a digit form a **bijection** between rows and blocks.
Let \(\sigma _d\in S_{3}\) be the permutation defined by

\[
\sigma _d(i)=\text{the block that contains }d\text{ in row }i .
\]

Thus each digit contributes one ‘1’ in the \(3\times3\) permutation matrix of \(\sigma _d\).

--------------------------------------------------------------------

**Step 2 –   The matrix condition**

If we add the nine permutation matrices of the nine digits we obtain a matrix
\(M\) whose entry \(M_{i,b}\) counts how many digits lie in row \(i\) and block \(b\).
The block condition says that each block must contain each digit once, i.e. each
cell \((i,b)\) must contain exactly three digits. Therefore

\[
M=3J,\qquad J=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}.
\]

So we must write \(3J\) as a sum of nine (not‑necessarily distinct) permutation
matrices of size \(3\).

--------------------------------------------------------------------

**Step 3 –   Types of permutations**

\(S_{3}\) has six elements:

\[
\begin{aligned}
&\text{id},\;(12),\;(13),\;(23),\;(123),\;(132).
\end{aligned}
\]

Let  

\[
a=x_{\text{id}}=x_{(123)}=x_{(132)},\qquad
b=x_{(12)}=x_{(13)}=x_{(23)} .
\]

Counting the contributions of each type to the nine entries of \(M\) gives the
system  

\[
\begin{cases}
a+b=3\\[2pt]
a+b=3\\[2pt]
a+b=3\\[2pt]
a+b=3,
\end{cases}
\qquad\Longrightarrow\qquad a+b=3 .
\]

Hence \(a\) may be \(0,1,2,3\) and \(b=3-a\).  For a fixed value of \(a\),

\[
\begin{aligned}
x_{\text{id}}&=a,\\
x_{(123)}&=a,\\
x_{(132)}&=a,\\
x_{(12)}&=x_{(13)}=x_{(23)}&=3-a .
\end{aligned}
\]

--------------------------------------------------------------------

**Step 4 –   Assigning the nine digits to the permutations**

The nine digits are distinct, so for a given \(a\) the number of ways to give the
six permutation types to the digits is the multinomial coefficient

\[
N_a=\frac{9!}{a!\,a!\,a!\,(3-a)!\,(3-a)!\,(3-a)!}
    =\frac{9!}{a!^{3}(3-a)!^{3}} .
\]

Summing over the four possible values of \(a\),

\[
\sum_{a=0}^{3}N_a
   =\frac{9!}{0!^{3}3!^{3}}+\frac{9!}{1!^{3}2!^{3}}
    +\frac{9!}{2!^{3}1!^{3}}+\frac{9!}{3!^{3}0!^{3}}
   =94\,080 .
\]

--------------------------------------------------------------------

**Step 5 –   Ordering inside each row‑block**

For a fixed assignment of the permutations, each of the nine
\( ( \text{row},\text{block} )\) cells contains exactly three different digits.
These three digits can be placed in the three columns of that cell in any order,
i.e. in \(3! = 6\) ways, independently of the other cells.
Hence we obtain an extra factor \((3!)^{9}=6^{9}\).

Thus the total number of admissible \(3\times9\) grids is

\[
N = 94\,080\cdot 6^{9}
   =9!\cdot 6^{6}\cdot 56 .
\]

--------------------------------------------------------------------

**Step 6 –   Prime factorisation**

\[
\begin{aligned}
9! &=2^{7}\,3^{4}\,5\,7,\\
6^{6}&=(2\cdot3)^{6}=2^{6}\,3^{6},\\
56 &=2^{3}\,7 .
\end{aligned}
\]

Multiplying gives  

\[
N=2^{7+6+3}\,3^{4+6}\,5^{1}\,7^{1+1}
  =2^{16}\,3^{10}\,5^{1}\,7^{2}.
\]

Thus we may write  

\[
N = p^{a}\,q^{b}\,r^{c}\,s^{d}
\quad\text{with}\quad
p=2,\;a=16;\;\; q=3,\;b=10;\;\; r=5,\;c=1;\;\; s=7,\;d=2 .
\]

--------------------------------------------------------------------

**Step 7 –   Required sum**

\[
p\cdot a+q\cdot b+r\cdot c+s\cdot d
 =2\cdot16+3\cdot10+5\cdot1+7\cdot2
 =32+30+5+14=81 .
\]

--------------------------------------------------------------------

\[
\boxed{81}
\]

Let the line containing \(A,B,C,D,E,F\) be the \(x\)-axis.

\[
A=(0,0),\qquad B=(b,0),\ C=(c,0),\ D=(d,0),\ E=(e,0),\ F=(f,0)
\]

From the given distances along the line  

\[
\begin{aligned}
AC&=c-a=26 \Longrightarrow c=26,\\
BD&=d-b=22,\\
CE&=e-c=31 \Longrightarrow e=26+31=57,\\
DF&=f-d=33,\\
AF&=f-a=73 \Longrightarrow f=73 .
\end{aligned}
\]

From \(DF=33\) we get \(d=f-33=73-33=40\).  
Then \(BD=22\) gives \(b=d-22=40-22=18\).

Thus the collinear points are

\[
A(0),\ B(18),\ C(26),\ D(40),\ E(57),\ F(73).
\]

---

### Coordinates of \(G\)

Let \(G=(x,y)\) with \(y\neq0\).  
We know

\[
\begin{cases}
(CG)^2=(x-26)^2+y^2=40^2=1600,\\
(DG)^2=(x-40)^2+y^2=30^2=900 .
\end{cases}
\]

Subtracting the equations:

\[
(x-26)^2-(x-40)^2 = 1600-900=700 .
\]

Compute the left side:

\[
\bigl(x^2-52x+676\bigr)-\bigl(x^2-80x+1600\bigr)=28x-924 .
\]

Hence \(28x-924=700\) giving  

\[
x=\frac{1624}{28}=58.
\]

Now use \((x-40)^2+y^2=900\):

\[
(58-40)^2+y^2=900\Longrightarrow 18^2+y^2=900\Longrightarrow y^2=576.
\]

Thus \(y=\pm24\).  The sign does not affect the area; we take \(y=24\).

Therefore  

\[
G=(58,24)\;(\text{or }(58,-24)).
\]

---

### Area of \(\triangle BGE\)

Points:
\[
B=(18,0),\qquad G=(58,24),\qquad E=(57,0).
\]

Since \(B\) and \(E\) lie on the \(x\)-axis, the base \(BE\) has length  

\[
BE = 57-18 = 39.
\]

The altitude from \(G\) to the \(x\)-axis is \(|y|=24\).

Hence

\[
\text{Area}=\frac12\cdot (\text{base})\cdot (\text{height})
          =\frac12\cdot 39\cdot 24
          =468.
\]

\[
\boxed{468}
\]

\[
x_{k+1}= \frac{x_k+\frac1{x_k}-1}{3},\qquad x_1=\frac{25}{11}
\]

---

### 1.  A convenient substitution  

Let  

\[
y_k=\frac{x_k-1}{x_k+1}\qquad\Longleftrightarrow\qquad 
x_k=\frac{1+y_k}{1-y_k}.
\]

Then a short computation gives  

\[
y_{k+1}= \frac{3y_k^{\,2}-1}{2}. \tag{1}
\]

Thus we only have to iterate the quadratic map (1).

---

### 2.  Scaling  

Define  

\[
u_k=\frac32\,y_k .
\]

From (1)

\[
u_{k+1}=u_k^{2}-\frac34 .\tag{2}
\]

Let  

\[
u_k=\frac{a_k}{12^{\,2^{k-1}}},\qquad a_1=7 .
\]

Because the denominator squares each step, indeed  

\[
12^{\,2^{k}}=(12^{\,2^{k-1}})^2 .
\]

Using (2) one checks that  

\[
a_{k+1}=a_k^{2}-\frac34\cdot12^{\,2^{k}} .
\]

Only the denominator grows; the exact value of the numerator is not needed later.

---

### 3.  Returning to \(x_k\)

Since \(y_k=\frac{2}{3}u_k\),

\[
x_k=\frac{1+y_k}{1-y_k}
     =\frac{3+2u_k}{3-2u_k}
     =\frac{3\cdot12^{\,2^{k-1}}+2a_k}
            {3\cdot12^{\,2^{k-1}}-2a_k}.
\]

Hence for any \(k\),

\[
x_k=\frac{M_k}{N_k},\qquad 
M_k+N_k =6\cdot12^{\,2^{k-1}} .
\]

In particular

\[
m+n = M_{2025}+N_{2025}= 6\cdot 12^{\,2^{2024}} .
\]

---

### 4.  Computing \(6\cdot12^{2^{2024}}\pmod{1000}\)

We need \(6\cdot12^{K}\pmod{1000}\) where \(K=2^{2024}\).

Because \(1000=8\cdot125\) we work modulo 8 and 125.

*Modulo 8*:  
\(12\equiv4\pmod8\), and \(4^{2}=0\pmod8\).  
Since \(K\ge2\), \(12^{K}\equiv0\pmod8\) ⇒ the whole expression is \(0\pmod8\).

*Modulo 125*:  \(\gcd(12,125)=1\) and \(\phi(125)=100\).  
Thus \(12^{100}\equiv1\pmod{125}\); we need \(K\bmod100\).

The powers of 2 modulo 100 repeat with period 20 for exponents \(\ge2\).  
\(2024\equiv4\pmod{20}\), so \(2^{2024}\equiv2^{4}=16\pmod{100}\).  
Hence \(12^{K}\equiv12^{16}\pmod{125}\).

Compute:
\[
12^{2}\equiv19,\;12^{4}\equiv111,\;12^{8}\equiv71,\;12^{16}\equiv71^{2}\equiv41\pmod{125}.
\]

Thus \(6\cdot12^{K}\equiv6\cdot41=246\equiv121\pmod{125}\).

Now solve the CRT:
\[
X\equiv0\pmod8,\qquad X\equiv121\pmod{125}.
\]

Let \(X=8t\). Then \(8t\equiv121\pmod{125}\Rightarrow t\equiv47\cdot121\equiv62\pmod{125}\) (since \(8^{-1}\equiv47\pmod{125}\)).  
Thus \(X=8\cdot62=496\pmod{1000}\).

Therefore

\[
m+n \equiv 496 \pmod{1000}.
\]

---

\[
\boxed{496}
\]

Let  

\[
m=n+2\qquad (m>2)
\]

Then  

\[
n+2=m,\qquad n+3=m+1,\qquad n^{2}+9=(m-2)^{2}+9=m^{2}-4m+13 .
\]

The condition  

\[
n+2\mid 3(n+3)(n^{2}+9)
\]

becomes  

\[
m \mid 3(m+1)(m^{2}-4m+13). \tag{1}
\]

--------------------------------------------------------------------
### 1.  Using the gcd with 3

Because  

\[
\gcd(m,m+1)=1,
\]

any prime divisor of \(m\) cannot come from the factor \(m+1\).  
The only possible common factor of \(m\) with the whole right‑hand side of (1) is the factor \(3\).

Let  

\[
d=\gcd(m,3),\qquad m=d\cdot m',\qquad \gcd(m',3)=1 .
\]

Then (1) is equivalent to  

\[
d m' \mid 3(m+1)(m^{2}-4m+13)
      = d\Bigl(\frac{3}{d}\Bigr)(m+1)(m^{2}-4m+13).
\]

Cancelling \(d\) gives  

\[
m' \mid \Bigl(\frac{3}{d}\Bigr)(m+1)(m^{2}-4m+13).\tag{2}
\]

Since \(\gcd(m',m+1)=1\) and also \(\gcd(m',3)=1\), the factor \((m+1)\) and the possible remaining factor \(\frac{3}{d}\) cannot help the divisibility. Hence (2) reduces to  

\[
m' \mid (m^{2}-4m+13). \tag{3}
\]

--------------------------------------------------------------------
### 2.  Reduce modulo \(m\)

\[
m^{2}-4m+13 \equiv 13\pmod{m},
\]

so (3) is equivalent to  

\[
m' \mid 13.
\]

Recall \(m' =\dfrac{m}{\gcd(m,3)}\).  Therefore

\[
\frac{m}{\gcd(m,3)}\mid 13.
\]

--------------------------------------------------------------------
### 3.  Solve the two possibilities

**(i)  \(\gcd(m,3)=1\) (i.e. \(3\nmid m\)).**  
Then \(\displaystyle m\mid13\).  
With \(m>2\), the only possibility is \(m=13\).

**(ii)  \(\gcd(m,3)=3\) (i.e. \(m=3k\)).**  
Then \(\displaystyle k\mid13\).  
Thus \(k=1\) or \(k=13\), giving \(m=3\) or \(m=39\).

--------------------------------------------------------------------
### 4.  Return to \(n\)

\[
n=m-2\; \Longrightarrow\; 
\begin{cases}
m=3 &\Rightarrow n=1,\\[2pt]
m=13 &\Rightarrow n=11,\\[2pt]
m=39 &\Rightarrow n=37.
\end{cases}
\]

All three values satisfy the original divisibility (direct verification shows \(3\mid120\), \(13\mid5460\), \(39\mid165360\)).

--------------------------------------------------------------------
### 5.  Sum of all such \(n\)

\[
1+11+37 = 49.
\]

\[
\boxed{49}
\]

First rewrite each factor using the change–of–base formula  

\[
\log_a b=\frac{\ln b}{\ln a}.
\]

\[
\frac{\log_k\bigl(5^{k^{2}-1}\bigr)}{\log_{k+1}\bigl(5^{k^{2}-4}\bigr)}
   =\frac{\dfrac{\ln5^{\,k^{2}-1}}{\ln k}}
          {\dfrac{\ln5^{\,k^{2}-4}}{\ln (k+1)}}
   =\frac{(k^{2}-1)\ln5}{\ln k}\cdot\frac{\ln(k+1)}{(k^{2}-4)\ln5}
   =\frac{k^{2}-1}{k^{2}-4}\cdot\frac{\ln(k+1)}{\ln k}.
\]

Hence

\[
P=\prod_{k=4}^{63}\frac{k^{2}-1}{k^{2}-4}
   \cdot\prod_{k=4}^{63}\frac{\ln(k+1)}{\ln k}.
\]

---

### 1.  The logarithmic product

\[
\prod_{k=4}^{63}\frac{\ln(k+1)}{\ln k}
   =\frac{\ln64}{\ln4}
   =\log_{4}64=3.
\]

(The terms cancel telescopically.)

---

### 2.  The rational product  

Factor each term:

\[
\frac{k^{2}-1}{k^{2}-4}
   =\frac{(k-1)(k+1)}{(k-2)(k+2)}
   =\frac{k-1}{k-2}\cdot\frac{k+1}{k+2}.
\]

Thus

\[
\prod_{k=4}^{63}\frac{k^{2}-1}{k^{2}-4}
 =\Bigl(\prod_{k=4}^{63}\frac{k-1}{k-2}\Bigr)
  \Bigl(\prod_{k=4}^{63}\frac{k+1}{k+2}\Bigr).
\]

The first product telescopes:

\[
\prod_{k=4}^{63}\frac{k-1}{k-2}
 =\frac{3}{2}\cdot\frac{4}{3}\cdot\cdots\cdot\frac{62}{61}
 =\frac{62}{2}=31.
\]

The second product also telescopes:

\[
\prod_{k=4}^{63}\frac{k+1}{k+2}
 =\frac{5}{6}\cdot\frac{6}{7}\cdot\cdots\cdot\frac{64}{65}
 =\frac{5}{65}=\frac{1}{13}.
\]

Therefore

\[
\prod_{k=4}^{63}\frac{k^{2}-1}{k^{2}-4}
   =31\cdot\frac{1}{13}=\frac{31}{13}.
\]

---

### 3.  Combine the two parts

\[
P=\frac{31}{13}\times3=\frac{93}{13}.
\]

The fraction is already in lowest terms (\(\gcd(93,13)=1\)).  
Thus \(m=93\), \(n=13\) and

\[
m+n=93+13=106.
\]

\[
\boxed{106}
\]

Let the four unit squares be  

\[
\begin{array}{c|c}
\text{TL}&\text{TR}\\\hline
\text{BL}&\text{BR}
\end{array}
\]

and denote the twelve unit edges as follows  

* Horizontal edges:  

\[
\begin{aligned}
&h_{1}= \text{top of TL},\qquad h_{2}= \text{top of TR},\\
&h_{3}= \text{middle left},\qquad h_{4}= \text{middle right},\\
&h_{5}= \text{bottom of BL},\qquad h_{6}= \text{bottom of BR}.
\end{aligned}
\]

* Vertical edges:  

\[
\begin{aligned}
&v_{1}= \text{left of TL},\qquad v_{2}= \text{left of BL},\\
&v_{3}= \text{middle top},\qquad v_{4}= \text{middle bottom},\\
&v_{5}= \text{right of TR},\qquad v_{6}= \text{right of BR}.
\end{aligned}
\]

For each square the sum of the red edges must be \(2\).
Writing a red edge as \(1\) and a blue edge as \(0\),

\[
\begin{aligned}
\text{TL}&:\;h_{1}+h_{3}+v_{1}+v_{3}=2,\\
\text{TR}&:\;h_{2}+h_{4}+v_{3}+v_{5}=2,\\
\text{BL}&:\;h_{3}+h_{5}+v_{2}+v_{4}=2,\\
\text{BR}&:\;h_{4}+h_{6}+v_{4}+v_{6}=2.
\end{aligned}
\]

The four **interior** edges are \(h_{3},h_{4},v_{3},v_{4}\); the other eight edges lie on the outer
boundary.  

Fix the colours of the interior edges:
\[
a=h_{3},\qquad b=h_{4},\qquad c=v_{3},\qquad d=v_{4}\in\{0,1\}.
\]

Then for each square we know the required sum of its two boundary edges:

\[
\begin{aligned}
h_{1}+v_{1}&=2-(a+c),\\
h_{2}+v_{5}&=2-(b+c),\\
h_{5}+v_{2}&=2-(a+d),\\
h_{6}+v_{6}&=2-(b+d).
\end{aligned}
\]

If the right‑hand side equals \(2\) or \(0\) the two boundary edges are forced to be
both red or both blue – exactly one choice.  
If it equals \(1\) the two edges must be of opposite colours – two choices.

Hence, for a fixed interior vector \((a,b,c,d)\) the number of admissible colourings of the
boundary edges is  

\[
2^{\,k(a,b,c,d)},
\]

where \(k\) is the number of squares whose interior‑edge sum equals \(1\); i.e.

\[
k=(a\oplus c)+(b\oplus c)+(a\oplus d)+(b\oplus d),
\]
(\(\oplus\) denotes XOR).

Therefore the total number of colourings is  

\[
\sum_{a,b,c,d\in\{0,1\}}2^{k}
= \sum_{c,d} \Bigl(\sum_{a}2^{\,a\oplus c + a\oplus d}\Bigr)^{2}.
\]

For the four possibilities of \((c,d)\) we have

\[
\begin{array}{c|c}
(c,d) & \displaystyle\sum_{a}2^{a\oplus c + a\oplus d} \\ \hline
(0,0) & 1+4 = 5\\
(1,1) & 4+1 = 5\\
(0,1),(1,0) & 2+2 = 4
\end{array}
\]

Thus

\[
\text{total}=2\cdot5^{2}+2\cdot4^{2}=2\cdot25+2\cdot16=50+32=82.
\]

Hence there are  

\[
\boxed{82}
\]

different colourings of the twelve unit edges so that each of the four unit squares has
exactly two red and two blue sides.

The divisors of  

\[
2025 = 3^{4}\cdot5^{2}
\]

are obtained by choosing the exponent of \(3\) from \(\{0,1,2,3,4\}\) and the exponent of \(5\) from \(\{0,1,2\}\).  
Hence  

\[
|A|=(4+1)(2+1)=15 .
\]

---

### Condition for the LCM to be \(2025\)

For a non‑empty subset \(B\subseteq A\),

\[
\operatorname{lcm}(B)=2025 \iff
\begin{cases}
\text{some element of }B\text{ has }3\text{-exponent }4,\\[2pt]
\text{some element of }B\text{ has }5\text{-exponent }2.
\end{cases}
\]

Define  

\[
S_3=\{81,\,405,\,2025\} \quad (\text{divisors with }3^4) ,
\]
\[
S_5=\{25,\,75,\,225,\,675,\,2025\} \quad (\text{divisors with }5^2).
\]

Thus \(B\) must intersect both \(S_3\) and \(S_5\).

---

### Counting favorable subsets

Let \(U=S_3\cup S_5\).  
\(|S_3|=3,\;|S_5|=5,\;|S_3\cap S_5|=1\) (the element \(2025\)), so

\[
|U|=3+5-1=7.
\]

The remaining divisors of \(2025\) (those not in \(U\)) are \(15-7=8\) in number and can be chosen freely.

Number of subsets of \(U\) that contain at least one element of \(S_3\) **and** at least one element of \(S_5\):

\[
\begin{aligned}
&2^{7}\;(\text{all subsets of }U)\\
&\quad-2^{4}\;(\text{subsets with no }S_3)\\
&\quad-2^{2}\;(\text{subsets with no }S_5)\\
&\quad+2^{0}\;(\text{subsets with neither }S_3\text{ nor }S_5)\\[2pt]
&=128-16-4+1=109 .
\end{aligned}
\]

For each of these \(109\) choices, any subset of the other 8 divisors may be added, giving a factor \(2^{8}=256\).

Hence the number of favorable subsets is  

\[
N=109\cdot256=27904.
\]

---

### Probability

All subsets of \(A\) (including the empty set) number  

\[
2^{15}=32768.
\]

Thus

\[
P=\frac{N}{2^{15}}=\frac{27904}{32768}.
\]

Dividing numerator and denominator by their greatest common divisor \(256\),

\[
P=\frac{27904/256}{32768/256}=\frac{109}{128}.
\]

So \(\displaystyle m=109,\;n=128\) and

\[
m+n=109+128=237.
\]

\[
\boxed{237}
\]

\[
\boxed{45}
\]

Let the centre of the large circle \(\omega _2\) be the origin \(O(0,0)\).
Since \(BC\) is a diameter of \(\omega _2\), take  

\[
B=(15,0),\qquad C=(-15,0),\qquad O=(0,0).
\]

Because \(\omega _1\) (radius \(6\)) is internally tangent to \(\omega _2\) at \(B\),
its centre \(A\) lies on \(OB\) with  

\[
OA=15-6=9\quad\Longrightarrow\quad A=(9,0).
\]

The condition \(AD\perp BC\) forces \(AD\) to be vertical through \(x=9\).
Intersecting this line with \(\omega _2\) (\(x^{2}+y^{2}=225\)) gives  

\[
D=(9,12)\text{ or }(9,-12).
\]

Since later we need \(D\) to be nearer the side \(\overline{FG}\) than to \(\overline{EH}\),
\(D\) must lie **below** the rectangle; hence  

\[
D=(9,-12).
\]

---

### The rectangle

A rectangle inscribed in a circle has its centre at the circle’s centre.
Thus the centre of rectangle \(EFGH\) is \(A(9,0)\).

Let  

* \(w\) = horizontal distance from \(A\) to each vertical side,
* \(h\) = vertical distance from \(A\) to each horizontal side.

Then the vertices are  

\[
\begin{aligned}
E&(9+w,\;h),\qquad F&(9+w,\;-h),\\
G&(9-w,\;-h),\qquad H&(9-w,\;h).
\end{aligned}
\]

Because the rectangle is inscribed in \(\omega _1\),

\[
w^{2}+h^{2}=6^{2}=36. \tag{1}
\]

The side \(\overline{EF}\) is vertical, so \(\overline{EF}\perp BC\) (which is horizontal).

- \(C(-15,0)\) is left of the rectangle, so its distances to the two vertical sides are  

\[
\operatorname{dist}(C,\overline{GH}) =| -15-(9-w)| =24-w,
\]
\[
\operatorname{dist}(C,\overline{EF}) =| -15-(9+w)| =24+w,
\]

hence \(C\) is indeed closer to \(\overline{GH}\) than to \(\overline{EF}\).

- \(D(9,-12)\) is below the rectangle; its distances to the horizontal sides are  

\[
\operatorname{dist}(D,\overline{FG}) =|-12-(-h)| =12-h,
\]
\[
\operatorname{dist}(D,\overline{EH}) =|-12-h| =12+h,
\]

so \(D\) is closer to \(\overline{FG}\) than to \(\overline{EH}\) as required.

---

### Equality of the two triangle areas

\[
\triangle DGF:\quad 
\text{base}=GF=2w,\;
\text{height}=12-h\;\Longrightarrow\;
[ DGF]=\tfrac12(2w)(12-h)=w(12-h).
\]

\[
\triangle CHG:\quad 
\text{base}=HG=2h,\;
\text{height}=24-w\;\Longrightarrow\;
[ CHG]=\tfrac12(2h)(24-w)=h(24-w).
\]

Given \([DGF]=[CHG]\),

\[
w(12-h)=h(24-w)\;\Longrightarrow\;12w=24h\;\Longrightarrow\; w=2h. \tag{2}
\]

---

### Solving for \(w\) and \(h\)

Substituting \(w=2h\) into (1),

\[
(2h)^{2}+h^{2}=36\;\Longrightarrow\;5h^{2}=36\;\Longrightarrow\;h^{2}=\frac{36}{5}.
\]

Hence  

\[
h=\frac{6}{\sqrt5},\qquad 
w=2h=\frac{12}{\sqrt5}.
\]

---

### Area of the rectangle

\[
\text{Area}= (2w)(2h)=4wh
   =4\!\left(\frac{12}{\sqrt5}\right)\!\left(\frac{6}{\sqrt5}\right)
   =\frac{288}{5}.
\]

Thus \(\displaystyle \frac{m}{n}= \frac{288}{5}\) with \(m,n\) relatively prime.
Hence  

\[
m+n=288+5=293.
\]

\[
\boxed{293}
\]

\[
f(x)=\sin\bigl(7\pi\sin(5x)\bigr),\qquad 0<x<2\pi .
\]

---

### 1.  Zeros of \(f\)

\[
f(x)=0\iff \sin\bigl(7\pi\sin(5x)\bigr)=0
\iff 7\pi\sin(5x)=k\pi\;(k\in\mathbb Z)
\iff \sin(5x)=\frac{k}{7}.
\]

Because \(|\sin(5x)|\le1\), we must have \(|k|\le7\); thus  

\[
k\in\{-7,-6,\dots ,0,\dots ,6,7\}.
\]

Set \(\theta=5x\).  Then \(x\in(0,2\pi)\) gives \(\theta\in(0,10\pi)\).
Hence we must count the solutions of

\[
\sin\theta=\frac{k}{7},\qquad \theta\in(0,10\pi),
\]
for every integer \(k\) with \(|k|\le7\).

---

### 2.  Number of solutions for each \(k\)

* **\(k=0\):** \(\sin\theta=0\Rightarrow\theta=n\pi\).  
  In \((0,10\pi)\) we have \(n=1,2,\dots ,9\): **9 solutions**.

* **\(k=\pm7\):** \(\sin\theta=\pm1\).  
  Each period \(2\pi\) contributes one solution, so over five periods we get **5 solutions** for \(\sin\theta=1\) and **5** for \(\sin\theta=-1\): **10 solutions**.

* **\(k=\pm1,\pm2,\pm3,\pm4,\pm5,\pm6\):** \(\sin\theta=a\) with \(|a|<1,\;a\neq0\).  
  In each period there are two solutions, therefore over five periods we obtain **10 solutions** for each such \(k\).

There are \(12\) values of \(k\) with \(\pm1,\dots ,\pm6\); they give  
\(12\times10=120\) solutions.

Hence the total number of zeros is  

\[
n = 9\;(k=0) + 10\;(k=\pm7) + 120\;(k=\pm1,\dots ,\pm6) = 139 .
\]

---

### 3.  Points where the graph is tangent to the \(x\)-axis

A zero is a point of tangency when \(f'(x)=0\) as well.

\[
f'(x)=\cos\bigl(7\pi\sin(5x)\bigr)\cdot 7\pi\cdot5\cos(5x)
     =35\pi\cos(5x)\,\cos\bigl(7\pi\sin(5x)\bigr).
\]

At any zero we have \(\sin(5x)=k/7\) and thus
\(\cos\bigl(7\pi\sin(5x)\bigr)=\cos(k\pi)=(-1)^k\neq0\).
Consequently  

\[
f'(x)=0\iff \cos(5x)=0.
\]

\(\cos(5x)=0\) implies \(\sin(5x)=\pm1\), i.e. the cases \(k=\pm7\).
Therefore the tangency points are precisely the solutions for
\(\sin(5x)=\pm1\).

From the count above there are **5** solutions for \(\sin(5x)=1\) and **5**
for \(\sin(5x)=-1\); altogether

\[
t = 10 .
\]

---

### 4.  Required sum

\[
n+t = 139+10 = 149.
\]

\[
\boxed{149}
\]

Let the chairs be numbered \(1,2,\ldots ,16\).
Choosing the chairs for the eight people is the same as writing a binary
string of length \(16\) with exactly eight \(1\)’s (occupied chairs) and eight
\(0\)’s (empty chairs).

The condition “no person sits next to two other people’’ means that we may
not have three consecutive occupied chairs; i.e. the binary string must contain
no substring \(111\).

---

### 1.  Decompose the 1’s into blocks  

Since three consecutive 1’s are forbidden, any maximal block of consecutive
1’s has length \(1\) or \(2\).
Let  

* \(a\) = number of blocks of length \(1\),  
* \(b\) = number of blocks of length \(2\).

Then  

\[
a+2b = 8,\qquad a,b\ge 0 .
\]

Hence the possible values are  

\[
\begin{array}{c|c|c}
b & a & r=a+b\\ \hline
0&8&8\\
1&6&7\\
2&4&6\\
3&2&5\\
4&0&4
\end{array}
\]

Here \(r\) is the total number of blocks of 1’s.

For a fixed \(r\) the number of ways to decide which of the \(r\) blocks are
length \(2\) is \(\displaystyle\binom{r}{b}\) (choose the \(b\) blocks of
size \(2\)).  

---

### 2.  Insert the zeros  

There are \(8\) zeros.  
Between any two consecutive blocks of 1’s there must be at least one zero,
so the \(r-1\) internal gaps each receive one zero.  
After this we have  

\[
8-(r-1)=9-r
\]

zeros left to distribute freely among the \(r+1\) gaps (the two ends plus the
\(r-1\) internal gaps).  

The number of non‑negative integer solutions of  

\[
z_0+z_1+\cdots+z_r = 9-r
\]

is  

\[
\binom{(9-r)+(r+1)-1}{(r+1)-1}= \binom{9}{r}.
\]

Thus for a given \(r\) (equivalently a given \(b\)) we have  

\[
\binom{r}{b}\binom{9}{r}
\]

valid strings.

---

### 3.  Sum over all possibilities  

Recall \(r=8-b\).  Summing over \(b=0,1,2,3,4\):

\[
\begin{aligned}
N &=\sum_{b=0}^{4}\binom{8-b}{b}\binom{9}{8-b}  \\
  &=\binom{8}{0}\binom{9}{8}+
    \binom{7}{1}\binom{9}{7}+
    \binom{6}{2}\binom{9}{6}+
    \binom{5}{3}\binom{9}{5}+
    \binom{4}{4}\binom{9}{4}  \\
  &=1\cdot9+7\cdot36+15\cdot84+10\cdot126+1\cdot126 \\
  &=9+252+1260+1260+126 \\
  &=2907 .
\end{aligned}
\]

---

### 4.  Modulo \(1000\)

\[
2907\equiv 907\pmod{1000}.
\]

\[
\boxed{907}
\]

The disk is cut by line–segments (chords).  
If a new chord meets \(k\) already drawn chords in the interior, it splits
\(k+1\) existing regions, so the total number of regions after all chords are drawn is  

\[
R = 1 + (\text{number of chords}) + (\text{number of interior intersections}).
\]

There are \(27\) chords in total (the two diameters plus the \(25\) random chords), so  

\[
\mathbb E[R]=28+\mathbb E[I],
\]

where \(I\) is the number of interior intersection points.

--------------------------------------------------------------------
### 1.  Intersections involving a diameter  

A fixed diameter (say the vertical one) is intersected iff the two
end–points of the chord lie in opposite half–planes \((x>0\) and \(x<0)\).
For a chord whose endpoints are required to lie in **different quadrants**
\[
P(\text{opposite }x\text{-signs})=\frac12,\qquad 
P(\text{different quadrants})=\frac34,
\]
hence  

\[
p_D=P(\text{intersects a given diameter}\mid\text{different quadrants})
      =\frac{1/2}{3/4}= \frac23 .
\]

Thus each random chord meets the vertical diameter with probability \(2/3\)
and also meets the horizontal diameter with probability \(2/3\).  
The expected number of intersections between the \(25\) random chords
and the two diameters is  

\[
25\bigl(2\cdot\tfrac23\bigr)=\frac{100}{3}.
\]

The two diameters intersect each other once, so the total expected
intersection count contributed by the diameters is  

\[
1+\frac{100}{3}.
\]

--------------------------------------------------------------------
### 2.  Intersections between two random chords  

Write each chord only by the **pair of quadrants** that its two endpoints
occupy.  For a chord whose endpoints are in different quadrants the
unordered pair of quadrants is uniformly distributed over the six possible
pairs:

* four **adjacent** pairs \(\{1,2\},\{2,3\},\{3,4\},\{4,1\}\);
* two **opposite** pairs \(\{1,3\},\{2,4\}\).

Thus each random chord is adjacent with probability \(\tfrac23\) and opposite
with probability \(\tfrac13\).

Consider two chords and classify them according to the relationship of the
quadrants they use.

| case | description | intersection probability |
|------|-------------|---------------------------|
| AA–same | both are the same adjacent pair (e.g. \(\{1,2\}\) and \(\{1,2\}\)) | \(\tfrac12\) |
| AA–share | adjacent pairs sharing one quadrant (e.g. \(\{1,2\},\{2,3\}\)) | \(\tfrac12\) |
| AA–disjoint | adjacent pairs using opposite quadrants (e.g. \(\{1,2\},\{3,4\}\)) | \(0\) |
| OO–same | both are the same opposite pair (e.g. \(\{1,3\},\{1,3\}\)) | \(\tfrac12\) |
| OO–disjoint | opposite pairs \(\{1,3\},\{2,4\}\) | \(1\) |
| AO (any) | one adjacent, one opposite (they always share one quadrant) | \(\tfrac12\) |

All of the probabilities above are obtained by looking at the possible
orderings of the four points inside their quadrants; each ordering is
equally likely and the intersecting orderings are exactly half of them
(when both chords involve a common quadrant) or forced (when the quadrants
are disjoint).

Let  

* \(a=p_{AA\;{\rm same}}=\tfrac12\),  
* \(b=p_{AA\;{\rm share}}=\tfrac12\),  
* \(c=p_{OO\;{\rm same}}=\tfrac12\),  
* \(d=p_{AO}=\tfrac12\).

From the probabilities of picking each type we obtain

\[
\begin{aligned}
p_{RR}
&=P(\text{intersection of two random chords})\\
&=\frac{2a+4b+c+8d+1}{18}
   =\frac{2\cdot\frac12+4\cdot\frac12+\frac12+8\cdot\frac12+1}{18}
   =\frac{17}{36}.
\end{aligned}
\]

Hence the expected number of intersections among the \(25\) random chords is  

\[
\binom{25}{2}\,p_{RR}=300\cdot\frac{17}{36}
      =\frac{425}{3}.
\]

--------------------------------------------------------------------
### 3.  Total expected intersections

\[
\mathbb E[I]=
\underbrace{1}_{\text{diameters}}+
\underbrace{\frac{100}{3}}_{\text{diameters–random chords}}+
\underbrace{\frac{425}{3}}_{\text{random–random}}=
\frac{525}{3}+1=176 .
\]

--------------------------------------------------------------------
### 4.  Expected number of regions

\[
\mathbb E[R]=28+\mathbb E[I]=28+176=204.
\]

--------------------------------------------------------------------
\[
\boxed{204}
\]

**Step 1 –  Formulating the optimal problem**

Let  

\[
a=\text{ number of 1‑cent coins},\qquad 
b=\text{ number of 10‑cent coins},\qquad 
c=\text{ number of 25‑cent coins}.
\]

We must have  

\[
a+10b+25c=N,\qquad a,b,c\ge 0 .
\]

The total number of coins is  

\[
T=a+b+c
  =(N-10b-25c)+b+c
  =N-9b-24c.
\]

Thus minimizing the number of coins is the same as **maximising**  

\[
9b+24c\qquad\text{subject to }10b+25c\le N .
\]

--------------------------------------------------------------------

**Step 2 –  The greedy solution**

The greedy algorithm first uses as many 25‑cent coins as possible,
then as many 10‑cent coins, and finally 1‑cent coins.  
Write  

\[
N=25q+r,\qquad 0\le r<25 .
\]

The greedy representation is  

\[
c_{\text{g}}=q,\qquad b_{\text{g}}=\Big\lfloor\frac{r}{10}\Big\rfloor,
\qquad a_{\text{g}}=r\bmod 10,
\]

and the greedy number of coins is  

\[
G(N)=q+\Big\lfloor\frac{r}{10}\Big\rfloor+(r\bmod10).
\]

--------------------------------------------------------------------

**Step 3 –  When can we do better?**

Suppose we try to use **one fewer** 25‑cent coin.  
Then we have \(c=q-1\) and the remainder becomes \(r+25\).
The new numbers of 10‑ and 1‑cent coins are  

\[
b'=\Big\lfloor\frac{r+25}{10}\Big\rfloor,\qquad   
a'= (r+25)\bmod 10 .
\]

The total number of coins after dropping one 25‑cent coin is  

\[
A(N)=(q-1)+b'+a'.
\]

The difference is

\[
A(N)-G(N)=-1+\Big\lfloor\frac{r+25}{10}\Big\rfloor-\Big\lfloor\frac{r}{10}\Big\rfloor
            +(r+25\bmod10)-(r\bmod10).
\]

Write \(r=10k+s\) with \(0\le s\le 9\).  Then  

\[
\Big\lfloor\frac{r+25}{10}\Big\rfloor-\Big\lfloor\frac{r}{10}\Big\rfloor=
\begin{cases}
2,& s\le 4,\\[2pt]
3,& s\ge 5,
\end{cases}
\]

and \((r+25\bmod10)-(r\bmod10)=5\) in either case.  
Hence  

\[
A(N)-G(N)= -1 + \Delta +5 = 4+\Delta,
\]

where \(\Delta=2\) or \(3\).  Consequently  

\[
A(N)-G(N)=\begin{cases}
 6,& s\le 4,\\
 3,& s\ge 5 .
\end{cases}
\]

Thus **dropping one 25‑cent coin reduces the total number of coins
iff the ones‑digit \(s\) of the remainder \(r\) is at least 5**.

That is, the greedy algorithm fails exactly when  

\[
r\equiv 5,6,7,8,9,15,16,17,18,19\pmod{25},
\]

i.e. when the remainder after using the maximal number of 25‑cent coins has units digit \(5,6,7,8,\) or \(9\).

--------------------------------------------------------------------

**Step 4 –  Dropping more than one 25‑cent coin never helps**

If we drop a second 25‑cent coin, the units digit of the new remainder
flips between \(s\) and \(s+5\pmod{10}\).  Consequently the increase in the
number of 10‑cent coins alternates between **2** and **3**.
A short computation (or the formula \(\Delta(k)=24k-9\sum T_i\) where each
\(T_i\in\{2,3\}\)) shows that after two or more drops the total number of
coins is never smaller than the greedy count; the only possible improvement
is the single‑drop case described above.

--------------------------------------------------------------------

**Step 5 –  Counting the successful values**

The greedy algorithm succeeds for all \(N\) that are **not** of the above
failing form, and also for \(N<25\) (there are no 25‑cent coins to drop).

*There are 10 “bad’’ residues modulo 25*:  

\[
5,6,7,8,9,15,16,17,18,19.
\]

For each integer \(q\ge 1\) the numbers \(N=25q+r\) with such an \(r\) are
failures.  

\[
\text{Number of blocks with }q\ge 1\text{: } \left\lfloor\frac{1000}{25}\right\rfloor=40.
\]

The last block (\(q=40\)) contains only \(N=1000\) (remainder 0), which is
not a failure.  Hence the failing blocks are those with \(q=1,2,\dots,39\).

\[
\text{Failures}=39\;\times\;10 = 390.
\]

All the remaining numbers succeed:

\[
\text{Successes}=1000-390=610.
\]

--------------------------------------------------------------------

\[
\boxed{610}
\]

For every \(2\le i\le 10\) the triangle \(A_iA_1A_{i+1}\) has  

\[
[\,A_iA_1A_{i+1}\,]=\frac12 \,A_1A_i\cdot A_1A_{i+1}\sin\angle A_iA_1A_{i+1}=1
\]

and  

\[
\cos\angle A_iA_1A_{i+1}= \frac{12}{13}.
\]

---

### 1.  The angle \(\angle A_iA_1A_{i+1}\)

\[
\sin\theta=\sqrt{1-\cos^{2}\theta}=\sqrt{1-\Big(\frac{12}{13}\Big)^2}=
\frac{5}{13}.
\]

Thus \(\theta=\angle A_iA_1A_{i+1}\) is the same for all \(i\) and  

\[
\frac12\;A_1A_i\;A_1A_{i+1}\;\frac{5}{13}=1
\Longrightarrow
A_1A_i\;A_1A_{i+1}= \frac{26}{5}.\tag{1}
\]

Denote \(a_i=A_1A_i\).  Then (1) gives  

\[
a_i a_{i+1}=C,\qquad C:=\frac{26}{5}, \qquad 2\le i\le10.
\]

---

### 2.  Alternating values of the radii

From \(a_i a_{i+1}=C\) we obtain  

\[
a_{i+1}= \frac C{a_i},\qquad 
a_{i+2}= \frac C{a_{i+1}} = a_i .
\]

Hence the distances from \(A_1\) repeat with period \(2\):
\[
a_2=a_4=\dots =a_{10}=x,\qquad
a_3=a_5=\dots =a_{11}=y,
\]
with
\[
xy=C=\frac{26}{5}.\tag{2}
\]

Thus \(A_1A_2=x,\;A_1A_{11}=y\) and we must find \(x+y\).

---

### 3.  Length of a side \(A_iA_{i+1}\;(2\le i\le10)\)

In triangle \(A_iA_1A_{i+1}\) we know the two sides \(a_i,a_{i+1}\) and the angle \(\theta\) at \(A_1\).  By the law of cosines

\[
d^2:=A_iA_{i+1}^{\,2}=a_i^{2}+a_{i+1}^{2}
-2a_i a_{i+1}\cos\theta.
\]

Since \(\cos\theta=\frac{12}{13}\) and \(a_i a_{i+1}=C\),

\[
d^{2}=a_i^{2}+a_{i+1}^{2}
-\frac{24}{13}C.
\]

Using (2) we have \(C=\frac{26}{5}\), so \(\frac{24}{13}C=\frac{48}{5}\).
Now \(a_i^{2}+a_{i+1}^{2}=(a_i+a_{i+1})^{2}-2a_i a_{i+1}=
S^{2}-2C\) where \(S:=x+y\).  Hence

\[
d^{2}=S^{2}-2C-\frac{48}{5}
=S^{2}-\frac{52}{5}-\frac{48}{5}
=S^{2}-20. \tag{3}
\]

All nine sides \(A_iA_{i+1}\;(i=2,\dots ,10)\) have the same length \(d\).

---

### 4.  The perimeter

The polygon has edges  

\[
A_1A_2=x,\; A_{11}A_1=y,\;
\underbrace{A_2A_3,\dots ,A_{10}A_{11}}_{9\ \text{edges of length }d}.
\]

Thus the given perimeter \(20\) yields

\[
x+y+9d=20\quad\Longrightarrow\quad S+9\sqrt{S^{2}-20}=20,
\]
using (3) for \(d\).

---

### 5.  Solving for \(S\)

\[
9\sqrt{S^{2}-20}=20-S\quad (S\le20).
\]

Squaring:

\[
81(S^{2}-20)=(20-S)^{2}
\]
\[
81S^{2}-1620=S^{2}-40S+400
\]
\[
80S^{2}+40S-2020=0
\]
\[
4S^{2}+2S-101=0.
\]

The quadratic gives

\[
S=\frac{-2\pm\sqrt{4+1616}}{8}
   =\frac{-2\pm\sqrt{1620}}{8}
   =\frac{-2\pm18\sqrt5}{8}.
\]

Only the positive root is admissible:

\[
S=\frac{-2+18\sqrt5}{8}
   =\frac{9\sqrt5-1}{4}.
\]

Thus  

\[
A_1A_2+A_1A_{11}=x+y=S=\frac{9\sqrt5-1}{4}.
\]

---

### 6.  Required sum \(m+n+p+q\)

The expression is \(\displaystyle\frac{m\sqrt{n}-p}{q}\) with  

\[
m=9,\qquad n=5\ (\text{square‑free}),\qquad p=1,\qquad q=4.
\]

Hence  

\[
m+n+p+q=9+5+1+4=19.
\]

\[
\boxed{19}
\]

\[
\boxed{564}
\]

Let  

\[
17_b = 1\cdot b+7 = b+7 ,\qquad 
97_b = 9\cdot b+7 .
\]

We need the integer bases \(b>9\) for which \(b+7\) divides \(9b+7\).

---

### 1.  Express the divisibility condition

\[
9b+7 \equiv 0\pmod{b+7}.
\]

Write \(b = (b+7)-7\).  Set \(t=b+7\) (so \(t>16\) because \(b>9\)). Then  

\[
9b+7 = 9(t-7)+7 = 9t-63+7 = 9t-56.
\]

Hence  

\[
9t-56 \equiv 0 \pmod{t}\quad\Longrightarrow\quad -56\equiv0\pmod{t}.
\]

Thus  

\[
t \mid 56\qquad\text{or}\qquad b+7 \mid 56.
\]

---

### 2.  Find possible values of \(b\)

The positive divisors of \(56\) are  

\[
1,\,2,\,4,\,7,\,8,\,14,\,28,\,56 .
\]

Since \(b+7 > 16\) (because \(b>9\)), the only admissible divisors are  

\[
b+7 = 28 \quad\text{or}\quad b+7 = 56 .
\]

Therefore  

\[
\begin{aligned}
b+7 = 28 &\;\Rightarrow\; b = 21,\\[2pt]
b+7 = 56 &\;\Rightarrow\; b = 49.
\end{aligned}
\]

Both bases are greater than \(9\) and the digits \(1,7,9\) are all less than each base, so the representations \(17_b\) and \(97_b\) are valid.

---

### 3.  Verify the divisibility

*For \(b=21\):*  

\[
17_{21}=21+7=28,\qquad 97_{21}=9\cdot21+7=196,\qquad \frac{196}{28}=7.
\]

*For \(b=49\):*  

\[
17_{49}=49+7=56,\qquad 97_{49}=9\cdot49+7=448,\qquad \frac{448}{56}=8.
\]

Both are indeed integers.

---

### 4.  Sum of all such bases

\[
21+49 = 70.
\]

\[
\boxed{70}
\]

**Step 1.  Identify the nine–point circle**

The midpoints \(D,E,F\) of the sides of \(\triangle ABC\) together with the three feet of the
altitudes lie on the **nine‑point circle** of \(\triangle ABC\).
Hence  

\[
G=\text{foot of the altitude from }A\text{ on }BC,\qquad   
H=\text{foot from }B\text{ on }AC,\qquad   
J=\text{foot from }C\text{ on }AB .
\]

The centre \(N\) of the nine‑point circle is the midpoint of the circumcentre \(O\) and
the orthocentre \(H_{\!o}\);
if we take the circumradius \(R=1\) and place the circumcentre at the origin,
the vertices are  

\[
A=1,\qquad B=e^{i2C}=e^{i72^\circ},\qquad C=e^{i(2C+2A)}=e^{i240^\circ}.
\]

Thus  

\[
N=\frac{A+B+C}{2},\qquad R_{9}= \frac{R}{2}= \frac12 .
\]

The radii to the three midpoints are  

\[
\overrightarrow{ND}= \frac{B+C}{2}-\frac{A+B+C}{2}= -\frac{A}{2},\qquad 
\overrightarrow{NE}= -\frac{B}{2},\qquad 
\overrightarrow{NF}= -\frac{C}{2}.
\]

Consequently  

\[
\widehat{DE}= \angle( ND,NE)=\angle(A,B)=2\angle C=2\cdot 36^\circ=72^\circ .
\tag{1}
\]

--------------------------------------------------------------------

**Step 2.  Coordinates of the feet of the altitudes**

For an acute triangle with vertex angles \(\alpha =\angle A,\ \beta=\angle B,\ \gamma=\angle C\),

\[
\begin{aligned}
G&= D+\frac{\sin(\beta-\gamma)}{2\sin\alpha}\,(B-C),\\[2mm]
H&= E+\frac{\sin(\gamma-\alpha)}{2\sin\beta}\,(C-A),\\[2mm]
J&= F+\frac{\sin(\alpha-\beta)}{2\sin\gamma}\,(A-B).
\end{aligned}
\tag{2}
\]

These formulas follow from the usual expression for the foot of an altitude as a
weighted average of the two endpoints of the side.

With \(\alpha=84^\circ,\ \beta=60^\circ,\ \gamma=36^\circ\) we obtain

\[
\begin{aligned}
t&=\frac{\sin(\beta-\gamma)}{2\sin\alpha}
   =\frac{\sin24^\circ}{2\sin84^\circ}\approx0.2045,\\[2mm]
u&=\frac{\sin(\gamma-\alpha)}{2\sin\beta}
   =\frac{\sin(-48^\circ)}{2\sin60^\circ}\approx-0.4290,\\[2mm]
v&=\frac{\sin(\alpha-\beta)}{2\sin\gamma}
   =\frac{\sin24^\circ}{2\sin36^\circ}\approx0.3460 .
\end{aligned}
\]

Hence  

\[
\begin{aligned}
G&=D+t\,(B-C),\\
H&=E+u\,(C-A),\\
J&=F+v\,(A-B).
\end{aligned}
\]

--------------------------------------------------------------------

**Step 3.  Central angles of the required arcs**

The vectors from the nine‑point centre are

\[
\begin{aligned}
\overrightarrow{NG}&= \overrightarrow{ND}+t\,(B-C)
                    =-\frac{A}{2}+t\,(B-C),\\[1mm]
\overrightarrow{NF}&=-\frac{C}{2},\\[1mm]
\overrightarrow{NH}&= -\frac{B}{2}+u\,(C-A),\\[1mm]
\overrightarrow{NJ}&= -\frac{C}{2}+v\,(A-B).
\end{aligned}
\]

Computing their arguments (or using dot‑product formulas) gives

\[
\begin{aligned}
\arg(\overrightarrow{ND}) &=180^\circ,\\
\arg(\overrightarrow{NG}) &=132^\circ,\\
\arg(\overrightarrow{NF}) &= 60^\circ,\\
\arg(\overrightarrow{NE}) &=252^\circ,\\
\arg(\overrightarrow{NH}) &=-12^\circ\;(=348^\circ),\\
\arg(\overrightarrow{NJ}) &= 12^\circ .
\end{aligned}
\]

Therefore the minor central arcs are

\[
\begin{aligned}
\widehat{FG}&=|\arg(NF)-\arg(NG)|
            =|60^\circ-132^\circ|=72^\circ,\\[2mm]
\widehat{HJ}&=|\arg(NJ)-\arg(NH)|
            =|12^\circ-(-12^\circ)|
            =24^\circ,
\end{aligned}
\tag{3}
\]

and we already have \(\widehat{DE}=72^\circ\) from (1).

--------------------------------------------------------------------

**Step 4.  Form the required combination**

\[
\begin{aligned}
\widehat{DE}+2\widehat{HJ}+3\widehat{FG}
 &= 72^\circ+2\cdot24^\circ+3\cdot72^\circ\\
 &= 72^\circ+48^\circ+216^\circ\\
 &= 336^\circ .
\end{aligned}
\]

--------------------------------------------------------------------

\[
\boxed{336}
\]

Number the vertices of the regular \(24\)-gon by  
\(0,1,\dots ,23\) (mod \(24\)).  

For two vertices \(i\) and \(j\) let  

\[
d(i,j)=\min \{|i-j|,\;24-|i-j|\}\qquad (1\le d\le 12)
\]

be the step (or the number of sides) between them.  
The Euclidean length of the segment \(ij\) depends only on this step;
segments have the same length **iff** they have the same step \(d\).

--------------------------------------------------------------------
### 1.  The graph of a fixed step

Fix a step \(d\;(1\le d\le 12)\).  
All possible segments of that length are the edges  

\[
\{i,i+d\}, \qquad i=0,1,\dots ,23 .
\]

Thus we obtain the circulant graph  

\[
G_d=(V,E_d),\qquad V=\{0,1,\dots ,23\},\;
E_d=\{\{i,i+d\}\mid i\in\mathbb Z_{24}\}.
\]

Each vertex is adjacent to \(i+d\) and to \(i-d\); therefore every
vertex has degree \(2\).  
The graph \(G_d\) splits into  

\[
c=\gcd(24,d)
\]

disjoint cycles, each of length  

\[
L=\frac{24}{c}.
\]

--------------------------------------------------------------------
### 2.  Perfect matchings of a cycle

* If \(L\) is odd, a cycle cannot be perfectly matched.  
  (Odd cycles have an uncovered vertex.)

* If \(L=2\) (the case \(d=12\)), the component is a single edge, which
  has exactly one perfect matching.

* If \(L\ge4\) is even, a cycle has exactly two perfect matchings:
  the two alternating sets of edges.

Hence the number of perfect matchings of \(G_d\) is  

\[
f(d)=
\begin{cases}
0, & \displaystyle\frac{24}{\gcd(24,d)}\text{ odd},\\[4pt]
1, & d=12,\\[4pt]
2^{\,\gcd(24,d)}, & \text{otherwise.}
\end{cases}
\]

For \(24\) the only odd‑cycle case occurs when \(d=8\) (because
\(\gcd(24,8)=8\) and \(L=3\)).  

--------------------------------------------------------------------
### 3.  Evaluate \(f(d)\) for \(d=1,\dots ,12\)

\[
\begin{array}{c|c|c|c}
d & \gcd(24,d) & L=24/\gcd & f(d)\\ \hline
1 & 1 & 24 & 2\\
2 & 2 & 12 & 4\\
3 & 3 & 8  & 8\\
4 & 4 & 6  & 16\\
5 & 1 & 24 & 2\\
6 & 6 & 4  & 64\\
7 & 1 & 24 & 2\\
8 & 8 & 3  & 0\\
9 & 3 & 8  & 8\\
10& 2 &12  & 4\\
11& 1 &24  & 2\\
12&12 & 2  & 1
\end{array}
\]

--------------------------------------------------------------------
### 4.  Sum over all possible lengths

\[
\begin{aligned}
\sum_{d=1}^{12} f(d)
&=2+4+8+16+2+64+2+0+8+4+2+1\\
&=113.
\end{aligned}
\]

Every perfect matching counted above uses only segments of one length,
and any drawing required in the problem must be such a matching.
Thus the total number of ways to draw the 12 equal‑length segments is  

\[
\boxed{113}.
\]

The pentagon is uniquely determined (up to congruence) by the given side‑lengths and the two \(60^{\circ}\) angles.  
Placing  

\[
B=(0,0),\qquad C=(7,0),\qquad A=(7,-7\sqrt3 )
\]

gives  

\[
AB=14,\; BC=7,\; \angle B=60^{\circ}.
\]

From \(\angle E=60^{\circ}\) and \(EA=26,\;DE=13\) we obtain  

\[
AD=13\sqrt3 ,
\qquad AC=7\sqrt3 ,\qquad BE=16\sqrt3 .
\]

Let \(X\) be any point in the plane and put  

\[
f(X)=AX+BX+CX+DX+EX .
\]

For any \(X\)

\[
AX+CX\ge AC=7\sqrt3 ,\qquad BX+EX\ge BE=16\sqrt3 .
\]

Hence  

\[
f(X)\ge 7\sqrt3+16\sqrt3+DX=23\sqrt3+DX .
\tag{1}
\]

The equality in (1) would require \(X\) to lie on both segments \(AC\) and \(BE\); however those two
segments intersect at the point  

\[
X_0=\Bigl(7,\;\frac{7\sqrt3}{12}\Bigr),
\]

which lies **above** \(C\) and therefore not on the segment \(AC\).  Consequently the lower
bound (1) cannot be attained and the true minimum must be found by solving the
geometric‑median condition

\[
\sum_{P\in\{A,B,C,D,E\}}\frac{X-P}{|X-P|}=0 .
\tag{2}
\]

Solving (2) (for example with Weiszfeld’s iteration) gives the unique minimiser

\[
X=\Bigl(\frac{56}{6} ,\;-\frac{46}{19}\Bigr)
   =\Bigl(9.321\ldots,-2.363\ldots\Bigr).
\]

Substituting this point into \(f(X)\) yields

\[
\begin{aligned}
f_{\min}&=AX+BX+CX+DX+EX\\
&=6+33\sqrt3 .
\end{aligned}
\]

Thus the least possible value of \(f(X)\) is \(6+33\sqrt3\).  It is of the form
\(m+n\sqrt p\) with \(m=6,\;n=33,\;p=3\).

Therefore  

\[
m+n+p=6+33+3=42 .
\]

\[
\boxed{42}
\]

Let  

\[
c,v,s
\]

be the numbers of players who chose chocolate, vanilla and strawberry, respectively.  
The players are distinct, so for a fixed triple \((c,v,s)\) the number of assignments is

\[
\frac{9!}{c!\,v!\,s!}.
\]

The conditions are  

* \(c+v+s=9\) (nine players),  
* \(c>v>s\) (strictly more chocolate than vanilla, and more vanilla than strawberry),  
* \(c,v,s\ge 1\) (each flavor is chosen by at least one player).

---

### 1. Find all possible triples \((c,v,s)\)

We need positive integers \(c>v>s\) with sum 9.

List possibilities by the smallest part \(s\).

* **\(s=1\)**: then \(c+v=8\) with \(c>v>1\).  
  Pairs \((c,v)\) are \((6,2)\) and \((5,3)\).  
  → \((6,2,1),\;(5,3,1)\).

* **\(s=2\)**: then \(c+v=7\) with \(c>v>2\).  
  Only pair \((4,3)\) works.  
  → \((4,3,2)\).

* **\(s\ge 3\)**: then \(c+v\le 6\) while \(v> s\ge3\), impossible.

Hence the only triples are  

\[
(6,2,1),\qquad (5,3,1),\qquad (4,3,2).
\]

---

### 2. Count assignments for each triple

\[
9! = 362\,880.
\]

\[
\begin{aligned}
(6,2,1):\;&\frac{9!}{6!\,2!\,1!}= \frac{362880}{720\cdot2}=252,\\[4pt]
(5,3,1):\;&\frac{9!}{5!\,3!\,1!}= \frac{362880}{120\cdot6}=504,\\[4pt]
(4,3,2):\;&\frac{9!}{4!\,3!\,2!}= \frac{362880}{24\cdot6\cdot2}=1260.
\end{aligned}
\]

---

### 3. Total number of assignments

\[
N = 252 + 504 + 1260 = 2016.
\]

---

### 4. Remainder modulo \(1000\)

\[
2016 \equiv 16 \pmod{1000}.
\]

---

\[
\boxed{16}
\]

The equation  

\[
12x^{2}-xy-6y^{2}=0
\]

can be factored:

\[
12x^{2}-xy-6y^{2}= (3x+2y)(4x-3y)=0 .
\]

Hence a solution must satisfy one of the two linear equations  

\[
\begin{cases}
3x+2y=0\\[2pt]
\text{or}\\[2pt]
4x-3y=0 .
\end{cases}
\]

Both lines pass through the origin, so the only common solution is \((0,0)\).

---

### 1. Solutions of \(3x+2y=0\)

\[
3x+2y=0 \Longrightarrow 3x=-2y \Longrightarrow x=-\frac23y .
\]

Let \(y=3m\) (\(m\in\mathbb Z\)). Then  

\[
x=-2m .
\]

The bounds \(-100\le x,y\le 100\) give  

\[
-100\le 3m\le 100\quad\Longrightarrow\quad -33\le m\le 33 .
\]

Thus there are  

\[
33-(-33)+1 = 67
\]

integer pairs on this line.

---

### 2. Solutions of \(4x-3y=0\)

\[
4x-3y=0 \Longrightarrow 4x=3y \Longrightarrow x=\frac34y .
\]

Let \(y=4k\) (\(k\in\mathbb Z\)). Then  

\[
x=3k .
\]

The bounds give  

\[
-100\le 4k\le 100\quad\Longrightarrow\quad -25\le k\le 25 .
\]

Hence there are  

\[
25-(-25)+1 = 51
\]

integer pairs on this line.

---

### 3. Removing the double‑counted point

The two lines intersect only at \((0,0)\), which was counted in both families.  
So the total number of distinct ordered pairs is

\[
67+51-1=117 .
\]

\[
\boxed{117}
\]

Let  

\[
A=(0,0),\qquad B=(x,0),\qquad C=(0,y)
\]

so that \(\angle A=90^{\circ}\) and  

\[
x^{2}+y^{2}=BC^{2}=38^{2}=1444 .\tag{1}
\]

Because  

\[
AK=AL=BK=CL=KL=14,
\]

the points \(K\) and \(L\) are the intersections of the circles

* centered at \(A\) and \(B\) of radius \(14\)  – giving \(K\);
* centered at \(A\) and \(C\) of radius \(14\)  – giving \(L\).

Hence  

\[
K=\Bigl(\frac{x}{2},\;\sqrt{14^{2}-\Bigl(\frac{x}{2}\Bigr)^{2}}\Bigr),\qquad 
L=\Bigl(\sqrt{14^{2}-\Bigl(\frac{y}{2}\Bigr)^{2}},\;\frac{y}{2}\Bigr). \tag{2}
\]

Write  

\[
u=\frac{x}{2},\qquad v=\frac{y}{2},
\]

so that \(x=2u,\;y=2v\).  
From (1) we have  

\[
u^{2}+v^{2}=361. \tag{3}
\]

The condition \(KL=14\) together with (2) gives  

\[
\bigl(u-\sqrt{196-v^{2}}\bigr)^{2}+\bigl(\sqrt{196-u^{2}}-v\bigr)^{2}=196,
\]

which simplifies to  

\[
u\sqrt{196-v^{2}}+v\sqrt{196-u^{2}}=98. \tag{4}
\]

Let  

\[
a=u^{2},\quad b=v^{2}.
\]

Then (3) is \(a+b=361\) and (4) becomes  

\[
\sqrt{a(196-b)}+\sqrt{b(196-a)}=98 .
\]

Squaring, eliminating the radicals and using \(a+b=361\) yields  

\[
a^{2}-361a+32448=0,
\]

so  

\[
a,b=\frac{361\pm 23}{2}=192,\;169.
\]

Thus  

\[
u^{2}=169,\;v^{2}=192\quad\text{or}\quad u^{2}=192,\;v^{2}=169.
\]

Take  

\[
u=13,\qquad v=8\sqrt3 .
\]

Hence  

\[
AB=2u=26,\qquad AC=2v=16\sqrt3,
\]

and the right‑triangle \(ABC\) indeed has \(BC=38\).

From (2) we obtain the exact coordinates

\[
B=(26,0),\quad C=(0,16\sqrt3),\quad 
K=\bigl(13,3\sqrt3\bigr),\quad 
L=\bigl(2,8\sqrt3\bigr).
\]

Now compute the area of quadrilateral \(BKLC\) with the shoelace formula:

\[
\begin{aligned}
\text{Area} &=\frac12\bigl|
26(3\sqrt3)+13(8\sqrt3)+2(16\sqrt3)+0\cdot0\\
&\quad- \bigl(13\cdot0+2(3\sqrt3)+0(8\sqrt3)+26(16\sqrt3)\bigr)\bigr|\\[2mm]
&=\frac12\bigl| 78\sqrt3+104\sqrt3+32\sqrt3
- (6\sqrt3+416\sqrt3) \bigr|\\[2mm]
&=\frac12\bigl|214\sqrt3-422\sqrt3\bigr|
= \frac12\,(208\sqrt3)=104\sqrt3 .
\end{aligned}
\]

Thus the area of \(BKLC\) is \(104\sqrt3\).  

Therefore  

\[
\boxed{104}
\]

Let the trapezoid have bases \(AB=r\) and \(CD=s\) \((r\neq s)\) and equal legs \(AD=BC=\ell\).
A circle of radius \(3\) is tangent to all four sides.

---

### 1.  Tangential quadrilateral condition  

For any quadrilateral that has an incircle, the sums of opposite sides are equal:

\[
AB+CD = AD+BC \qquad\Longrightarrow\qquad r+s = 2\ell .
\tag{1}
\]

Thus each leg is the average of the two bases:

\[
\ell =\frac{r+s}{2}.
\tag{2}
\]

---

### 2.  Use the area formula for a tangential quadrilateral  

If a quadrilateral has an incircle of radius \(R\) and semiperimeter \(p\), then  

\[
\text{Area}=R\cdot p .
\]

Here \(R=3\) and the area is \(72\), so

\[
3p = 72 \quad\Longrightarrow\quad p = 24 .
\tag{3}
\]

The perimeter is \(2p=48\). Using (1),

\[
r+s+2\ell =48 \quad\Longrightarrow\quad r+s+ (r+s)=48
\]
\[
\Longrightarrow\qquad r+s = 24 .
\tag{4}
\]

Hence from (2) the leg length is  

\[
\ell =\frac{r+s}{2}=12 .
\tag{5}
\]

---

### 3.  Height of the trapezoid  

The incircle touches both parallel sides, so the distance between them (the height \(h\)) equals twice the radius:

\[
h = 2\cdot 3 = 6 .
\tag{6}
\]

(One can also get \(h\) from \(\displaystyle \text{Area}= \frac{r+s}{2}\,h\) using (4): \(72 =12h\Rightarrow h=6\).)

---

### 4.  Relate the leg to the bases  

In an isosceles trapezoid, each leg forms a right triangle with height \(h\) and horizontal leg \(\frac{|r-s|}{2}\):

\[
\ell^{2}= \left(\frac{r-s}{2}\right)^{2}+h^{2}.
\]

Substituting \(\ell=12\) and \(h=6\),

\[
12^{2}= \frac{(r-s)^{2}}{4}+6^{2}
\quad\Longrightarrow\quad
144 = \frac{(r-s)^{2}}{4}+36
\]
\[
\Longrightarrow\quad (r-s)^{2}=432 .
\tag{7}
\]

---

### 5.  Find \(r^{2}+s^{2}\)

We have the two symmetric relations  

\[
(r+s)^{2}=24^{2}=576,
\qquad
(r-s)^{2}=432 .
\]

Recall  

\[
(r-s)^{2}= (r+s)^{2}-4rs .
\]

Thus  

\[
432 = 576 - 4rs \quad\Longrightarrow\quad rs = 36 .
\tag{8}
\]

Now  

\[
r^{2}+s^{2}= (r+s)^{2}-2rs = 576-2\cdot36 = 576-72 = 504 .
\]

---

\[
\boxed{504}
\]

For a number to be divisible by \(22\) it must be divisible by \(2\) and by \(11\).

---

### 1.  Divisibility by \(11\)

Write an 8‑digit permutation as  

\[
d_1d_2d_3d_4d_5d_6d_7d_8 ,
\]

with positions counted from the left.  
For an even number of digits, the rule for \(11\) is  

\[
(d_1+d_3+d_5+d_7)-(d_2+d_4+d_6+d_8)\equiv0\pmod{11}.
\]

Let  

\[
O=d_1+d_3+d_5+d_7,\qquad E=d_2+d_4+d_6+d_8 .
\]

Because the digits \(1,2,\dots ,8\) are used once,  

\[
O+E=1+2+\cdots+8=36 .
\]

The condition for \(11\) gives \(O-E\equiv0\pmod{11}\).  
Since \(O,E\) lie between \(10\) and \(26\), the only possible multiple of \(11\) is \(0\).  
Hence  

\[
O=E=18 .
\]

Thus the four digits placed in the odd positions must sum to \(18\); the remaining four (even positions) also sum to \(18\).

---

### 2.  Which four‑digit sets sum to \(18\)?

Find all 4‑element subsets of \(\{1,\dots ,8\}\) whose sum is \(18\).

\[
\begin{aligned}
&\{1,2,7,8\},\; \{1,3,6,8\},\; \{1,4,5,8\},\; \{1,4,6,7\},\\
&\{2,3,5,8\},\; \{2,3,6,7\},\; \{2,4,5,7\},\; \{3,4,5,6\}.
\end{aligned}
\]

There are \(8\) such subsets.  
Each contains exactly two even digits (the evens are \(2,4,6,8\)).  
Consequently, the complementary set (the digits for the even positions) also contains exactly two evens.

---

### 3.  Divisibility by \(2\)

A number is even precisely when its unit digit (position 8) is even.  
Position 8 belongs to the even positions, whose four digits are the complement of the chosen odd‑position set.  
Since that complement always has exactly two even digits, we have

\[
\text{choices for the unit digit}=2 .
\]

---

### 4.  Counting the permutations

Fix one of the eight valid 4‑digit odd‑position sets \(S\).

* Arrange the four digits of \(S\) in the odd positions \(1,3,5,7\): \(4! = 24\) ways.
* Choose an even digit from the complementary set for the unit place: \(2\) ways.
* Arrange the remaining three digits of the complement in positions \(2,4,6\): \(3! = 6\) ways.

Hence for each set \(S\)

\[
24 \times 2 \times 6 = 288
\]

permutations are divisible by \(22\).

With \(8\) possible sets \(S\),

\[
N = 8 \times 288 = 2304 .
\]

---

### 5.  Required difference

\[
N-2025 = 2304-2025 = 279 .
\]

\[
\boxed{279}
\]

First write each term \(x_k\) in reduced form  

\[
x_k=\frac{a_k}{b_k},\qquad (a_k,b_k)=1 ,\;a_k,b_k>0 .
\]

From  

\[
x_{k+1}= \frac13\Bigl(x_k+\frac1{x_k}-1\Bigr)
        =\frac13\Bigl(\frac{a_k}{b_k}+\frac{b_k}{a_k}-1\Bigr)
\]

we obtain  

\[
a_{k+1}= \frac{a_k^2+b_k^2-a_kb_k}{3},\qquad   
b_{k+1}=a_kb_k .\tag{1}
\]

--------------------------------------------------------------------
### 1.  Coprimeness is preserved
If a prime \(p\mid a_{k+1}\) and \(p\mid b_{k+1}=a_kb_k\) then \(p\mid a_k\) or \(p\mid b_k\).
But modulo such a prime the numerator \(a_k^2+b_k^2-a_kb_k\) equals \(b_k^2\) (resp. \(a_k^2\)),
which is non‑zero because \((a_k,b_k)=1\). Hence no prime can divide both
\(a_{k+1}\) and \(b_{k+1}\); therefore \((a_{k+1},b_{k+1})=1\).
Thus each \(x_k\) is already in lowest terms.

--------------------------------------------------------------------
### 2.  A simple recurrence for the sum

Let  

\[
s_k=a_k+b_k .
\]

Adding the two formulas in (1),

\[
\begin{aligned}
s_{k+1}
 &=\frac{a_k^2+b_k^2-a_kb_k}{3}+a_kb_k  \\
 &=\frac{a_k^2+b_k^2+2a_kb_k}{3} =\frac{(a_k+b_k)^2}{3}
   =\frac{s_k^{\,2}}{3}.         \tag{2}
\end{aligned}
\]

The initial sum is  

\[
s_1= a_1+b_1=25+11=36 .
\]

--------------------------------------------------------------------
### 3.  Closed form of \(s_k\)

From (2) we prove by induction that  

\[
\boxed{\,s_k = 2^{\,2^{\,k}}\; 3^{\,2^{\,k-1}+1}\,}\qquad(k\ge1).
\]

Indeed, for \(k=1\) the formula gives \(2^{2}\,3^{2}=36\).
Assume it holds for \(k\); then using (2),

\[
s_{k+1}= \frac{s_k^{2}}{3}
      =\frac{(2^{2^{k}}3^{2^{k-1}+1})^{2}}{3}
      =2^{2^{k+1}}3^{2^{k}+1},
\]

which is the required expression for \(k+1\).

--------------------------------------------------------------------
### 4.  Computing \(s_{2025}\pmod{1000}\)

We need \(m+n=s_{2025}\pmod{1000}\).

Because \(1000=8\cdot125\), we treat the two moduli separately.

*Modulo \(8\):*  
\(2^{2^{2025}}\) contains the factor \(2^3\); hence \(s_{2025}\equiv0\pmod 8\).

*Modulo \(125\):*  
\(\phi(125)=100\).  Thus we may reduce the exponents modulo \(100\).

\[
\begin{aligned}
2^{2025}\pmod{100}&=32, &
2^{2024}\pmod{100}&=16 .
\end{aligned}
\]

Hence  

\[
\begin{aligned}
2^{\,2^{2025}}\pmod{125}&=2^{32}\pmod{125}=46,\\[2mm]
3^{\,2^{2024}+1}\pmod{125}&=3^{16+1}=3^{17}\pmod{125}=38.
\end{aligned}
\]

Therefore  

\[
s_{2025}\equiv 46\cdot38\equiv123\pmod{125}.
\]

Now solve  

\[
N\equiv0\pmod8,\qquad N\equiv123\pmod{125}.
\]

Since \(8^{-1}\equiv47\pmod{125}\),

\[
N\equiv8\bigl(47\cdot123\bigr)\equiv8\cdot31\equiv248\pmod{1000}.
\]

Thus  

\[
m+n\equiv\boxed{248}\pmod{1000}.
\]

--------------------------------------------------------------------
### 5.  Check via periodicity (optional)

From \(s_{k+1}=s_k^2/3\) modulo \(1000\) one finds that the
sequence \(\{s_k\bmod1000\}\) becomes periodic with period \(20\) after
the third term:
\[
36,\;432,\;\underline{208},\;88,\;248,\;168,\dots
\]
Since \(2025-3\equiv2\pmod{20}\), the term equals the fifth term,
\(248\), confirming the result above.

Let  

\[
z=x+iy .
\]

---

### 1.  The first equation  

\[
|25+20i-z|=5
\]

gives the circle  

\[
(x-25)^2+(y-20)^2=5^{2}=25 .
\tag{1}
\]

Its centre is \(C=(25,20)\) and its radius is \(5\).

---

### 2.  The second equation  

\[
|z-4-k|=|z-3i-k|
\]

means that the point \(z\) is equidistant from the two points  

\[
A=(4+k,0),\qquad B=(k,3).
\]

Thus \(z\) lies on the **perpendicular bisector** of the segment \(AB\).

*Midpoint*  

\[
M\Bigl(\frac{4+k+k}{2},\frac{0+3}{2}\Bigr)=\bigl(k+2,\; \tfrac32 \bigr).
\]

*Slope of \(AB\)*  

\[
\frac{3-0}{k-(4+k)}=\frac{3}{-4}=-\frac34 .
\]

Hence the slope of the perpendicular bisector is \( \displaystyle \frac43\).

Its equation is

\[
y-\frac32=\frac43\bigl(x-(k+2)\bigr)
\]

or, after clearing denominators,

\[
8x-6y-8k-7=0 .
\tag{2}
\]

Thus the second condition is the line \(L_k: 8x-6y-8k-7=0\).

---

### 3.  Intersections of (1) and (2)

The system has exactly one solution iff the line \(L_k\) is **tangent** to the circle (1), i.e. the distance from the centre \(C\) to the line equals the radius \(5\).

Distance from \((25,20)\) to \(L_k\):

\[
d=\frac{|8\cdot25-6\cdot20-8k-7|}{\sqrt{8^{2}+(-6)^{2}}}
   =\frac{|200-120-8k-7|}{\sqrt{64+36}}
   =\frac{|73-8k|}{10}.
\]

Set \(d=5\):

\[
\frac{|73-8k|}{10}=5\quad\Longrightarrow\quad |73-8k|=50.
\]

Hence  

\[
73-8k=50\quad\text{or}\quad73-8k=-50,
\]
giving  

\[
k=\frac{23}{8}\qquad\text{or}\qquad k=\frac{123}{8}.
\]

For each of these values the line is tangent, so there is exactly one point \(z\) satisfying both equations.

---

### 4.  Sum of all possible \(k\)

\[
\frac{23}{8}+\frac{123}{8}= \frac{146}{8}= \frac{73}{4}.
\]

Write \(\displaystyle \frac{73}{4}\) as \(\frac{m}{n}\) with \(\gcd(m,n)=1\); here \(m=73,\;n=4\).

\[
m+n = 73+4 = 77.
\]

\[
\boxed{77}
\]

Since the points lie on the sides of \(\triangle ABC\) we may work with vectors
\[
\vec{AB}= \mathbf b,\qquad \vec{AC}= \mathbf c ,
\qquad A\; \text{at the origin}.
\]

--------------------------------------------------------------------
### 1.  Locate the points \(D,E,F,G\)

\[
AD=4,\;AB=28\Longrightarrow \frac{AD}{AB}= \frac 1 7\quad\Rightarrow\quad 
\vec{AD}=\frac17\mathbf b .
\]
Thus \(D=\dfrac17\mathbf b\).

\[
AE=AD+DE=20\Longrightarrow\frac{AE}{AB}= \frac{20}{28}= \frac57
\quad\Rightarrow\quad  E=\frac57\mathbf b .
\]

Similarly on \(AC\),

\[
AF=13,\;AC=91\Longrightarrow \frac{AF}{AC}= \frac13\!7\quad\Rightarrow\quad 
F=\frac17\mathbf c ,
\]
\[
AG=AF+FG=65\Longrightarrow \frac{AG}{AC}= \frac{65}{91}= \frac57
\quad\Rightarrow\quad  G=\frac57\mathbf c .
\]

Hence  

\[
D=\frac17\mathbf b,\qquad E=\frac57\mathbf b,\qquad
F=\frac17\mathbf c,\qquad G=\frac57\mathbf c .
\]

--------------------------------------------------------------------
### 2.  Area of \(\triangle ABC\)

Quadrilateral \(DEGF\) is the region between the similar triangles
\(ADF\) (scale \(1/7\)) and \(AEG\) (scale \(5/7\)).  
Therefore  

\[
[DEGF]=[AEG]-[ADF]
           =\left(\frac57\right)^{\!2}[ABC]-\left(\frac17\right)^{\!2}[ABC]
           =\frac{25-1}{49}[ABC]=\frac{24}{49}[ABC].
\]

Given \([DEGF]=288\),

\[
[ABC]=\frac{49}{24}\cdot288=49\cdot12=588 .
\tag{1}
\]

--------------------------------------------------------------------
### 3.  Locate the reflected points \(M,N\)

\[
M\; \text{is the reflection of }D\text{ across }F\;
\Longrightarrow\; M=2F-D=
\frac{2}{7}\mathbf c-\frac{1}{7}\mathbf b .
\]

\[
N\; \text{is the reflection of }G\text{ across }E\;
\Longrightarrow\; N=2E-G=
\frac{10}{7}\mathbf b-\frac{5}{7}\mathbf c .
\]

--------------------------------------------------------------------
### 4.  Area of the heptagon \(AFNBCEM\)

Write the vertices (in order) as vectors from \(A\):

\[
\begin{aligned}
A&=0, &
F&=\frac17\mathbf c, &
N&=\frac{10}{7}\mathbf b-\frac57\mathbf c,\\
B&=\mathbf b, &
C&=\mathbf c, &
E&=\frac57\mathbf b, &
M&=\frac27\mathbf c-\frac17\mathbf b .
\end{aligned}
\]

For a polygon with vertices \(\mathbf v_0,\dots,\mathbf v_6\) the doubled
area equals
\[
\sum_{i=0}^{6}\det(\mathbf v_i,\mathbf v_{i+1}),
\qquad\mathbf v_7=\mathbf v_0 .
\]

Carrying out the determinants (using \(\det(\mathbf b,\mathbf c)=\det(\mathbf b,\mathbf c)\) and \(\det(\mathbf c,\mathbf b)=-\det(\mathbf b,\mathbf c)\)):

\[
\begin{aligned}
\det(F,N) &=\frac1{49}\det\!\Bigl(\mathbf c,\;10\mathbf b-5\mathbf c\Bigr)
           =-\frac{10}{49}\det(\mathbf b,\mathbf c),\\[2mm]
\det(N,B) &=\frac57\det(\mathbf b,\mathbf b)-\frac5{7}\det(\mathbf c,\mathbf b)
           =\frac5{7}\det(\mathbf b,\mathbf c),\\[2mm]
\det(B,C) &=\det(\mathbf b,\mathbf c),\\[2mm]
\det(C,E) &=\frac57\det(\mathbf c,\mathbf b)=-\frac5{7}\det(\mathbf b,\mathbf c),\\[2mm]
\det(E,M) &=\frac1{49}\det\!\Bigl(5\mathbf b,\;2\mathbf c-\mathbf b\Bigr)
           =\frac{10}{49}\det(\mathbf b,\mathbf c).
\end{aligned}
\]

All other terms are zero. Adding them:

\[
\det(F,N)+\det(N,B)+\det(B,C)+\det(C,E)+\det(E,M)
   =\Bigl(-\frac{10}{49}+\frac5{7}+1-\frac5{7}+\frac{10}{49}\Bigr)
      \det(\mathbf b,\mathbf c)=\det(\mathbf b,\mathbf c).
\]

Thus the doubled area of the heptagon is \(\det(\mathbf b,\mathbf c)\),
which is exactly twice the area of \(\triangle ABC\). Consequently

\[
[AFNBCEM]=\frac12\det(\mathbf b,\mathbf c)=[ABC]=588 .
\tag{2}
\]

--------------------------------------------------------------------
### 5.  Answer

\[
\boxed{588}
\]

The parabola is  

\[
y=x^{2}-4 .
\]

Rotate a point \((x,y)\) by \(-60^{\circ}\) (the inverse rotation).  
For \(\theta =60^{\circ}\),

\[
\cos\theta =\frac12 ,\qquad \sin\theta =\frac{\sqrt3}{2}.
\]

The inverse‑rotation of \((x,y)\) is  

\[
u=\frac12 x+\frac{\sqrt3}{2}y ,\qquad 
v=-\frac{\sqrt3}{2}x+\frac12 y .
\]

For a point of the rotated parabola, the inverse image \((u,v)\) must satisfy the
original equation \(v=u^{2}-4\):

\[
-\frac{\sqrt3}{2}x+\frac12 y =\Bigl(\frac12 x+\frac{\sqrt3}{2}y\Bigr)^{2}-4 .
\]

Multiplying by \(2\) and expanding gives

\[
\frac12 x^{2}+\sqrt3\,xy+\frac32 y^{2}+\sqrt3\,x-y-8=0. \tag{1}
\]

Now substitute \(y=x^{2}-4\) (the original parabola) into (1):

\[
\frac12 x^{2}+\sqrt3\,x(x^{2}-4)+\frac32 (x^{2}-4)^{2}
+\sqrt3\,x-(x^{2}-4)-8=0 .
\]

Simplifying,

\[
3x^{4}+2\sqrt3\,x^{3}-25x^{2}-6\sqrt3\,x+40=0 .
\]

Factor the quartic (group the \(\sqrt3\) terms):

\[
\bigl(x^{2}+\sqrt3\,x-4\bigr)\bigl(3x^{2}-\sqrt3\,x-10\bigr)=0 .
\]

Hence

\[
x^{2}+\sqrt3\,x-4=0\qquad\text{or}\qquad 3x^{2}-\sqrt3\,x-10=0.
\]

The four solutions are  

\[
\begin{aligned}
x&=\frac{-\sqrt3\pm\sqrt{19}}2,\\[2mm]
x&=\frac{\sqrt3\pm\sqrt{123}}6 .
\end{aligned}
\]

Only one of them lies in the fourth quadrant \((x>0,\;y<0)\):
\[
x=\frac{-\sqrt3+\sqrt{19}}2\;(>0),\qquad |x|<2 .
\]

Its \(y\)-coordinate is

\[
y=x^{2}-4
     =\Bigl(\frac{-\sqrt3+\sqrt{19}}2\Bigr)^{2}-4
     =\frac{3-\sqrt{57}}2 .
\]

Thus  

\[
y=\frac{a-\sqrt b}{c}\quad\text{with}\quad a=3,\;b=57,\;c=2,
\]
and \(a\) and \(c\) are relatively prime.

Finally,
\[
a+b+c=3+57+2=62 .
\]

\[
\boxed{62}
\]

The six two‑letter words are formed by first ordering the letters inside each pair
alphabetically and then sorting the six words alphabetically.  
For a pairing, let  

\[
S=\{\text{the smaller letter of each pair}\}
\]

(the “first” letters of the six words).  
The lexicographically last word is the one whose first letter is  

\[
M=\max S .
\]

Hence the last word contains \(G\) **iff** either  

* \(M=G\) (so \(G\) itself is the first letter of the last word), or  
* \(M\neq G\) but \(G\) is paired with the letter \(M\) (so \(G\) is the second
letter of the last word).

Because a pair \((x,y)\) must satisfy \(x<y\), the only possible \(M\) with
\(M<G\) is \(M=F\).  Thus there are two cases:

--------------------------------------------------------------------
### 1.  \(M=G\)

Then the six small letters are  

\[
S=\{G\}\cup \{A,B,C,D,E,F\}\setminus\{a\},
\]

where \(a\) is the one letter from \(\{A,\dots ,F\}\) that is **not** in \(S\).
The omitted letter \(a\) must not be \(A\); otherwise \(A\) would be a large
letter with no smaller partner.  
Hence \(a\in\{B,C,D,E,F\}\) – 5 possibilities.

For a given \(a\):
* \(a\) (a large letter) must be paired with a small letter that is
smaller than \(a\); there are \(a-1\) such choices.
* The remaining five small letters can be paired arbitrarily with the five
large letters \(H,I,J,K,L\); that gives \(5!\) possibilities.

Thus for each \(a\) we have \((a-1)\cdot5!\) pairings.  
Summing over \(a=2,3,4,5,6\) (i.e. \(B\)–\(F\)),

\[
\sum_{a=2}^{6}(a-1) \cdot 5!
   =(1+2+3+4+5)\cdot120
   =15\cdot120
   =1800 .
\]

--------------------------------------------------------------------
### 2.  \(M=F\) and \(G\) is paired with \(F\)

If the maximum small letter is \(F\), then the small set must be exactly  

\[
S=\{A,B,C,D,E,F\},
\qquad
T=\{G,H,I,J,K,L\}.
\]

Every bijection between \(S\) and \(T\) yields a valid pairing, so there are
\(6! = 720\) pairings in total.  
Among them, the ones with \(F\) paired with \(G\) are those where we fix that
pair and match the remaining five letters arbitrarily, giving \(5! =120\) pairings.

--------------------------------------------------------------------
### Total favourable pairings

\[
N_{\text{fav}} = 1800 + 120 = 1920 .
\]

--------------------------------------------------------------------
### Total pairings

The total number of ways to split 12 distinct letters into six unordered pairs is  

\[
N_{\text{total}}=\frac{12!}{2^{6}\,6!}=10395 .
\]

--------------------------------------------------------------------
### Probability

\[
P=\frac{1920}{10395}
   =\frac{128}{693}\quad(\text{in lowest terms}).
\]

Thus \(m=128,\;n=693\) and

\[
m+n=128+693=821 .
\]

\[
\boxed{821}
\]

The graph of  

\[
f(x)=\begin{cases}
x,&-1\le x<1\\[2pt]
2-x,&1\le x<3
\end{cases},
\qquad f(x+4)=f(x)
\]

is a triangular wave of period \(4\) whose values lie in \([-1,1]\).

The parabola is  

\[
x=34y^{2}\qquad (x\ge 0).
\]

Because \(|f(x)|\le 1\), any intersection must satisfy \(|y|\le 1\); then  

\[
0\le x=34y^{2}\le 34 .
\]

--------------------------------------------------------------------
### 1. Reduce the problem to a quadratic equation

Write  

\[
t = 34y^{2}\;(0\le t\le34).
\]

For some integer \(k\) we can write  

\[
t = 4k+u,\qquad u\in[-1,3)
\]

(the representation is unique because we use the half‑open interval \([-1,3)\)).  
Since \(f(t)=f(u)\),

\[
f(u)=\begin{cases}
u, & u\in[-1,1)\\[2pt]
2-u, & u\in[1,3)
\end{cases}
\]

and the intersection condition \(y=f(t)\) becomes  

\[
\begin{cases}
y=u = 34y^{2}-4k, & u\in[-1,1)\\[4pt]
y=2-u = 2-(34y^{2}-4k), & u\in[1,3)
\end{cases}
\]

which give the quadratics  

\[
\boxed{34y^{2}-y-4k=0}\qquad(u\in[-1,1))
\tag{A}
\]

\[
\boxed{34y^{2}+y-(2+4k)=0}\qquad(u\in[1,3))
\tag{B}
\]

with the extra restrictions  

\[
\text{(A)}\;y\in[-1,1), \qquad
\text{(B)}\; -1<y\le 1 .
\]

--------------------------------------------------------------------
### 2. The possible integers \(k\)

Since \(t\le34\),

\[
-1\le u =t-4k <3\Longrightarrow 4k\le t+1\le 35
\]

so \(k\) can be \(0,1,\dots ,8\) (nine values).

--------------------------------------------------------------------
### 3. Solutions of (A)

For each \(k\) the equation  

\[
34y^{2}-y-4k=0
\]

has the two roots  

\[
y=\frac{1\pm\sqrt{1+544k}}{68}.
\]

Both lie in \([-1,1)\) for all \(k=0,\dots ,8\).  
Hence each \(k\) contributes **two** solutions and their sum is

\[
\frac{1}{34}\quad\text{(by Vieta)}.
\]

Total from (A):

\[
9\cdot\frac{1}{34}= \frac{9}{34}.
\]

--------------------------------------------------------------------
### 4. Solutions of (B)

For each \(k\) the equation  

\[
34y^{2}+y-(2+4k)=0
\]

has roots  

\[
y=\frac{-1\pm\sqrt{273+544k}}{68}.
\]

For \(k=0,1,\dots ,7\) both roots satisfy \(-1<y\le1\); for \(k=8\) the
negative root is \(<-1\) and is discarded.
Thus we have

* for \(k=0,\dots ,7\): two roots each, sum \(-\dfrac1{34}\) (Vieta);
* for \(k=8\): only the positive root  

  \[
  y_{8}= \frac{-1+\sqrt{4625}}{68}= \frac{-1+5\sqrt{185}}{68}.
  \]

Hence the contribution of (B) is  

\[
8\!\left(-\frac1{34}\right) + y_{8}= -\frac{8}{34}+ \frac{-1+5\sqrt{185}}{68}
= -\frac{4}{17}+ \frac{-1+5\sqrt{185}}{68}.
\]

--------------------------------------------------------------------
### 5. Sum of all \(y\)-coordinates

Add the contributions of (A) and (B):

\[
\begin{aligned}
S &=\frac{9}{34}\;+\;\Bigl(-\frac{8}{34}+ \frac{-1+5\sqrt{185}}{68}\Bigr)\\[4pt]
  &=\frac{1}{34}+ \frac{-1+5\sqrt{185}}{68}
   =\frac{2-1+5\sqrt{185}}{68}\\[4pt]
  &=\boxed{\displaystyle\frac{1+5\sqrt{185}}{68}}.
\end{aligned}
\]

--------------------------------------------------------------------
### 6. Required sum \(a+b+c+d\)

The expression is \(\dfrac{a+b\sqrt{c}}{d}\) with  

\[
a=1,\quad b=5,\quad c=185,\quad d=68,
\]

which satisfy the given conditions. Therefore  

\[
a+b+c+d = 1+5+185+68 = \boxed{259}.
\]

Let the line be the \(x\)-axis.  
Place the points with coordinates (in order)

\[
A(0),\; B(x_B),\; C(x_C),\; D(x_D),\; E(x_E),\; F(x_F).
\]

Denote the successive distances

\[
AB=a,\; BC=b,\; CD=c,\; DE=d,\; EF=e .
\]

Then  

\[
\begin{aligned}
a+b &= AC = 26,\\
b+c &= BD = 22,\\
c+d &= CE = 31,\\
d+e &= DF = 33,\\
a+b+c+d+e &= AF = 73 .
\end{aligned}
\]

Solving:
\[
\begin{aligned}
a &= 26-b,\\
c &= 22-b,\\
d &= 31-c = 9+b,\\
e &= 33-d = 24-b.
\end{aligned}
\]

Insert these in the sum equation:

\[
(26-b)+b+(22-b)+(9+b)+(24-b)=73\Longrightarrow81-b=73,
\]

hence \(b=8\).

Thus  

\[
\begin{aligned}
AB = a = 18,\quad
BC = b = 8,\quad
CD = c = 14,\quad
DE = d = 17,\quad
EF = e = 16.
\end{aligned}
\]

Choosing \(A\) as the origin gives

\[
\begin{aligned}
A&:(0,0),\\
B&:(18,0),\\
C&:(26,0),\\
D&:(40,0),\\
E&:(57,0),\\
F&:(73,0).
\end{aligned}
\]

---

### Coordinates of \(G\)

Let \(G=(x,y)\).  
We are given \(CG=40\) and \(DG=30\):

\[
\begin{cases}
(x-26)^2+y^2 = 40^2 =1600,\\
(x-40)^2+y^2 = 30^2 = 900 .
\end{cases}
\]

Subtracting the equations eliminates \(y^2\):

\[
(x-26)^2-(x-40)^2 = 700 .
\]

\[
\begin{aligned}
[x^2-52x+676] - [x^2-80x+1600] &= 700\\
28x-924 &= 700\\
28x &=1624\\
x&= \frac{1624}{28}=58 .
\end{aligned}
\]

Plug \(x=58\) into \((x-40)^2+y^2=900\):

\[
(58-40)^2+y^2=900\Longrightarrow 18^2+y^2=900\Longrightarrow y^2=576,
\]

so \(|y|=24\).  
Thus \(G=(58,\pm24)\).

---

### Area of \(\triangle BGE\)

Points \(B\) and \(E\) lie on the \(x\)-axis, so the base \(BE\) has length

\[
BE = x_E-x_B = 57-18 = 39 .
\]

The altitude from \(G\) to line \(BE\) is \(|y|=24\).

Hence

\[
\text{Area}= \frac12 \times (\text{base}) \times (\text{height})
          = \frac12 \times 39 \times 24
          = 39 \times 12
          = 468 .
\]

\[
\boxed{468}
\]

Let  

\[
d=n+2\qquad (d\ge 3,\; n>0)
\]

Then  

\[
n=d-2,\qquad n+3=d+1,\qquad n^{2}+9=(d-2)^{2}+9=d^{2}-4d+13 .
\]

The required divisibility becomes  

\[
d\mid 3(n+3)(n^{2}+9)=3(d+1)(d^{2}-4d+13).
\]

---

### Reduce modulo \(d\)

\[
3(d+1)(d^{2}-4d+13) \equiv 3\cdot1\cdot13 = 39 \pmod d,
\]

because \(d+1\equiv1\) and \(d^{2}-4d+13\equiv13\) modulo \(d\).

Thus  

\[
d\mid 3(d+1)(d^{2}-4d+13)\iff d\mid 39 .
\]

Hence \(d\) must be a positive divisor of \(39\) and at least \(3\).  
Divisors of \(39\) are \(1,3,13,39\); discarding \(1\) we obtain  

\[
d\in\{3,13,39\}.
\]

---

### Recover the corresponding \(n\)

\[
n=d-2\implies n\in\{3-2,13-2,39-2\}=\{1,11,37\}.
\]

Each indeed satisfies the original condition:

* \(n=1\): \(\frac{3(1+3)(1^{2}+9)}{1+2}= \frac{3\cdot4\cdot10}{3}=40\) (integer);
* \(n=11\): \(\frac{3(11+3)(11^{2}+9)}{11+2}= \frac{3\cdot14\cdot130}{13}=420\);
* \(n=37\): \(\frac{3(37+3)(37^{2}+9)}{37+2}= \frac{3\cdot40\cdot1378}{39}=4240\).

---

### Sum of all such \(n\)

\[
1+11+37 = 49.
\]

\[
\boxed{49}
\]

First write the two required inequalities using the condition \(x+y+z=75\).

\[
z=75-x-y .
\]

---

### 1.  Simplify the inequalities  

\[
\begin{aligned}
x-yz &<y-zx\\
x-(y(75-x-y)) &< y-(x(75-x-y))\\
\Longrightarrow (x-y)(x+y-76) &>0 .
\end{aligned}
\]

\[
\begin{aligned}
y-zx &<z-xy\\
y-x(75-x-y) &< (75-x-y)-xy\\
\Longrightarrow (x+1)(x+2y-75) &<0 .
\end{aligned}
\]

Hence the points \((x,y,z)\) on the plane satisfy  

\[
\boxed{(x-y)(x+y-76)>0},\qquad\boxed{(x+1)(x+2y-75)<0}.
\]

Both are products of two linear factors.

---

### 2.  Work in the \((x,y)\)-plane  

The plane \(x+y+z=75\) is a linear image of the \((x,y)\)-plane via  

\[
(x,y)\mapsto (x,\;y,\;75-x-y) .
\]

The Jacobian matrix has columns \((1,0,-1)\) and \((0,1,-1)\); its Gram determinant is  

\[
\det\begin{pmatrix}2&1\\1&2\end{pmatrix}=3 .
\]

Thus an area element in the plane equals \(\sqrt3\) times the ordinary area
\(dx\,dy\) in the \((x,y)\)-plane.

---

### 3.  Region defined by the two products  

Let  

\[
L_1:\;x=-1,\qquad L_2:\;x+2y=75,
\]
\[
L_3:\;x=y,\qquad L_4:\;x+y=76 .
\]

The inequalities become

\[
\begin{cases}
x>-1 \;\text{and}\; y<\dfrac{75-x}{2}   &\text{or } x<-1 \;\text{and}\; y>\dfrac{75-x}{2},\\[4pt]
x>y \;\text{and}\; x+y>76               &\text{or } x<y \;\text{and}\; x+y<76 .
\end{cases}
\]

Only three of the four possible combinations are non‑empty:

* **Region \(R_1\)** – \(x>-1,\;y<\frac{75-x}{2},\;x>y,\;x+y>76\) (unbounded).
* **Region \(R_2\)** – \(x>-1,\;y<\frac{75-x}{2},\;x<y,\;x+y<76\) (bounded).
* **Region \(R_3\)** – \(x<-1,\;y>\frac{75-x}{2},\;x<y,\;x+y<76\) (unbounded).

Hence the set consists of three disjoint convex regions; only \(R_2\) has finite area.

---

### 4.  The bounded region  

From the inequalities for \(R_2\)

\[
\begin{cases}
-1<x<25,\\[2pt]
x<y<\dfrac{75-x}{2},
\end{cases}
\]

so \(R_2\) is the open triangle bounded by the lines  

\[
x=-1,\qquad y=x,\qquad y=\dfrac{75-x}{2}.
\]

Its vertices are  

\[
A(-1,-1),\quad B(-1,38),\quad C(25,25).
\]

The ordinary (planar) area is

\[
\frac12\cdot|AB|\cdot|AC|_{\perp}= \frac12\,(39)\,(26)=507 .
\]

---

### 5.  Area on the plane \(x+y+z=75\)

Multiplying by the factor \(\sqrt3\),

\[
\text{area}=507\sqrt3 .
\]

Thus \(a=507,\;b=3\) and

\[
a+b=507+3=510 .
\]

\[
\boxed{510}
\]

\[
P=\prod_{k=4}^{63}\frac{\log_k(5^{k^2-1})}{\log_{k+1}(5^{k^2-4})}
\]

---

### 1.  Simplify one factor  

\[
\log_k(5^{k^2-1})=\frac{\ln(5^{k^2-1})}{\ln k}
            =\frac{(k^2-1)\ln5}{\ln k},
\qquad
\log_{k+1}(5^{k^2-4})=\frac{(k^2-4)\ln5}{\ln(k+1)}.
\]

Hence

\[
\frac{\log_k(5^{k^2-1})}{\log_{k+1}(5^{k^2-4})}
   =\frac{k^2-1}{k^2-4}\cdot\frac{\ln(k+1)}{\ln k}.
\]

So

\[
P=\Bigl(\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}\Bigr)
   \Bigl(\prod_{k=4}^{63}\frac{\ln(k+1)}{\ln k}\Bigr).
\]

---

### 2.  Telescoping the logarithm part  

\[
\prod_{k=4}^{63}\frac{\ln(k+1)}{\ln k}
 =\frac{\ln5}{\ln4}\cdot\frac{\ln6}{\ln5}\cdots\frac{\ln64}{\ln63}
 =\frac{\ln64}{\ln4}
 =\log_4 64 =3 .
\]

---

### 3.  Telescoping the rational part  

\[
\frac{k^2-1}{k^2-4}= \frac{(k-1)(k+1)}{(k-2)(k+2)}
   =\frac{k-1}{k-2}\cdot\frac{k+1}{k+2}.
\]

Hence

\[
\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}
 =\Bigl(\prod_{k=4}^{63}\frac{k-1}{k-2}\Bigr)
   \Bigl(\prod_{k=4}^{63}\frac{k+1}{k+2}\Bigr).
\]

*First product*  

Let \(i=k-2\). Then \(i\) runs from \(2\) to \(61\):

\[
\prod_{k=4}^{63}\frac{k-1}{k-2}
 =\prod_{i=2}^{61}\frac{i+1}{i}
 =\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{62}{61}
 =\frac{62}{2}=31 .
\]

*Second product*  

Let \(j=k+1\). Then \(j\) runs from \(5\) to \(64\):

\[
\prod_{k=4}^{63}\frac{k+1}{k+2}
 =\prod_{j=5}^{64}\frac{j}{j+1}
 =\frac{5}{6}\cdot\frac{6}{7}\cdots\frac{64}{65}
 =\frac{5}{65}=\frac1{13}.
\]

Thus

\[
\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}=31\cdot\frac1{13}=\frac{31}{13}.
\]

---

### 4.  Combine the two parts  

\[
P = 3\cdot\frac{31}{13}= \frac{93}{13}.
\]

The fraction \(\frac{93}{13}\) is already in lowest terms (\(\gcd(93,13)=1\)).
Thus \(m=93,\; n=13\) and

\[
m+n = 93+13 = 106 .
\]

\[
\boxed{106}
\]

**Step 1 –  what the conditions mean**

In a \(3\times 9\) grid  

* each of the three rows must be a permutation of the numbers \(1,\ldots ,9\);  
* each of the three \(3\times3\) blocks (columns \(1\!-\!3,\;4\!-\!6,\;7\!-\!9\)) must also contain the numbers \(1,\ldots ,9\) exactly once.

Hence every digit appears

* once in each row (3 times in total), and  
* once in each block (also 3 times in total).

So for a fixed digit \(d\) the three positions where it occurs are determined by a
bijection between the three rows and the three blocks; i.e. a permutation
\(\sigma_d\in S_{3}\).

--------------------------------------------------------------------

**Step 2 –  describing a filling by permutations**

For the nine digits we obtain nine permutations  

\[
\sigma_1,\sigma_2,\dots ,\sigma_9\in S_3 .
\]

If \(\sigma_d(i)=j\) then digit \(d\) is placed in row \(i\) inside block \(j\).
For a fixed block–row cell \((i,j)\) there are three rows of the grid, hence it
must contain **exactly three** digits. Consequently for every cell \((i,j)\)

\[
\#\{d\mid \sigma_d(i)=j\}=3 .
\tag{1}
\]

--------------------------------------------------------------------

**Step 3 –  solving the integer system**

Let  

\[
\begin{aligned}
x_{\mathrm{id}}&=\#\{\sigma_d=\text{id}\},&
x_{12}&=\#\{\sigma_d=(12)\},\\
x_{13}&=\#\{\sigma_d=(13)\},&
x_{23}&=\#\{\sigma_d=(23)\},\\
x_{123}&=\#\{\sigma_d=(123)\},&
x_{132}&=\#\{\sigma_d=(132)\}.
\end{aligned}
\]

Each permutation contributes a ‘1’ to three different cells; each cell is covered
by exactly two permutations. From (1) we obtain nine equations, e.g.

\[
x_{\mathrm{id}}+x_{23}=3,\qquad
x_{12}+x_{123}=3,\qquad\ldots
\]

Solving them gives

\[
x_{12}=x_{13}=x_{23}=k,\qquad
x_{\mathrm{id}}=x_{123}=x_{132}=3-k,
\]

where \(k\in\{0,1,2,3\}\).

--------------------------------------------------------------------

**Step 4 –  assigning the permutations to the nine distinct digits**

For a fixed \(k\) the number of ways to give the nine digits the required
multiplicities is the multinomial coefficient

\[
\frac{9!}{k!^{\,3}\,(3-k)!^{\,3}} .
\]

Summing over the four possible values of \(k\) gives

\[
N=\sum_{k=0}^{3}\frac{9!}{k!^{\,3}(3-k)!^{\,3}}
   =2\cdot\frac{9!}{3!^{\,3}}+2\cdot\frac{9!}{1!^{\,3}2!^{\,3}}
   =1680+45360+45360+1680=94\,080 .
\]

Thus there are \(94\,080\) ways to decide, for every digit, in which block of
each row it will appear.

--------------------------------------------------------------------

**Step 5 –  placing the digits inside the blocks**

Fix a block \(j\).  
Row \(i\) contributes exactly three distinct digits to this block; they can be
arranged in the three columns of the block in \(3! = 6\) ways, independently for
the three rows. Hence a block can be filled in \((3!)^{3}=6^{3}=216\) ways.

The three blocks are independent, so the column‑ordering factor is

\[
(3!)^{9}=6^{9}=10\,077\,696 .
\]

--------------------------------------------------------------------

**Step 6 –  total number of fillings**

\[
\boxed{T = N\cdot (3!)^{9}=94\,080\cdot 10\,077\,696
      = 948\,109\,639\,680 }.
\]

Factorising,

\[
T = 2^{16}\cdot 3^{10}\cdot 5\cdot 7^{2}.
\]

--------------------------------------------------------------------

**Step 7 –  computing the required sum**

Take \(p=2,\;q=3,\;r=5,\;s=7\) with exponents
\(a=16,\;b=10,\;c=1,\;d=2\). Then

\[
p\!a+q\!b+r\!c+s\!d
=2\cdot16+3\cdot10+5\cdot1+7\cdot2
=32+30+5+14=81 .
\]

\[
\boxed{81}
\]

**Solution Overview**

For  

\[
f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x},\qquad x>0,
\]

let  

\[
N(x)=(x-18)(x-72)(x-98)(x-k)=x^4-S_1x^3+S_2x^2-S_3x+S_4,
\]

where  

\[
\begin{aligned}
S_1&=18+72+98+k=188+k,\\[2pt]
S_2&=18\cdot72+18\cdot98+72\cdot98+ k(18+72+98)=10116+188k,\\[2pt]
S_4&=18\cdot72\cdot98\cdot k=127008\,k .
\end{aligned}
\]

Then  

\[
f(x)=\frac{N(x)}{x}=x^{3}-S_1x^{2}+S_2x-S_3+\frac{S_4}{x},
\]

and  

\[
f'(x)=\frac{3x^{4}-2S_1x^{3}+S_2x^{2}-S_4}{x^{2}} .
\]

Hence the critical points are the (positive) roots of  

\[
P(x)=3x^{4}-2S_1x^{3}+S_2x^{2}-S_4=0\tag{1}
\]

(the denominator $x^{2}>0$ for $x>0$).

Because $f(x)\to +\infty$ as $x\to0^{+}$ and as $x\to\infty$, the graph must
first decrease, then increase, then decrease, and finally increase again.
Thus (1) has three positive roots:

* $x_1$ – a local **minimum** in the first negative interval,
* $x_2$ – a local **maximum** in the positive interval,
* $x_3$ – a second local **minimum** in the last negative interval.

The global minimum is achieved at the lower of the two minima.
For the minimum to be attained **exactly at two points** we need  

\[
f(x_1)=f(x_3)\qquad(\text{the two minima have the same value}).
\tag{2}
\]

--------------------------------------------------------------------
### 1.  Translating the condition

At a critical point $x$ we have $f'(x)=0$, i.e. $P(x)=0$.
From $f(x)=\dfrac{N(x)}{x}$ and $P(x)=0$ it follows that  

\[
f(x)=\frac{N(x)}{x}=N'(x)\qquad\text{for any critical point}.
\tag{3}
\]

Thus (2) is equivalent to  

\[
N'(x_1)=N'(x_3).\tag{4}
\]

Writing $x_1+ x_3=s$ and $x_1x_3=p$, the two equations $P(x_1)=P(x_3)=0$
give after elimination  

\[
\begin{cases}
4(s^{2}-p)-3S_1s+2S_2=0,\\[2pt]
3(s^{3}-2ps)-2S_1(s^{2}-p)+S_2s=0.
\end{cases}\tag{5}
\]

Equation (5) yields  

\[
(2s-S_1)\Bigl(3s(s-S_1)+2S_2\Bigr)=0 .
\]

Hence either  

\[
\boxed{s=\dfrac{S_1}{2}} \qquad\text{or}\qquad
3s^{2}-3S_1s+2S_2=0. \tag{6}
\]

--------------------------------------------------------------------
### 2.  The case $s=S_1/2$

From the first possibility in (6) we obtain  

\[
p=\frac{4S_2-S_1^{2}}{8}.
\]

Using $x_1x_2=p$ and $x_1+x_2=S_1/2$ the two minima are  

\[
x_{1,3}= \frac{S_1\pm 2\sqrt{\,\Delta\,}}{4},
\qquad 
\Delta=\frac34k^{2}-94k+6276 .
\]

Now we must also satisfy $P(x_1)=0$, which after substitution gives

\[
(S_1^{2}-4S_2)^{2}=64S_4 .
\]

With the explicit expressions for $S_1,S_2,S_4$ this becomes

\[
(k^{2}-376k-5120)^{2}=8128512\,k,
\]

or  

\[
k^{4}-752k^{3}+131136k^{2}-4278272k+26214400=0.
\tag{7}
\]

Factoring (7) gives  

\[
(k-8)(k-32)(k-200)(k-512)=0 .
\]

For a root to be admissible we need the two minima to lie in the
two negative intervals of $f$.  
The left minimum equals  

\[
x_1=\frac{k+188-2\sqrt{\Delta}}{4},
\]

which is **positive** only when  

\[
k^{2}-376k-5120<0\qquad\Longleftrightarrow\qquad k<389\ldots .
\]

Thus $k=512$ is rejected, while $k=8,\,32,\,200$ are valid.

--------------------------------------------------------------------
### 3.  The second possibility in (6)

The alternative equation $3s^{2}-3S_1s+2S_2=0$ together with the condition
$P(x_1)=0$ leads, after a completely analogous elimination, to the same
quartic (7).  Its admissible positive solutions are again $8,32,200$,
so no new $k$ appear.

--------------------------------------------------------------------
### 4.  Conclusion

The function $f$ has its global minimum attained at exactly two positive
points precisely for  

\[
k\in\{8,\;32,\;200\}.
\]

Therefore the required sum is  

\[
8+32+200 = \boxed{240}.
\]

Let the four unit squares be  

\[
\begin{array}{cc}
A& B\\
C& D
\end{array}
\]

and denote by  

* \(a=h_{0,1}\)   – the middle horizontal segment between \(A\) and \(C\);  
* \(b=h_{1,1}\)   – the middle horizontal segment between \(B\) and \(D\);  
* \(c=v_{1,0}\)   – the middle vertical segment between \(A\) and \(B\);  
* \(d=v_{1,1}\)   – the middle vertical segment between \(C\) and \(D\).

The eight outer unit‑segments are written as  

\[
\begin{aligned}
&e_1=h_{0,0},\qquad e_2=v_{0,0},\\
&e_3=v_{2,0},\qquad e_4=h_{1,0},\\
&e_5=h_{0,2},\qquad e_6=v_{0,1},\\
&e_7=v_{2,1},\qquad e_8=h_{1,2}.
\end{aligned}
\]

All variables take the value 1 for a red side and 0 for a blue side.

--------------------------------------------------------------------
### 1.  Equations for the squares  

Each unit square must have exactly two red sides, so we obtain

\[
\begin{aligned}
A:&\;e_1+e_2+a+c=2, \\
B:&\;e_3+e_4+b+c=2, \\
C:&\;e_5+e_6+a+d=2, \\
D:&\;e_7+e_8+b+d=2 .\qquad (1)
\end{aligned}
\]

--------------------------------------------------------------------
### 2.  Fix the interior edges  

The four interior edges \(a,b,c,d\) are independent; there are \(2^4=16\) possible
choices.
For a fixed quadruple \((a,b,c,d)\) the right‑hand side of each equation in (1)
becomes  

\[
s_A=2-(a+c),\; s_B=2-(b+c),\; s_C=2-(a+d),\; s_D=2-(b+d).
\]

The numbers \(s_A,s_B,s_C,s_D\) are the required sums of the two
outer edges belonging to each square.

*If \(s_i=0\) or \(s_i=2\):* the two outer edges are forced to be
\((0,0)\) or \((1,1)\) – exactly **one** possibility.

*If \(s_i=1\):* the outer edges must be \((0,1)\) or \((1,0)\) – **two**
possibilities.

Hence, for a given \((a,b,c,d)\) the number of admissible colourings of the
outer edges equals  

\[
2^{\,N},\qquad N=\#\{i\in\{A,B,C,D\}:s_i=1\}.
\]

But \(s_i=1\) occurs precisely when the two interior edges of that square sum to
\(1\); i.e.

\[
N=[a\neq c]+[b\neq c]+[a\neq d]+[b\neq d],\tag{2}
\]

where \([P]=1\) if the statement \(P\) is true and \(0\) otherwise.

--------------------------------------------------------------------
### 3.  Summing over the interior edges  

Write the sum we need as  

\[
\sum_{a,b,c,d\in\{0,1\}} 2^{[a\neq c]+[b\neq c]+[a\neq d]+[b\neq d]}.
\]

For fixed \(a,b\) the contributions of \(c\) and \(d\) factor:

\[
\bigl(\sum_{c\in\{0,1\}}2^{[a\neq c]+[b\neq c]}\bigr)^2 .
\]

Now compute the inner sum.

*If \(a=b\) (either \((0,0)\) or \((1,1)\)):  

\[
c=0\!:\;[a\neq c]+[b\neq c]=0,\qquad
c=1\!:\;[a\neq c]+[b\neq c]=2,
\]
so the sum equals \(2^{0}+2^{2}=1+4=5\).

*If \(a\neq b\) (either \((0,1)\) or \((1,0)\)):  

\[
c=0\!:\;[a\neq c]+[b\neq c]=1,\qquad
c=1\!:\;[a\neq c]+[b\neq c]=1,
\]
so the sum equals \(2^{1}+2^{1}=2+2=4\).

Thus

\[
\begin{aligned}
\text{total}&=
2\bigl(5^2\bigr)+2\bigl(4^2\bigr)        \\
&=2\cdot25+2\cdot16\\
&=50+32\\
&=82 .
\end{aligned}
\]

--------------------------------------------------------------------
### 4.  Answer  

There are  

\[
\boxed{82}
\]

different colourings of the twelve unit segments such that each of the four
unit squares has exactly two red sides and two blue sides.

Let  

* \(O\) be the centre of \(\omega _2\) (radius \(15\)),
* \(A\) be the centre of \(\omega _1\) (radius \(6\)),
* \(B\) the internal point of tangency.

Place \(O\) at the origin and take the diameter \(BC\) on the \(x\)-axis.

\[
O=(0,0),\qquad B=(15,0)
\]

Because \(\omega_1\) is tangent internally at \(B\),

\[
OA=15-6=9\qquad\Longrightarrow\qquad A=(9,0).
\]

Since \(BC\) is a diameter of \(\omega _2\),

\[
C=(-15,0).
\]

\(AD\perp BC\) forces \(AD\) to be vertical through \(A\); intersecting this line with \(\omega _2\),

\[
x=9,\qquad x^{2}+y^{2}=225\Longrightarrow y=\pm 12 .
\]

Because the problem states that \(D\) is nearer to the side \(FG\) than to \(EH\), we take the point above the \(x\)-axis:

\[
D=(9,12).
\]

---------------------------------------------------------------------

### 1.  The rectangle \(EFGH\)

\(EF\perp BC\); with \(BC\) horizontal this makes \(EF\) vertical, so the rectangle is axis‑aligned.
Let its half–width be \(d>0\) and its half–height be \(h>0\).  
Since the rectangle is inscribed in \(\omega _1\) (centre \(A\)), its centre must coincide with \(A\).  
Thus the vertices are  

\[
\begin{aligned}
E&=(9+d,\,-h),  &F&=(9+d,\,h),\\
G&=(9-d,\,h),   &H&=(9-d,\,-h).
\end{aligned}
\]

Each vertex lies on \(\omega _1\):  

\[
(x-9)^2+y^2=6^2\quad\Longrightarrow\quad d^{\,2}+h^{\,2}=36. \tag{1}
\]

---------------------------------------------------------------------

### 2.  Equality of the two triangle areas

*Triangle \(DGF\)*  
Its base \(GF\) is horizontal with length \(2d\); the height is the vertical distance from \(D\) to the line \(y=h\):
\[
[\,DGF\,]=\frac12\cdot (2d)\bigl(12-h\bigr)=d(12-h).
\]

*Triangle \(CHG\)*  
Its base \(GH\) is vertical with length \(2h\); the height is the horizontal distance from \(C\) to the line \(x=9-d\):
\[
[\,CHG\,]=\frac12\cdot (2h)\bigl( (9-d)-(-15)\bigr)=h(24-d).
\]

The condition \([DGF]=[CHG]\) gives  

\[
d(12-h)=h(24-d)\;\Longrightarrow\;12d=24h\;\Longrightarrow\;d=2h. \tag{2}
\]

---------------------------------------------------------------------

### 3.  Solving for \(d\) and \(h\)

Substituting \(d=2h\) into (1):

\[
(2h)^2+h^{2}=36\quad\Longrightarrow\quad5h^{2}=36
\quad\Longrightarrow\quad h^{2}=\frac{36}{5},\qquad 
h=\frac{6}{\sqrt5},\qquad d=2h=\frac{12}{\sqrt5}.
\]

---------------------------------------------------------------------

### 4.  Area of the rectangle

\[
\text{Area}= (2d)(2h)=4dh=4(2h)h=8h^{2}=8\cdot\frac{36}{5}
=\frac{288}{5}.
\]

Thus the area equals \(\dfrac{m}{n}\) with \(m=288,\;n=5\) (coprime).  

\[
m+n=288+5=293.
\]

\[
\boxed{293}
\]

Factor  

\[
2025=3^{4}\cdot5^{2}.
\]

A divisor of \(2025\) has the form \(3^{a}5^{b}\) with \(0\le a\le4,\ 0\le b\le2\).  
Hence  

\[
|A|=(4+1)(2+1)=15,
\qquad\text{total subsets}=2^{15}=32768 .
\]

---

### When does a subset have lcm \(=2025\)?

The lcm of a set of divisors is \(3^{\max a}\cdot5^{\max b}\).  
Thus the lcm equals \(2025=3^{4}\cdot5^{2}\) **iff**

* at least one element has exponent \(a=4\) (a divisor containing \(3^{4}\));
* at least one element has exponent \(b=2\) (a divisor containing \(5^{2}\)).

---

### Counting the desired subsets  

Divisors with \(a=4\): \(3^{4}5^{b}\ (b=0,1,2)\) – 3 of them.  
Divisors with \(b=2\): \(3^{a}5^{2}\ (a=0,\dots ,4)\) – 5 of them.  

Let  

* \(X\) = subsets containing **no** divisor with \(a=4\);  
* \(Y\) = subsets containing **no** divisor with \(b=2\).

\[
|X|=2^{12}\quad (\text{choose any of the other }12\text{ divisors}),\qquad
|Y|=2^{10}\quad (\text{choose any of the other }10\text{ divisors}).
\]

Divisors lacking both \(a=4\) and \(b=2\) have \(a\le3,\ b\le1\); there are \(4\cdot2=8\) of them, so

\[
|X\cap Y|=2^{8}.
\]

Using inclusion–exclusion, the number of subsets that contain **both** an \(a=4\) divisor and a \(b=2\) divisor is  

\[
N = 2^{15} - 2^{12} - 2^{10} + 2^{8}
    = 32768 - 4096 - 1024 + 256
    = 27904 .
\]

These subsets are precisely the non‑empty subsets whose lcm equals \(2025\); the empty set is automatically excluded by the inclusion–exclusion count.

---

### Probability

\[
\text{Probability} = \frac{N}{2^{15}} = \frac{27904}{32768}
                     = \frac{2^{8}\cdot 109}{2^{15}}
                     = \frac{109}{128}.
\]

The fraction \(\frac{109}{128}\) is already in lowest terms, so  

\[
m=109,\qquad n=128,\qquad m+n=237.
\]

\[
\boxed{237}
\]

Let  

\[
\text{chairs}=1,2,\dots ,16,\qquad 8\ \text{people sit in 8 of them.}
\]

The condition *no person sits next to two other people* means that no three
occupied chairs are consecutive.  
Hence we must count binary strings of length \(16\) with exactly eight 1’s
and no substring \(111\).

---

### 1.  Blocks of occupied chairs  

Occupied chairs appear in blocks of length \(1\) or \(2\) (a block of length
\(3\) would give a person with two neighbours).  

Let  

* \(t\) = number of blocks of length \(2\);
* \(s\) = number of blocks of length \(1\).

Then  

\[
s+2t=8,\qquad r=s+t \text{ (total number of blocks)} .
\]

Hence  

\[
t=0,1,2,3,4,\qquad r=8-t\;(=8,7,6,5,4).
\]

For a fixed \(t\) (hence fixed \(r\)) we must choose which of the \(r\) blocks
are the double ones:

\[
\text{ways to choose the double blocks}= \binom{r}{t}= \binom{r}{8-r}.
\]

---

### 2.  Placing the blocks among the empty chairs  

There are \(16-8=8\) empty chairs.  
Between the \(r\) blocks there must be at least one empty chair, using
\(r-1\) chairs.  

The remaining empty chairs are  

\[
8-(r-1)=9-r .
\]

These \(9-r\) chairs can be placed arbitrarily in the \(r+1\) gaps
(before the first block, between successive blocks, after the last block).  
By stars‑and‑bars the number of ways is  

\[
\binom{(9-r)+(r+1)-1}{(r+1)-1}= \binom{9}{r}.
\]

---

### 3.  Total number of admissible subsets  

\[
\begin{aligned}
N&=\sum_{t=0}^{4} \binom{r}{t}\binom{9}{r},
\qquad r=8-t \\
  &=\sum_{r=4}^{8}\binom{r}{\,8-r\,}\binom{9}{r}.
\end{aligned}
\]

Compute each term:

\[
\begin{array}{c|c|c|c}
r & \binom{r}{8-r} & \binom{9}{r} & \text{product}\\ \hline
8 & 1 & 9   & 9\\
7 & 7 & 36  & 252\\
6 & 15& 84  & 1260\\
5 & 10& 126 & 1260\\
4 & 1 & 126 & 126
\end{array}
\]

Adding them,

\[
N = 9+252+1260+1260+126 = 2907.
\]

---

### 4.  Remainder modulo \(1000\)

\[
2907 \equiv 907 \pmod{1000}.
\]

\[
\boxed{907}
\]

To solve  

\[
f(x)=\sin \bigl(7\pi \,\sin(5x)\bigr)=0 ,\qquad 0<x<2\pi 
\]

let  

\[
g(x)=7\pi\sin(5x).
\]

Then  

\[
f(x)=\sin(g(x))=0\iff g(x)=k\pi,\;k\in\mathbb Z .
\]

Hence  

\[
7\pi\sin(5x)=k\pi\Longrightarrow \sin(5x)=\frac{k}{7}.
\]

Because \(|\sin(5x)|\le 1\), the integer \(k\) can be  
\(-7\le k\le 7\).

---

### 1. Counting the zeros (\(n\))

Put \(y=5x\).  
When \(x\) runs from \(0\) to \(2\pi\), \(y\) runs from \(0\) to \(10\pi\).

For a fixed value \(c=\frac{k}{7}\) we count the solutions of \(\sin y=c\) in the interval \((0,10\pi)\).

* **\(c=0\) \;(i.e. \(k=0\)):**  
  \(\sin y=0\) ⇒ \(y=m\pi\).  
  In \((0,10\pi)\) we have \(m=1,2,\dots ,9\) → **9 solutions**.

* **\(c=\pm1\) \;(i.e. \(k=\pm7\)):**  
  \(\sin y=1\) gives \(y=\frac{\pi}{2}+2\pi n\)  
  \(\sin y=-1\) gives \(y=\frac{3\pi}{2}+2\pi n\).  
  For \(n=0,1,2,3,4\) each set yields 5 points → **5+5 = 10 solutions**.

* **\(c\) with \(|c|<1\) and \(c\neq0\) \;(i.e. \(k=\pm1,\dots ,\pm6\)):**  
  For any such \(c\) there are two solutions per period of sine.  
  The interval \((0,10\pi)\) contains five full periods, so we obtain **\(5\times2=10\) solutions** for each such \(k\).

Now sum:

\[
\begin{aligned}
n & =\underbrace{9}_{k=0}
   +\underbrace{12\times10}_{k=\pm1,\dots ,\pm6}
   +\underbrace{10}_{k=\pm7}\\[2mm]
  & = 9+120+10 = 139 .
\end{aligned}
\]

Thus **\(n=139\)**.

---

### 2. When is the graph tangent to the \(x\)-axis? (\(t\))

The graph is tangent at a zero when the derivative also vanishes:

\[
f'(x)=\cos\bigl(7\pi\sin(5x)\bigr)\cdot 35\pi\cos(5x)=0 .
\]

At the zeros we already have \(\sin(7\pi\sin(5x))=0\), i.e. \(7\pi\sin(5x)=k\pi\).  
Hence \(\cos\bigl(7\pi\sin(5x)\bigr)=\cos(k\pi)=(-1)^k\neq0\).

Consequently \(f'(x)=0\) can only occur through \(\cos(5x)=0\).

\[
\cos(5x)=0 \Longrightarrow 5x=\frac{\pi}{2}+m\pi
            \Longrightarrow x=\frac{(2m+1)\pi}{10},\qquad m\in\mathbb Z .
\]

Within \(0<x<2\pi\) we have \(m=0,1,\dots ,9\), giving **10 distinct \(x\)** values.

At each of these points \(\sin(5x)=\pm1\), so \(\sin(7\pi\sin(5x))=\sin(\pm7\pi)=0\); thus they are indeed zeros, and because the derivative also vanishes the graph is tangent there.

Hence **\(t=10\)**.

---

### 3. Required sum

\[
n+t = 139 + 10 = 149 .
\]

\[
\boxed{149}
\]

**Answer:** \(\boxed{637}\)

Let  

\[
x_i=A_1A_i\qquad (i=2,3,\ldots ,11),\qquad 
\theta=\angle A_iA_1A_{i+1}\;(2\le i\le10).
\]

---

### 1.  Using the area condition  

\[
\text{Area}\bigl(A_iA_1A_{i+1}\bigr)=\frac12 x_i x_{i+1}\sin\theta =1 .
\]

Because \(\cos\theta=\dfrac{12}{13}\),

\[
\sin\theta=\sqrt{1-\Bigl(\frac{12}{13}\Bigr)^2}= \frac{5}{13}.
\]

Thus  

\[
\frac12 x_i x_{i+1}\cdot\frac{5}{13}=1
\Longrightarrow x_i x_{i+1}=\frac{26}{5}\qquad (2\le i\le10).
\tag{1}
\]

Denote \(c=\dfrac{26}{5}\).  From (1) we have  

\[
x_{i+1}= \frac{c}{x_i}.
\]

Consequently  

\[
x_{i+2}=x_i\qquad\Longrightarrow\qquad
\begin{cases}
x_{2}=x_{4}=x_{6}=x_{8}=x_{10}=a,\\[2pt]
x_{3}=x_{5}=x_{7}=x_{9}=x_{11}=b,
\end{cases}
\]

with  

\[
ab=c=\frac{26}{5}. \tag{2}
\]

So the distances from \(A_1\) alternate between two values \(a\) and \(b\).

---

### 2.  Length of each side \(A_iA_{i+1}\;(2\le i\le10)\)

In \(\triangle A_iA_1A_{i+1}\) the side opposite \(\theta\) is \(A_iA_{i+1}\).
By the law of cosines

\[
A_iA_{i+1}^2 = x_i^2+x_{i+1}^2-2x_ix_{i+1}\cos\theta .
\]

Because \(x_i\) and \(x_{i+1}\) are always one \(a\) and one \(b\),

\[
A_iA_{i+1}^2 = a^2+b^2-2ab\cdot\frac{12}{13}.
\]

Hence all nine sides \(A_iA_{i+1}\;(i=2,\dots ,10)\) have the same length; denote

\[
d^2=a^2+b^2-\frac{24}{13}ab.
\tag{3}
\]

Write \(s=a+b\).  Using \(ab=c\),

\[
a^2+b^2=s^2-2ab=s^2-2c .
\]

Substituting into (3),

\[
d^2=s^2-2c-\frac{24}{13}c
    =s^2-c\Bigl(2+\frac{24}{13}\Bigr)
    =s^2-\frac{50}{13}c.
\]

Since \(c=\dfrac{26}{5}\),

\[
d^2=s^2-\frac{50}{13}\cdot\frac{26}{5}=s^2-20,
\qquad\text{so}\qquad d=\sqrt{s^{2}-20}.
\tag{4}
\]

---

### 3.  Use the perimeter

The polygon has \(11\) sides: the two incident to \(A_1\) of lengths \(a\) and \(b\), and the nine equal sides of length \(d\).  
Given perimeter \(20\),

\[
a+b+9d=s+9\sqrt{s^{2}-20}=20.
\tag{5}
\]

Solve (5) for \(s\).

\[
9\sqrt{s^{2}-20}=20-s\;(s\le20).
\]

Square:

\[
81(s^{2}-20)=(20-s)^{2}
\Longrightarrow 80s^{2}+40s-2020=0
\Longrightarrow 4s^{2}+2s-101=0.
\]

Hence

\[
s=\frac{-2\pm\sqrt{4+1616}}{8}
   =\frac{-2\pm18\sqrt5}{8}
   =\frac{-1\pm9\sqrt5}{4}.
\]

Only the positive root is admissible:

\[
\boxed{s=\frac{9\sqrt5-1}{4}}.
\]

Thus  

\[
A_1A_2+A_1A_{11}=a+b=s=\frac{9\sqrt5-1}{4}
      =\frac{m\sqrt n-p}{q},
\]

with \(m=9,\;n=5,\;p=1,\;q=4\).

---

### 4.  Required sum

\[
m+n+p+q=9+5+1+4=19.
\]

\[
\boxed{19}
\]

**Greedy representation**

For a value \(N\) let  

\[
q=\Big\lfloor\frac{N}{25}\Big\rfloor ,\qquad 
r=N-25q\;(0\le r\le 24).
\]

After taking \(q\) quarters the greedy algorithm uses  

\[
a=\Big\lfloor\frac{r}{10}\Big\rfloor ,\qquad 
b=r-10a\;(0\le b\le 9)
\]

dimes and ones.  
Thus the greedy collection contains  

\[
G(N)=q+a+b
\]

coins.

--------------------------------------------------------------------

**When can a better collection exist?**

Let a different collection use \(x\) quarters and \(y\) dimes
(\(z\) pennies fill the rest).  
Its total number of coins is  

\[
C=x+y+z = x+y+(N-25x-10y)=N-(24x+9y).
\]

Hence a representation is better than the greedy one iff it makes the
“saving”

\[
S=24x+9y
\]

larger than the greedy saving \(S_G = 24q+9a\).

Suppose we start from the greedy solution and **remove** \(d\ge1\) quarters
(\(d\le q\)).  
The value we free is \(25d\).
With this extra value we can add as many dimes as possible:

\[
y' = a+\Big\lfloor\frac{r+25d}{10}\Big\rfloor
      = a+\Big\lfloor 2.5d+\frac{b}{10}\Big\rfloor .
\]

The new saving is  

\[
S'=24(q-d)+9y'
   =24q+9a +\bigl[-24d+9\big\lfloor2.5d+\tfrac{b}{10}\big\rfloor\bigr].
\]

Thus the new collection beats the greedy one precisely when  

\[
-24d+9\Big\lfloor2.5d+\frac{b}{10}\Big\rfloor >0
\qquad\Longleftrightarrow\qquad
\Big\lfloor2.5d+\frac{b}{10}\Big\rfloor \ge
\Big\lfloor\frac{8}{3}d\Big\rfloor+1 .\tag{1}
\]

--------------------------------------------------------------------

**Only one quarter can ever help**

For even \(d\) we have \(\big\lfloor2.5d\big\rfloor =2.5d\) and
\(\big\lfloor\frac{8}{3}d\big\rfloor =2.5d+\big\lfloor\frac{d}{6}\big\rfloor\).
Hence the left–hand side of (1) never exceeds the right–hand side,
so even \(d\) never improves the greedy solution.

For odd \(d=2k+1\) we obtain  

\[
\Big\lfloor2.5d+\frac{b}{10}\Big\rfloor
   =5k+2+{\bf 1}_{\{b\ge5\}},\qquad
\Big\lfloor\frac{8}{3}d\Big\rfloor
   =5k+2+\Big\lfloor\frac{k+2}{3}\Big\rfloor .
\]

Consequently  

\[
\Big\lfloor2.5d+\frac{b}{10}\Big\rfloor-
\Big\lfloor\frac{8}{3}d\Big\rfloor
= {\bf 1}_{\{b\ge5\}}-\Big\lfloor\frac{k+2}{3}\Big\rfloor .
\]

The right‑hand side is \(\ge 1\) only when \(k=0\) (that is, \(d=1\))
and \(b\ge5\).
Thus the **only** way to improve upon the greedy solution is to
replace **exactly one** quarter and have \(b\ge5\).

--------------------------------------------------------------------

**What does \(b\ge5\) mean?**

Recall \(b\) is the number of pennies left after taking all possible
quarters and then all possible dimes.  Hence

\[
b = N-25q-10a\in\{0,1,\dots ,9\}.
\]

The condition \(b\ge5\) is equivalent to

\[
N-25q\equiv 5,6,7,8,9\pmod{10},
\]

or, more simply,

\[
N\equiv 5,6,7,8,9\pmod{25}.
\]

If \(q=0\) (i.e. \(N<25\)) there is no quarter to replace, so the greedy
algorithm is always optimal for \(1\le N\le24\).

Hence **the greedy algorithm fails exactly for the numbers**

\[
N\ge 25 \quad\text{with}\quad N\equiv 5,6,7,8,9\pmod{25}.
\]

--------------------------------------------------------------------

**Counting the successes**

All integers \(1\le N\le1000\): 1000 of them.

Numbers that cause failure:

* The residues \(5,6,7,8,9\) each occur
  \(\bigl\lfloor\frac{1000-r}{25}\bigr\rfloor+1 = 40\) times for
  \(r=5,6,7,8,9\).
  Hence \(5\times40=200\) such numbers in \([1,1000]\).

* The five smallest ones, \(5,6,7,8,9\), are \(<25\); they do **not**
  cause failure because no quarter can be replaced.

Thus the genuine failures are \(200-5 = 195\) numbers.

Consequently the number of values for which the greedy algorithm
succeeds is  

\[
1000-195 = 805.
\]

\[
\boxed{805}
\]

Let the vertices of the regular \(24\)-gon be labelled  

\[
0,1,2,\dots ,23\qquad (\text{mod }24).
\]

If a segment connects vertex \(i\) to vertex \(j\) then its length is determined only by the
step  

\[
k\equiv j-i\pmod{24},\qquad 1\le k\le 12 .
\]

All \(12\) segments must have the same length, therefore they all have the same step \(k\) (the
step \(24-k\) gives the same length, so it is enough to consider \(k=1,2,\dots ,12\)).

--------------------------------------------------------------------
### 1.  The graph for a fixed step \(k\)

For a fixed \(k\) join each vertex \(i\) with the two vertices
\(i\pm k\).  The resulting graph \(G_k\) is a union of cycles:

* Let \(d=\gcd(k,24)\).
* Each cycle has length \(\displaystyle L=\frac{24}{d}\) and there are \(d\) such
  cycles.

Indeed the permutation \(i\mapsto i+k\) on \(\{0,\dots ,23\}\) consists of
\(d\) cycles of length \(L\).

--------------------------------------------------------------------
### 2.  Perfect matchings of \(G_k\)

A required drawing is a perfect matching of \(G_k\).

* If a cycle has even length, it has exactly **2** perfect matchings
  (alternating edges).
* If a cycle has odd length, a perfect matching does **not** exist.
* The only exceptional case is \(k=12\); then each vertex has only one
  neighbour (its opposite) and the whole graph is already a perfect
  matching.

Thus, for \(k\neq 12\),

\[
\text{# matchings for }G_k=
\begin{cases}
2^{\,d} & \text{if }L\text{ is even},\\[2mm]
0       & \text{if }L\text{ is odd}.
\end{cases}
\]

Since \(L\) is odd exactly when \(L=3\), i.e. when \(d=8\) (the only such
\(k\) is \(k=8\)), we have

\[
N(k)=
\begin{cases}
2^{\gcd(k,24)} & \gcd(k,24)\neq 8,\;k\neq 12,\\[2mm]
0 & k=8,\\[2mm]
1 & k=12 .
\end{cases}
\]

--------------------------------------------------------------------
### 3.  Evaluate for \(k=1,\dots ,12\)

\[
\begin{array}{c|c|c|c}
k & \gcd(k,24) & \text{cycles} & N(k)\\ \hline
1 & 1 & 1\ \text{cycle of length }24 & 2\\
2 & 2 & 2\ \text{cycles of length }12 & 4\\
3 & 3 & 3\ \text{cycles of length }8  & 8\\
4 & 4 & 4\ \text{cycles of length }6  & 16\\
5 & 1 & 1\ \text{cycle of length }24 & 2\\
6 & 6 & 6\ \text{cycles of length }4  & 64\\
7 & 1 & 1\ \text{cycle of length }24 & 2\\
8 & 8 & 8\ \text{cycles of length }3  & 0\\
9 & 3 & 3\ \text{cycles of length }8  & 8\\
10& 2 & 2\ \text{cycles of length }12 & 4\\
11& 1 & 1\ \text{cycle of length }24 & 2\\
12&12 & 12\ \text{edges (pairs)}       & 1
\end{array}
\]

--------------------------------------------------------------------
### 4.  Sum over all possible lengths

\[
\begin{aligned}
\sum_{k=1}^{12} N(k)
&=2+4+8+16+2+64+2+0+8+4+2+1\\
&=113 .
\end{aligned}
\]

--------------------------------------------------------------------
\[
\boxed{113}
\]

For a base \(b>9\),

\[
17_{b}=1\cdot b+7=b+7 ,\qquad   
97_{b}=9\cdot b+7=9b+7 .
\]

We need  

\[
b+7 \mid 9b+7 .
\]

Let \(d=b+7\;(>0)\). Then \(b=d-7\) and

\[
9b+7 = 9(d-7)+7 = 9d-56 .
\]

Hence  

\[
d \mid (9d-56) \Longrightarrow d\mid 56 .
\]

So \(d\) must be a positive divisor of \(56\).  
Since \(b>9\), we have \(d=b+7>9+7=16\).  
The divisors of \(56\) larger than \(16\) are

\[
d=28,\;56 .
\]

Corresponding bases are  

\[
b=d-7:
\begin{cases}
d=28 \Rightarrow b=21,\\[2mm]
d=56 \Rightarrow b=49.
\end{cases}
\]

Both satisfy \(b>9\) and indeed  

\[
\frac{9b+7}{b+7}=7\;(b=21),\qquad 
\frac{9b+7}{b+7}=8\;(b=49).
\]

Thus the integer bases are \(b=21\) and \(b=49\). Their sum is  

\[
21+49=70.
\]

\[
\boxed{70}
\]

**Solution Overview**

The circle through the midpoints \(D,E,F\) of the sides of \(\triangle ABC\) is the **nine‑point circle** of \(\triangle ABC\).
Hence  

* \(D,E,F\) are the vertices of the medial triangle,
* \(\angle D= \angle A=84^{\circ},\ \angle E= \angle B=60^{\circ},\ \angle F=\angle C=36^{\circ}\).

Since the nine‑point circle is the circumcircle of \(\triangle DEF\),

\[
\widehat{DE}=2\angle F=2\cdot 36^{\circ}=72^{\circ}.
\tag{1}
\]

We must also determine the arcs \(\widehat{FG}\) and \(\widehat{HJ}\).

---------------------------------------------------------------------

### 1.  The arc \(\widehat{FG}\)

\(G\) is the second intersection of the median \(BD\) with the nine‑point circle.
At point \(D\) the chord \(DG\) lies on the line \(BD\), while the chord \(DF\) is parallel to side \(AC\) (mid‑segment theorem).

Therefore  

\[
\angle FDG=\angle(\text{line } DF,\text{line } DG)
          =\angle(\,AC ,\,BD\,).
\]

The angle between side \(AC\) and the median from \(B\) can be read directly from the triangle:
\( \angle(\,AC ,\,BD\,)=\angle C=36^{\circ}\).

Consequently  

\[
\widehat{FG}=2\angle FDG=2\cdot36^{\circ}=72^{\circ}.
\tag{2}
\]

---------------------------------------------------------------------

### 2.  The arc \(\widehat{HJ}\)

\(H\) and \(J\) are the second intersections of the sides \(AC\) and \(AB\) with the nine‑point circle.
Thus the two secants through the exterior point \(A\) intersect the circle at the pairs \(\{F,J\}\) and \(\{E,H\}\).

For an exterior point the angle formed by two secants equals half the **difference** of the intercepted arcs:

\[
\angle BAC=\frac12\bigl(\widehat{JH}-\widehat{FE}\bigr).
\tag{3}
\]

We already know the arc \(\widehat{FE}\) of the nine‑point circle:
\[
\widehat{FE}=2\angle D =2\cdot 84^{\circ}=168^{\circ}.
\tag{4}
\]

Plugging \(\angle BAC=84^{\circ}\) and (4) into (3) gives  

\[
84^{\circ}=\frac12\bigl(\widehat{JH}-168^{\circ}\bigr)
\Longrightarrow \widehat{JH}=336^{\circ}.
\]

The **minor** arc \(HJ\) is the complement of \(336^{\circ}\) on the circle:

\[
\widehat{HJ}=360^{\circ}-336^{\circ}=24^{\circ}.
\tag{5}
\]

---------------------------------------------------------------------

### 3.  Computing the required sum

From (1), (2) and (5)

\[
\begin{aligned}
\widehat{DE}+2\widehat{HJ}+3\widehat{FG}
&=72^{\circ}+2\cdot24^{\circ}+3\cdot72^{\circ}  \\
&=72^{\circ}+48^{\circ}+216^{\circ}\\
&=336^{\circ}.
\end{aligned}
\]

---------------------------------------------------------------------

\[
\boxed{336}
\]

The nine players are distinct, so for a given triple \((c,v,s)\) (numbers of chocolate, vanilla, and strawberry cones) the number of assignments is  

\[
\frac{9!}{c!\,v!\,s!}.
\]

We must count all triples of positive integers satisfying  

\[
c+v+s = 9,\qquad c>v>s\ge 1 .
\]

---

### Find the possible triples \((c,v,s)\)

Because the three numbers are distinct and sum to 9, we list the partitions of 9 into three distinct positive parts:

\[
\begin{aligned}
9 &= 6+2+1,\\
9 &= 5+3+1,\\
9 &= 4+3+2 .
\end{aligned}
\]

These are the only triples with strict descending order, so  

\[
(c,v,s) \in \{(6,2,1),\;(5,3,1),\;(4,3,2)\}.
\]

---

### Count assignments for each triple

\[
\begin{aligned}
(6,2,1):\quad &\frac{9!}{6!\,2!\,1!}= \frac{362880}{720\cdot2}=252,\\[4pt]
(5,3,1):\quad &\frac{9!}{5!\,3!\,1!}= \frac{362880}{120\cdot6}=504,\\[4pt]
(4,3,2):\quad &\frac{9!}{4!\,3!\,2!}= \frac{362880}{24\cdot6\cdot2}=1260.
\end{aligned}
\]

---

### Total number of assignments

\[
N = 252 + 504 + 1260 = 2016.
\]

---

### Remainder modulo \(1000\)

\[
2016 \equiv 16 \pmod{1000}.
\]

\[
\boxed{16}
\]

**Step 1.  What determines the number of regions?**  

If we draw line‑segments in a disc and no three of them meet in a common interior point, then each new segment splits one existing region for every interior intersection it makes, plus one more region for the segment itself.  
Hence  

\[
\#\text{regions}=1+(\#\text{segments})+(\#\text{interior intersection points}).
\]

We have  

* 2 diameters,
* 25 random chords,
* total segments \(=27\).

So we need the **expected number of interior intersections**.

---------------------------------------------------------------------

**Step 2.  Intersections that are certain**

The two diameters intersect at the centre ⇒ one interior intersection.

---------------------------------------------------------------------

**Step 3.  Intersections of a chord with the diameters**

A chord is drawn by picking two points on the circle that lie in **different quadrants**.

*If the two quadrants are adjacent* (e.g. quadrant I and II), the chord crosses **exactly one** diameter.  
*If the two quadrants are opposite* (e.g. quadrant I and III), the chord crosses **both** diameters.

The unordered pair of distinct quadrants is uniformly chosen among the \(\binom{4}{2}=6\) possibilities:

* 4 adjacent pairs → probability \(4/6=2/3\);
* 2 opposite pairs → probability \(2/6=1/3\).

Hence for one random chord

\[
E[\hbox{diameter‑intersections}]
  =\frac23\cdot1+\frac13\cdot2=\frac43 .
\]

For the 25 chords  

\[
E[I_{\text{chord–diameter}}]=25\cdot\frac43=\frac{100}{3}.
\]

---------------------------------------------------------------------

**Step 4.  Intersections between two random chords**

Let the two chords be \(AB\) and \(CD\).  
Write \(L\) for the clockwise length of the arc from \(A\) to \(B\) (so \(0\le L\le2\pi\)).  
Let \(L_i^{(1)}\) be the length of that arc inside quadrant \(i\) (\(i=1,\dots ,4\)), and
\(L_i^{(2)}=\frac{\pi}{2}-L_i^{(1)}\) the length of the complementary arc inside the same quadrant.

For a given chord \(AB\)

* the probability that a random chord \(CD\) meets \(AB\) **and** has its endpoints in different quadrants is  

\[
p_{\text{int}}(A,B)=
\frac{L(2\pi-L)-\displaystyle\sum_{i=1}^{4}L_i^{(1)}L_i^{(2)}}{2\pi^{2}} .
\tag{1}
\]

(The numerator is the area of the product set
\(\{(C,D):C\in\text{arc}_1,D\in\text{arc}_2\}\) minus the part where \(C\) and \(D\) fall in the same quadrant.)

Define  

\[
Q(A,B)=L(2\pi-L)-\sum_{i=1}^{4}L_i^{(1)}L_i^{(2)} .
\]

Then \(p_{\text{int}}(A,B)=Q(A,B)/(2\pi^{2})\).

---------------------------------------------------------------------

**Step 5.  Averaging \(Q\)**  

Put the circle’s total length as \(4d\) with a quadrant length \(d=\pi/2\).
Write the clockwise length as a multiple of \(d\): \(t=L/d\in[0,4]\).

For a fixed \(t\) and a uniformly random starting point of the arc,
the expected value of \(\sum_i (L_i^{(1)})^{2}\) (the sum of squares of the pieces of the arc) is

\[
h(t)=
\begin{cases}
t^{2}-\dfrac{t^{3}}{3}, & 0\le t\le 1,\\[4pt]
t-\dfrac13,               & 1\le t\le 4 .
\end{cases}
\]

Consequently  

\[
E\!\left[\sum_i L_i^{(1)}L_i^{(2)}\right]
      =\frac{\pi}{2}E[L]-E\!\left[\sum_i(L_i^{(1)})^{2}\right]
      =\frac{\pi^{2}}{2}-\frac{27\pi^{2}}{64}
      =\frac{5\pi^{2}}{64}.
\]

From this we obtain the unconditional expectation

\[
E[Q]=E\!\bigl[L(2\pi-L)\bigr]-E\!\Bigl[\sum_i L_i^{(1)}L_i^{(2)}\Bigr]
      =\frac{2}{3}\pi^{2}-\frac{5}{64}\pi^{2}
      =\frac{113}{192}\pi^{2}.
\tag{2}
\]

---------------------------------------------------------------------

**Step 6.  Conditioning on “different quadrants’’ for the first chord**

If the first chord’s endpoints lie in the *same* quadrant (probability \(1/4\)),
then \(Q\) simplifies to \(Q=\frac{3\pi}{2}L\).  
Because the absolute difference of two points inside a quadrant has mean \(d/3=\pi/6\),

\[
E[Q\mid\text{same quadrant}]=\frac{3\pi}{2}\cdot\frac{\pi}{6}
                              =\frac{\pi^{2}}{4}.
\]

Hence  

\[
E[Q\;{\bf 1}_{\text{different quadrants}}]
   =E[Q]-\frac14\cdot\frac{\pi^{2}}{4}
   =\frac{113}{192}\pi^{2}-\frac{\pi^{2}}{16}
   =\frac{101}{192}\pi^{2}. \tag{3}
\]

---------------------------------------------------------------------

**Step 7.  Probability that two *valid* chords intersect**

From (1) and (3),

\[
\boxed{
P(\text{intersection } \&\; AB\text{ valid}\; \&\; CD\text{ valid})
    =\frac{E[Q\,{\bf 1}_{\text{different quadrants}}]}{2\pi^{2}}
    =\frac{101}{384}.
}
\]

The event “both chords are valid’’ occurs with probability \((3/4)^{2}=9/16\).  
Therefore the conditional probability that two random valid chords intersect is

\[
P_{\text{int}}=
\frac{101/384}{9/16}= \frac{101}{216}\approx0.4676.
\tag{4}
\]

---------------------------------------------------------------------

**Step 8.  Expected number of chord–chord intersections**

\[
E[I_{\text{chord–chord}}]
   =\binom{25}{2}\;P_{\text{int}}
   =300\cdot\frac{101}{216}
   =\frac{2525}{18}.
\]

---------------------------------------------------------------------

**Step 9.  Total expected interior intersections**

\[
\begin{aligned}
E[I] &=
\underbrace{1}_{\text{diameters}}+
\underbrace{\frac{100}{3}}_{\text{chord–diameter}}+
\underbrace{\frac{2525}{18}}_{\text{chord–chord}} \\[4pt]
&= \frac{18+600+2525}{18}
   =\frac{3143}{18}.
\end{aligned}
\]

---------------------------------------------------------------------

**Step 10.  Expected number of regions**

\[
\begin{aligned}
E[\#\text{regions}]
&=1+(\#\text{segments})+E[I] \\
&=1+27+\frac{3143}{18}\\[4pt]
&=28+\frac{3143}{18}
   =\frac{504+3143}{18}
   =\boxed{\frac{3647}{18}}\; \approx 202.6 .
\end{aligned}
\]

Thus, on average Alex’s 27 line segments partition the disk into \(\displaystyle \frac{3647}{18}\) regions.

\[
12x^{2}-xy-6y^{2}=0
\]

---

### 1.  Factor the quadratic  

Look for integers \(a,b,c,d\) such that  

\[
(ax+by)(cx+dy)=12x^{2}-xy-6y^{2}.
\]

We need  

\[
ac=12,\qquad bd=-6,\qquad ad+bc=-1 .
\]

Taking \(a=4,\;c=3\) (since \(4\cdot 3=12\)) and trying \(b=-3,\;d=2\) gives  

\[
bd=(-3)(2)=-6,\qquad ad+bc=4\cdot2+3\cdot(-3)=8-9=-1.
\]

Hence  

\[
12x^{2}-xy-6y^{2}=(4x-3y)(3x+2y).
\]

So the original equation is equivalent to  

\[
(4x-3y)(3x+2y)=0.
\]

Thus any integer solution must satisfy one of the two linear equations  

\[
\boxed{4x-3y=0}\qquad\text{or}\qquad\boxed{3x+2y=0}.
\]

---

### 2.  Solve each linear equation in integers  

**(i) \(4x-3y=0\).**  

\(4x=3y\Rightarrow y= \frac{4}{3}x\).  
Since \(\gcd(4,3)=1\), write \(x=3k,\;y=4k\) for some integer \(k\).

Both coordinates must satisfy \(-100\le x,y\le 100\):

\[
-100\le 3k\le 100 \Longrightarrow -\frac{100}{3}\le k\le \frac{100}{3},
\qquad
-100\le 4k\le 100 \Longrightarrow -25\le k\le 25.
\]

The tighter bound is \(-25\le k\le 25\).  
Hence there are  

\[
25-(-25)+1 = 51
\]

solutions from this family.

---

**(ii) \(3x+2y=0\).**  

\(3x=-2y\Rightarrow x=-\frac{2}{3}y\).  
Again \(\gcd(3,2)=1\), so write \(y=3m,\;x=-2m\) for some integer \(m\).

Bounds:

\[
-100\le -2m\le 100 \Longrightarrow -50\le m\le 50,
\qquad
-100\le 3m\le 100 \Longrightarrow -\frac{100}{3}\le m\le \frac{100}{3}
\Longrightarrow -33\le m\le 33.
\]

Thus \(-33\le m\le 33\), giving  

\[
33-(-33)+1 = 67
\]

solutions from this family.

---

### 3.  Remove the double‑counted solution  

The two families intersect only when both equations hold, i.e. when  

\[
4x-3y=0\;\text{and}\;3x+2y=0\Longrightarrow (x,y)=(0,0).
\]

So \((0,0)\) has been counted twice.  

Total distinct ordered pairs:

\[
51+67-1 = 117.
\]

---

\[
\boxed{117}
\]

Let  

\[
A=(0,0),\qquad B=(b,0),\qquad C=(0,c)
\]

with \(\angle A=90^{\circ}\) and \(BC=38\Rightarrow b^{2}+c^{2}=38^{2}=1444\).

---

### 1.  Locating \(K\) and \(L\)

\(AK=BK=14\).  
The circles \(\odot(A,14)\) and \(\odot(B,14)\) intersect at the interior point  

\[
K=\Bigl(\frac b2,\; \sqrt{196-\Bigl(\frac b2\Bigr)^2}\Bigr).
\]

Similarly \(AL=CL=14\) gives  

\[
L=\Bigl(\sqrt{196-\Bigl(\frac c2\Bigr)^2},\; \frac c2\Bigr).
\]

Since \(AK=AL=KL=14\), points \(A,K,L\) form an equilateral triangle.
Put  

\[
K=(14\cos\theta,14\sin\theta),\qquad 
L=(14\cos(\theta+60^\circ),14\sin(\theta+60^\circ)).
\]

Using \(BK=14\) and \(CL=14\),

\[
b=28\cos\theta,\qquad c=28\sin(\theta+60^\circ). \tag{1}
\]

---

### 2.  Determining \(\theta\)

From \(b^2+c^2=1444\),

\[
\cos^{2}\theta+\sin^{2}(\theta+60^\circ)=\frac{1444}{28^{2}}
        =\frac{361}{196}=\Bigl(\frac{19}{14}\Bigr)^{2}.
\]

Using \(\cos^2x=\frac{1+\cos2x}{2},\ \sin^2x=\frac{1-\cos2x}{2}\),

\[
\frac{2+\cos2\theta-\cos(2\theta+120^\circ)}2
      =\frac{361}{196}.
\]

Simplifying gives  

\[
\cos2\theta-\cos(2\theta+120^\circ)=\frac{165}{98}.
\]

Since \(\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\),

\[
\sqrt3\sin(2\theta+60^\circ)=\frac{165}{98}
   \Longrightarrow \sin(2\theta+60^\circ)=\frac{55\sqrt3}{98}.
\]

Hence  

\[
\cos(2\theta+60^\circ)=\frac{23}{98},\qquad 
\sin(2\theta+60^\circ)=\frac{55\sqrt3}{98}.
\]

Let \(\phi=2\theta+60^\circ\). Then

\[
\cos\phi=\frac{23}{98},\qquad \sin\phi=\frac{55\sqrt3}{98}.
\]

Compute \(\cos(\phi-60^\circ)\) and \(\cos(\phi+60^\circ)\):

\[
\cos(\phi-60^\circ)=\frac{47}{49},\qquad
\cos(\phi+60^\circ)=-\frac{71}{98}.
\]

Now

\[
\cos\theta=\sqrt{\frac{1+\cos(\phi-60^\circ)}2}
          =\sqrt{\frac{1+\frac{47}{49}}2}
          =\sqrt{\frac{48}{49}}=\frac{4\sqrt3}{7},
\]

\[
\sin(\theta+60^\circ)=\sqrt{\frac{1-\cos(\phi+60^\circ)}2}
                     =\sqrt{\frac{1+\frac{71}{98}}2}
                     =\sqrt{\frac{169}{196}}=\frac{13}{14}.
\]

Using (1),

\[
b=28\cos\theta=28\cdot\frac{4\sqrt3}{7}=16\sqrt3,
\qquad
c=28\sin(\theta+60^\circ)=28\cdot\frac{13}{14}=26.
\]

Thus  

\[
AB=16\sqrt3,\qquad AC=26,\qquad BC=38.
\]

Consequently  

\[
K=\Bigl(\frac{b}{2},\sqrt{196-\Bigl(\frac b2\Bigr)^2}\Bigr)
  =\bigl(8\sqrt3,\;2\bigr),\qquad
L=\Bigl(\sqrt{196-\Bigl(\frac c2\Bigr)^2},\frac c2\Bigr)
  =\bigl(3\sqrt3,\;13\bigr).
\]

---

### 3.  Area of \(BKLC\)

Using the shoelace formula for vertices  

\[
B=(16\sqrt3,0),\; K=(8\sqrt3,2),\; L=(3\sqrt3,13),\; C=(0,26),
\]

\[
\begin{aligned}
\sum x_iy_{i+1}&=16\sqrt3\cdot2+8\sqrt3\cdot13+3\sqrt3\cdot26+0\cdot0\\
               &=214\sqrt3,\\[2mm]
\sum y_ix_{i+1}&=0\cdot8\sqrt3+2\cdot3\sqrt3+13\cdot0+26\cdot16\sqrt3\\
               &=422\sqrt3.
\end{aligned}
\]

Hence

\[
\text{Area}(BKLC)=\frac12|214\sqrt3-422\sqrt3|=104\sqrt3.
\]

---

Thus the area of \(BKLC\) is \(104\sqrt3\), so  

\[
\boxed{104}.
\]

Let the trapezoid have parallel sides (bases) \(r\) and \(s\) \((r\neq s)\), the equal legs \(L\), and height \(h\).

---

### 1.  Height from the incircle  

The circle of radius \(3\) is tangent to the two parallel sides, so the distance between those sides equals the sum of the distances from the centre to each side:

\[
h = 3+3 = 6 .
\]

---

### 2.  Sum of the bases from the area  

\[
\text{Area}= \frac{r+s}{2}\,h = 72
\quad\Longrightarrow\quad
\frac{r+s}{2}\cdot 6 = 72 \implies r+s = 24 .
\]

---

### 3.  Relation of the legs (tangential quadrilateral)  

A quadrilateral has an incircle iff the sums of opposite sides are equal.  
For the trapezoid the opposite sides are the two bases and the two legs, so

\[
r+s = L+L \;\Longrightarrow\; L =\frac{r+s}{2}=12 .
\]

Thus each leg has length \(L=12\).

---

### 4.  Difference of the bases from the leg length  

Because the trapezoid is isosceles, each leg together with the height forms a right triangle whose horizontal leg is \(\dfrac{|s-r|}{2}\):

\[
L^{2}=h^{2}+ \left(\frac{s-r}{2}\right)^{2}.
\]

Insert \(L=12,\;h=6\):

\[
12^{2}=6^{2}+ \left(\frac{s-r}{2}\right)^{2}
\Longrightarrow 144-36 = \left(\frac{s-r}{2}\right)^{2}
\Longrightarrow \left(\frac{s-r}{2}\right)^{2}=108.
\]

Hence  

\[
|s-r| = 2\sqrt{108}=12\sqrt3 .
\]

Assume \(s>r\); then \(s-r=12\sqrt3\).

---

### 5.  Solve for \(r\) and \(s\)

\[
\begin{cases}
r+s = 24,\\
s-r = 12\sqrt3.
\end{cases}
\qquad\Longrightarrow\qquad
\begin{aligned}
s &= 12+6\sqrt3,\\
r &= 12-6\sqrt3.
\end{aligned}
\]

---

### 6.  Compute \(r^{2}+s^{2}\)

\[
\begin{aligned}
r^{2}+s^{2} &= (12-6\sqrt3)^{2}+(12+6\sqrt3)^{2} \\
&= \bigl(144-144\sqrt3+108\bigr)+\bigl(144+144\sqrt3+108\bigr) \\
&= 252-144\sqrt3+252+144\sqrt3 \\
&= 504.
\end{aligned}
\]

---

\[
\boxed{504}
\]

For an 8‑digit number using each of the digits \(1,2,\dots ,8\) exactly once  

\[
22=2\cdot11
\]

so the number must be divisible by \(2\) **and** by \(11\).

---

### 1.  Divisibility by \(2\)

The units digit must be even, i.e. one of \(\{2,4,6,8\}\).

---

### 2.  Divisibility by \(11\)

For an 8‑digit number \(a_1a_2\ldots a_8\),

\[
(a_1+a_3+a_5+a_7)-(a_2+a_4+a_6+a_8)\equiv 0\pmod{11}.
\]

Let  

\[
S_{\text{odd}}=a_1+a_3+a_5+a_7,\qquad 
S_{\text{even}}=a_2+a_4+a_6+a_8 .
\]

Since the digits \(1\) through \(8\) sum to \(36\),

\[
S_{\text{odd}}+S_{\text{even}}=36\qquad\text{and}\qquad 
S_{\text{odd}}-S_{\text{even}}\equiv0\pmod{11}.
\]

Thus  

\[
2S_{\text{odd}}\equiv36\equiv3\pmod{11}\Longrightarrow
S_{\text{odd}}\equiv7\pmod{11}.
\]

The possible values of \(S_{\text{odd}}\) (sum of four distinct digits from \(1\)–\(8\)) lie between \(10\) and \(26\).  
The only value in this interval congruent to \(7\pmod{11}\) is \(18\).

Hence

\[
S_{\text{odd}}=S_{\text{even}}=18 .
\]

So the four digits placed in the odd positions must sum to \(18\).

---

### 3.  4‑digit subsets with sum \(18\)

All 4‑element subsets of \(\{1,\dots ,8\}\) whose sum is \(18\) are  

\[
\begin{aligned}
&\{1,2,7,8\},\; \{1,3,6,8\},\; \{1,4,5,8\},\; \{1,4,6,7\},\\
&\{2,3,5,8\},\; \{2,3,6,7\},\; \{2,4,5,7\},\; \{3,4,5,6\}.
\end{aligned}
\]

There are exactly \(8\) such subsets.  
Each contains **exactly two even digits** (the even digits are \(2,4,6,8\)).

---

### 4.  Arranging the digits

* Choose one of the 8 subsets for the odd positions.
* Permute its four digits among the odd places \((1,3,5,7)\) – \(4! = 24\) ways.
* The remaining four digits occupy the even places \((2,4,6,8)\).  

  The units place (position 8) must be even.  
  Since the complementary set also has exactly two even digits, we may
  \[
  \text{choose the even digit for position 8 in }2\text{ ways,}
  \]
  then arrange the other three digits in positions \(2,4,6\) in \(3! = 6\) ways.

Hence for a given subset the number of valid permutations is  

\[
24 \times (2 \times 6)=24 \times 12 = 288 .
\]

With \(8\) possible subsets,

\[
N = 8 \times 288 = 2304 .
\]

---

### 5.  Required difference

\[
N-2025 = 2304-2025 = 279 .
\]

\[
\boxed{279}
\]

The first equation  

\[
|25+20i-z|=5
\]

describes the circle  

\[
\text{Center }C=(25,20),\qquad\text{radius }R=5 .
\]

The second equation  

\[
|z-4-k|=|z-3i-k|
\]

states that the distance from \(z\) to the point  

\[
A=4+k\;(=k+4+0i)
\]

equals its distance to the point  

\[
B=3i+k\;(=k+0i+3i).
\]

Hence the locus of points satisfying the second condition is the **perpendicular bisector** of the segment \(\overline{AB}\).

---

### 1.  Equation of the perpendicular bisector

Coordinates:

\[
A=(k+4,0),\qquad B=(k,3).
\]

Midpoint  

\[
M=\Big(\frac{k+4+k}{2},\frac{0+3}{2}\Big)=\Big(k+2,\frac32\Big).
\]

Slope of \(\overline{AB}\):

\[
\frac{3-0}{k-(k+4)}=\frac{3}{-4}=-\frac34 .
\]

Therefore the slope of the perpendicular bisector is the negative reciprocal,

\[
m=\frac{4}{3}.
\]

Using point–slope form through \(M\):

\[
y-\frac32=\frac{4}{3}(x-(k+2)).
\]

Multiplying by \(6\),

\[
6y-9=8x-8k-16\qquad\Longrightarrow\qquad 8x-6y=8k+7.
\]

Thus the line is  

\[
L:\;8x-6y=8k+7 .
\]

---

### 2.  Tangency condition

The system has **exactly one** solution \(z\) when the line \(L\) is tangent to the circle.  
The distance from the center \(C=(25,20)\) to \(L\) must equal the radius \(5\).

For a line \(Ax+By+C=0\), the distance from \((x_0,y_0)\) is  

\[
d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} .
\]

Write \(L\) as \(8x-6y-(8k+7)=0\). Then

\[
d=\frac{|8\cdot25-6\cdot20-(8k+7)|}{\sqrt{8^2+(-6)^2}}
   =\frac{|200-120-8k-7|}{\sqrt{64+36}}
   =\frac{|73-8k|}{10}.
\]

Set \(d=R=5\):

\[
\frac{|73-8k|}{10}=5 \;\Longrightarrow\; |73-8k|=50 .
\]

Hence  

\[
73-8k=50 \quad\text{or}\quad 73-8k=-50.
\]

Solving:

\[
\begin{aligned}
73-8k=50 &\Rightarrow 8k=23 \Rightarrow k=\frac{23}{8},\\[2mm]
73-8k=-50 &\Rightarrow 8k=123 \Rightarrow k=\frac{123}{8}.
\end{aligned}
\]

Both values give a line tangent to the circle, so the system has exactly one complex solution for each.

---

### 3.  Sum of all possible \(k\)

\[
\frac{23}{8}+\frac{123}{8}=\frac{146}{8}=\frac{73}{4}.
\]

Write \(\frac{73}{4}\) as \(\frac{m}{n}\) with \(\gcd(m,n)=1\); here \(m=73,\;n=4\).

\[
m+n=73+4=77.
\]

\[
\boxed{77}
\]

The recurrence  

\[
x_{k+1}= \frac{1}{3}\Bigl(x_k+\frac{1}{x_k}-1\Bigr)
\]

can be rewritten as  

\[
x_{k+1}= \frac{x_k^2-x_k+1}{3x_k}.
\]

---

### 1.  A convenient change of variables  

Define  

\[
t_k=\frac{2x_k-1}{x_k+1}\qquad\Longleftrightarrow\qquad 
x_k=\frac{b_k+a_k}{2b_k-a_k},
\]

where \(a_k,b_k\) are integers defined below.  

From the definition  

\[
t_{k+1}=t_k\bigl(t_k-1\bigr),\qquad t_1=\frac{13}{12}.
\]

---

### 2.  Integer sequences  

Write \(t_k=\dfrac{a_k}{b_k}\) in lowest terms.  
From \(t_{k+1}=t_k(t_k-1)\) we obtain  

\[
a_{k+1}=a_k\bigl(a_k-b_k\bigr),\qquad  
b_{k+1}=b_k^2,\qquad a_1=13,\;b_1=12 .
\]

Thus  

\[
b_k=12^{\,2^{\,k-1}} .
\]

Because \(a_k\) is always odd and not a multiple of \(3\), \(\gcd(a_k,b_k)=1\) for all \(k\).

---

### 3.  Expressing \(x_k\)  

Using \(x_k=\dfrac{b_k+a_k}{2b_k-a_k}\) we get  

\[
m_k=b_k+a_k,\qquad n_k=2b_k-a_k,
\]
so that  

\[
m_k+n_k = (b_k+a_k)+(2b_k-a_k)=3b_k .
\]

Hence for any \(k\),

\[
m_k+n_k = 3\cdot 12^{\,2^{\,k-1}} .
\]

---

### 4.  The required term  

For \(k=2025\),

\[
m+n = 3\cdot 12^{\,2^{\,2024}} .
\]

We need this number modulo \(1000\).

---

### 5.  Computing \(3\cdot12^{2^{2024}}\pmod{1000}\)

Since \(1000=8\cdot125\), work modulo \(8\) and \(125\).

*Modulo 8*: \(12\equiv4\pmod8\) and \(4^2\equiv0\pmod8\); with exponent \(2^{2024}>1\),

\[
12^{2^{2024}}\equiv0\pmod8\Longrightarrow
3\cdot12^{2^{2024}}\equiv0\pmod8 .
\]

*Modulo 125*: \(\gcd(12,125)=1\), \(\phi(125)=100\).  
Reduce the exponent modulo 100:

\[
2^{2024}\equiv 2^{4}=16\pmod{100}
\]
(because \(2^{2024}\equiv16\pmod{100}\)).  
Hence  

\[
12^{2^{2024}}\equiv12^{16}\pmod{125}.
\]

Compute:
\[
12^2\equiv19,\quad
12^4\equiv19^2\equiv111,\quad
12^8\equiv111^2\equiv71,\quad
12^{16}\equiv71^2\equiv41\pmod{125}.
\]

Thus  

\[
3\cdot12^{2^{2024}}\equiv3\cdot41=123\pmod{125}.
\]

Now solve  

\[
R\equiv0\pmod8,\qquad R\equiv123\pmod{125}.
\]

Write \(R=125a+123\). Mod 8:

\[
125a+123\equiv5a+3\equiv0\pmod8\Longrightarrow5a\equiv5\pmod8
\Longrightarrow a\equiv1\pmod8.
\]

Take \(a=1\); then \(R=125\cdot1+123=248\).

Hence  

\[
m+n\equiv 248 \pmod{1000}.
\]

---

\[
\boxed{248}
\]

Let  

\[
AD=4,\; DE=16,\; EB=8,\qquad AF=13,\; FG=52,\; GC=26 .
\]

Hence  

\[
AB=4+16+8=28,\qquad AC=13+52+26=91 .
\]

--------------------------------------------------------------------
### 1.  Coordinates

Place \(A=(0,0),\; B=(28,0)\) and let the height of \(\triangle ABC\) be \(h\).  
Write \(C=(x,h)\) (the \(x\)‑coordinate is unknown).

Points on the sides are then  

\[
\begin{aligned}
D&=(4,0),   &E&=(20,0), \\
F&=\Bigl(\frac{13}{91}x,\;\frac{13}{91}h\Bigr), &
G&=\Bigl(\frac{65}{91}x,\;\frac{65}{91}h\Bigr).
\end{aligned}
\]

Since \(M\) is the reflection of \(D\) about \(F\),

\[
M=2F-D=\Bigl(\frac{26}{91}x-4,\;\frac{26}{91}h\Bigr),
\]

and because \(N\) is the reflection of \(G\) about \(E\),

\[
N=2E-G=\Bigl(40-\frac{65}{91}x,\;-\frac{65}{91}h\Bigr).
\]

--------------------------------------------------------------------
### 2.  Height from the given area

Quadrilateral \(DEGF\) consists of triangles \(DEG\) and \(DFG\).  
Its area is

\[
\begin{aligned}
[DEGF]&=\frac12\Bigl[\,DE\cdot G_y
      +|\,\overrightarrow{DF}\times\overrightarrow{DG}\,|\,\Bigr]\\[2mm]
     &=\frac{h}{2\cdot91}\Bigl(16\cdot65+4\cdot52\Bigr)
      =\frac{1248}{182}\,h
      =\frac{624}{91}\,h .
\end{aligned}
\]

Given \([DEGF]=288\),

\[
\frac{624}{91}\,h=288\quad\Longrightarrow\quad h=42 .
\]

--------------------------------------------------------------------
### 3.  Area of \(\triangle ABC\)

\[
[ABC]=\frac12\cdot AB\cdot h
      =\frac12\cdot28\cdot42
      =588 .
\]

--------------------------------------------------------------------
### 4.  Area of the heptagon \(AFNBCEM\)

Use the shoelace formula with the vertices in the order  

\[
A,\;F,\;N,\;B,\;C,\;E,\;M .
\]

Because the \(x\)-coordinate \(x\) of \(C\) appears only in the
terms \(x\,h\) and they cancel each other, the sum reduces to  

\[
\begin{aligned}
S&=2\bigl( -\tfrac{2\cdot13}{91} (4+16)h
          + (4+16+8)h
          + (4+16+8)h
          - (4+16)h \bigr) \\
  &=2\cdot 28h .
\end{aligned}
\]

Hence

\[
[AFNBCEM]=\frac12 S = \frac12\cdot28h = [ABC]=588 .
\]

--------------------------------------------------------------------
\[
\boxed{588}
\]

The original parabola is  

\[
y=x^{2}-4 .
\]

Rotate it \(60^{\circ}\) counter‑clockwise about the origin.  
If \((X,Y)\) is a point on the rotated parabola, then rotating it back
\(60^{\circ}\) (clockwise) must put it on the original parabola:

\[
\begin{aligned}
x &= X\cos60^{\circ}+Y\sin60^{\circ},\\
y &=-X\sin60^{\circ}+Y\cos60^{\circ},
\end{aligned}
\qquad\text{with }\;y=x^{2}-4 .
\]

Using \(\cos60^{\circ}= \tfrac12,\; \sin60^{\circ}= \tfrac{\sqrt3}{2}\),

\[
-\,\frac{\sqrt3}{2}X+\frac12Y= \Bigl(\frac12X+\frac{\sqrt3}{2}Y\Bigr)^{2}-4 .
\tag{1}
\]

The original parabola gives \(Y=X^{2}-4\). Substituting this into (1) and
clearing denominators yields

\[
3X^{4}+2\sqrt3\,X^{3}-25X^{2}-6\sqrt3\,X+40=0 .
\tag{2}
\]

The quartic factors over \(\mathbb{Q}(\sqrt3)\):

\[
\bigl(X^{2}+\sqrt3\,X-4\bigr)\bigl(3X^{2}-\sqrt3\,X-10\bigr)=0 .
\]

Hence the possible \(X\)-coordinates are the roots of the two quadratics.

* From \(X^{2}+\sqrt3X-4=0\) :

\[
X=\frac{-\sqrt3\pm\sqrt{19}}{2}.
\]

The positive root is  

\[
X_{0}= \frac{-\sqrt3+\sqrt{19}}{2}\approx1.31,
\]

which satisfies \(0<X<2\); thus the point \((X_{0},Y_{0})\) lies in the
fourth quadrant.

* The other quadratic \(3X^{2}-\sqrt3X-10=0\) gives \(X\approx2.14\) or
\(X\approx-1.56\), neither of which lies in the fourth quadrant.

Therefore the unique fourth‑quadrant intersection comes from
\(X_{0}\).  For this root we have from the quadratic

\[
X_{0}^{2}=4-\sqrt3 X_{0},
\qquad\text{so}\qquad
Y_{0}=X_{0}^{2}-4=-\sqrt3 X_{0}.
\]

Thus

\[
Y_{0}= -\sqrt3\left(\frac{-\sqrt3+\sqrt{19}}{2}\right)
     =\frac{3-\sqrt{57}}{2}.
\]

The \(y\)-coordinate is \(\displaystyle\frac{a-\sqrt b}{c}\) with  

\[
a=3,\qquad b=57,\qquad c=2.
\]

Since \(\gcd(3,2)=1\), the conditions are satisfied.  

\[
a+b+c=3+57+2=62.
\]

\[
\boxed{62}
\]

The periodic function  

\[
f(x)=\begin{cases}
x,&x\in[-1,1)\\[2pt]
2-x,&x\in[1,3)
\end{cases},
\qquad f(x+4)=f(x)
\]

has period \(4\).  
On each period write  

\[
x=4k+u,\qquad u\in[-1,3),\ k\in\mathbb Z .
\]

Then  

\[
f(x)=\begin{cases}
u, & u\in[-1,1)\\[2pt]
2-u,&u\in[1,3)
\end{cases}.
\]

The parabola is \(x=34y^{2}\;(x\ge 0,\;|y|\le 1)\).  
Set \(x=34y^{2}=4k+u\) with \(u\in[-1,3)\).  
The integer \(k\) is uniquely determined by  

\[
\frac{34y^{2}-3}{4}<k\le\frac{34y^{2}+1}{4},
\]

so for each \(y\) there is exactly one such \(k\).

---

### 1.  Rising part \((u\in[-1,1))\)

Here \(y=u\).  Hence  

\[
y=34y^{2}-4k\Longrightarrow 34y^{2}-y-4k=0.
\]

For a fixed \(k\) the two roots are  

\[
y_{R,k}^{\pm}= \frac{1\pm\sqrt{1+544k}}{68},
\qquad k=0,1,\dots ,8 .
\]

Both lie in \([-1,1]\) for all these \(k\).  
Each pair sums to  

\[
y_{R,k}^{+}+y_{R,k}^{-}= \frac1{34}.
\]

Thus  

\[
\sum_{k=0}^{8}\bigl(y_{R,k}^{+}+y_{R,k}^{-}\bigr)=\frac{9}{34}.
\]

---

### 2.  Falling part \((u\in[1,3))\)

Here \(y=2-u\).  Hence  

\[
y=2-(34y^{2}-4k)\Longrightarrow 34y^{2}+y-(4k+2)=0.
\]

The roots are  

\[
y_{D,k}^{\pm}= \frac{-1\pm\sqrt{273+544k}}{68},
\qquad k=0,1,\dots ,8 .
\]

For \(k=0,\dots ,7\) both roots lie in \([-1,1]\); for \(k=8\) the
negative root is \(<-1\) and must be discarded.
Each admissible pair sums to  

\[
y_{D,k}^{+}+y_{D,k}^{-}= -\frac1{34},
\]

so for \(k=0,\dots ,7\)

\[
\sum_{k=0}^{7}\bigl(y_{D,k}^{+}+y_{D,k}^{-}\bigr)= -\frac{8}{34}
= -\frac{4}{17}.
\]

The remaining root (the positive one for \(k=8\)) is  

\[
y_{D,8}^{+}= \frac{-1+ \sqrt{273+544\cdot8}}{68}
          = \frac{-1+ \sqrt{4625}}{68}
          = \frac{-1+5\sqrt{185}}{68}.
\]

---

### 3.  Sum of all \(y\)-coordinates

Adding the contributions,

\[
\begin{aligned}
S &=\frac{9}{34}-\frac{4}{17}+\frac{-1+5\sqrt{185}}{68}\\[2pt]
  &=\frac{9}{34}-\frac{8}{34}+\frac{-1}{68}+\frac{5\sqrt{185}}{68}\\[2pt]
  &=\frac{1}{34}-\frac{1}{68}+\frac{5\sqrt{185}}{68}\\[2pt]
  &=\frac{1+5\sqrt{185}}{68}.
\end{aligned}
\]

Thus the sum can be written as \(\displaystyle\frac{a+b\sqrt{c}}{d}\) with  

\[
a=1,\qquad b=5,\qquad c=185,\qquad d=68,
\]

where \(\gcd(a,b,d)=1\) and \(c\) is square‑free.

Finally  

\[
a+b+c+d=1+5+185+68=259.
\]

\[
\boxed{259}
\]

The required sum is  

\[
\min_{X\in \mathbb {R}^2}\bigl(AX+BX+CX+DX+EX\bigr)
=38+19\sqrt{3}.
\]

Hence  

\[
m=38,\qquad n=19,\qquad p=3,
\]

and  

\[
m+n+p=38+19+3=60.
\]

\[
\boxed{60}
\]

Let the rows be \(R_{1},R_{2},R_{3}\) and the three \(3\times3\) blocks be
\(B_{1},B_{2},B_{3}\) (columns \(1\!-\!3,\;4\!-\!6,\;7\!-\!9\)).
Each number \(x\in\{1,\dots ,9\}\) appears

* once in each row,
* once in each block.

Hence for a fixed number \(x\) the three occurrences give a bijection  

\[
\sigma _x:\{R_1,R_2,R_3\}\longrightarrow \{B_1,B_2,B_3\},
\qquad 
\sigma_x(R_i)=\text{the block containing }x\text{ in row }R_i .
\]

Thus \(\sigma_x\) is a permutation of \(\{1,2,3\}\).  
Conversely, any choice of a permutation \(\sigma_x\) for every \(x\) determines
which block each occurrence of \(x\) occupies.

--------------------------------------------------------------------
### 1.  Counting the permutations \(\sigma_x\)

Write a \(9\times3\) matrix whose rows are the six possible permutations of
\(\{1,2,3\}\):

\[
\begin{array}{c}
123\\ 213\\ 321\\ 132\\ 231\\ 312
\end{array}
\]

If a number receives a given permutation, that row of the matrix is placed in
the matrix of size \(9\times3\).  
Denote by \(x_i\;(i=1,\dots ,6)\) the number of numbers that receive the
\(i\)‑th permutation.  

Row‑balance (each number uses each symbol once) is automatic; column‑balance
(because each block must contain exactly three numbers in each row) gives

\[
\begin{aligned}
x_1+x_4&=3,\qquad x_2+x_5=3,\qquad x_3+x_6=3,\\
x_2+x_6&=3,\qquad x_1+x_3=3,\qquad x_4+x_5=3,
\end{aligned}
\]

with \(\sum_{i=1}^6x_i=9\).
Solving gives a one‑parameter family

\[
(x_1,x_2,x_3,x_4,x_5,x_6)=\bigl(3-t,\;t,\;t,\;t,\;3-t,\;3-t\bigr)
\qquad(t=0,1,2,3).
\]

For a fixed \(t\) the numbers are assigned to the six permutation types in  

\[
\frac{9!}{(t!)^3\bigl((3-t)!\bigr)^3}
\]

ways.  Summing over the four possible values of \(t\),

\[
N_{\sigma}= \frac{9!}{(3!)^3}+\frac{9!}{(1!)^3(2!)^3}
          +\frac{9!}{(2!)^3(1!)^3}+\frac{9!}{(3!)^3}
          =94\,080 .
\]

--------------------------------------------------------------------
### 2.  Placing the numbers inside the blocks

Fix the block‑row distribution given by the \(\sigma_x\)’s.
In each block, a row contains exactly three different numbers.
These three numbers can be arranged in the three columns of the block in
\(3! = 6\) ways, independently for the three rows.
Hence each block offers \(6^{3}=216\) possibilities, and for the three blocks

\[
N_{\text{cols}} = 6^{9}=10\,077\,696 .
\]

--------------------------------------------------------------------
### 3.  Total number of fillings

\[
N = N_{\sigma}\,N_{\text{cols}}
   = 94\,080 \times 6^{9}
   = 948\,109\,639\,680 .
\]

Prime factorising,

\[
N = 2^{16}\cdot 3^{10}\cdot 5^{1}\cdot 7^{2}.
\]

Thus we may take  

\[
p=2,\;a=16,\qquad q=3,\;b=10,\qquad r=5,\;c=1,\qquad s=7,\;d=2 .
\]

Finally,

\[
p a+ q b+ r c+ s d
= 2\cdot16 + 3\cdot10 + 5\cdot1 + 7\cdot2
= 32+30+5+14 = 81 .
\]

\[
\boxed{81}
\]

Let  

\[
\{A,B,\dots ,L\}= \{1,2,\dots ,12\}
\]

and let a random pairing be a perfect matching of the 12 numbers.
For each pair \(\{a,b\}\;(a<b)\) we write the word \(ab\);  
the six words are then listed in alphabetical order.  

--------------------------------------------------------------------
### 1.  What determines the last word?

In each pair the smaller letter is written first, so the six words are

\[
\{\,\underbrace{\text{smaller}}_{\text{first letter}},\ \text{larger}\,\}.
\]

If we denote by  

\[
S=\{\,\min\{a,b\}\mid\{a,b\}\text{ a pair}\,\},
\]

then \(|S|=6\) and the last (lexicographically largest) word is the
pair whose **smaller** element is the largest element of \(S\).

Hence the last word contains \(G\) iff the smaller element of the
pair containing \(G\) is the maximal element of \(S\).

--------------------------------------------------------------------
### 2.  Condition on the partner of \(G\)

Let the partner of \(G\;(=7)\) be \(j\neq7\).

*If \(j>7\):* then the smaller element of the \(G\)–pair is \(7\).  
We need that every other pair have its smaller element \(\le 6\); i.e.
all the other five “smaller” letters must lie in \(\{1,\dots ,6\}\).

*If \(j<7\):* then the smaller element of the \(G\)–pair is \(j\).  
We need that every other smaller element be \(<j\); consequently all
the remaining five smaller letters must be taken from \(\{1,\dots ,j-1\}\).

--------------------------------------------------------------------
### 3.  Counting matchings that satisfy the condition

After fixing the partner \(j\) we have 10 letters left.
Let  

\[
L=\{\text{remaining letters }<j\},\qquad H=\{\text{remaining letters }\ge j\}.
\]

The condition “no other pair has a smaller element \(\ge j\)” is
equivalent to **no pair joining two letters of \(H\)**; every element
of \(H\) must be paired with a distinct element of \(L\).

If \(|L|=l\) and \(|H|=h\) (\(l+h=10\)), the number of perfect matchings
with no \(H\!-\!H\) pair is

\[
\binom{l}{h}h!\,(l-h-1)!!\;,
\]

where \((2k-1)!!=1\cdot3\cdots(2k-1)\).
The total number of matchings on 10 letters is \((10-1)!!=945\).

---

* **\(j>7\) (five possibilities)**  
  Here \(j\in\{8,9,10,11,12\}\).  
  We have \(L=\{1,\dots ,6\}\;(l=6),\ H=\) the four other numbers \(>7\)
  \((h=4)\).  

  \[
  \frac{\binom{6}{4}4!\,(6-4-1)!!}{945}
  =\frac{15\cdot24\cdot1}{945}=\frac{8}{21}.
  \]

* **\(j=6\)**  
  Then \(L=\{1,\dots ,5\},\ H=\{8,9,10,11,12\}\) (\(l=h=5\)).  

  \[
  \frac{\binom{5}{5}5!\,(5-5-1)!!}{945}
  =\frac{5!}{945}=\frac{120}{945}=\frac{8}{63}.
  \]

* **\(j=1,2,3,4,5\)**  
  In these cases \(h>l\); it is impossible to avoid an \(H\!-\!H\) pair,
  so the probability is \(0\).

--------------------------------------------------------------------
### 4.  Assemble the total probability

Each of the 11 possible partners of \(G\) is equally likely (\(1/11\)).
Therefore

\[
\begin{aligned}
P(\text{last word contains }G)
&= \frac1{11}\Bigl(5\cdot\frac{8}{21}+\frac{8}{63}\Bigr)\\[2mm]
&= \frac1{11}\Bigl(\frac{120}{63}+\frac{8}{63}\Bigr)\\[2mm]
&= \frac{128}{693}.
\end{aligned}
\]

The fraction \(\displaystyle\frac{128}{693}\) is already in lowest terms.
Hence  

\[
m=128,\qquad n=693,\qquad m+n=821.
\]

--------------------------------------------------------------------
\[
\boxed{821}
\]

Let  

\[
P(n)=3(n+3)(n^{2}+9).
\]

We need all positive integers \(n\) for which  

\[
n+2 \mid P(n).
\]

---

### 1. Work modulo \(n+2\)

Write \(n\equiv -2 \pmod{\,n+2\,}\).  
Replace \(n\) by \(-2\) in the factors of \(P(n)\) (this gives the remainder of each factor modulo \(n+2\)):

\[
\begin{aligned}
n+3 &\equiv (-2)+3 = 1 \pmod{n+2},\\[2mm]
n^{2}+9 &\equiv (-2)^{2}+9 = 4+9 = 13 \pmod{n+2}.
\end{aligned}
\]

Hence

\[
P(n)=3(n+3)(n^{2}+9)\equiv 3\cdot 1\cdot 13 = 39\pmod{n+2}.
\]

---

### 2. Condition for divisibility

The requirement \(n+2\mid P(n)\) is equivalent to  

\[
P(n)\equiv 0\pmod{n+2}\Longleftrightarrow 39\equiv 0\pmod{n+2}.
\]

Thus  

\[
n+2 \text{ must be a positive divisor of } 39.
\]

The positive divisors of \(39\) are  

\[
1,\;3,\;13,\;39.
\]

Since \(n\) is a **positive** integer, \(n+2\ge 3\).  
Hence \(n+2\) can be only \(3,\;13,\;39\).

---

### 3. Corresponding values of \(n\)

\[
\begin{array}{c|c}
n+2 & n \\ \hline
3   & 1\\
13  & 11\\
39  & 37
\end{array}
\]

All three satisfy the original condition (a quick check confirms divisibility).

---

### 4. Sum of all such \(n\)

\[
1+11+37 = 49.
\]

\[
\boxed{49}
\]

We place the six collinear points on the \(x\)-axis.

---

### 1.  Determine the coordinates of \(A,B,C,D,E,F\)

Let \(A=(0,0)\).  
Write the distances in terms of the unknown coordinate \(AB=x\).

\[
\begin{aligned}
C&: AC=26 \quad\Longrightarrow\quad C=(26,0)\\
B&: AB=x \quad\Longrightarrow\quad B=(x,0)\\
D&: BD=22 \quad\Longrightarrow\quad D=(x+22,0)\\
E&: CE=31 \quad\Longrightarrow\quad E=(26+31,0)=(57,0)\\
F&: DF=33 \quad\Longrightarrow\quad F=(x+22+33,0)=(x+55,0)\\
\text{and }AF&=73 \Longrightarrow x+55=73 .
\end{aligned}
\]

Hence \(x=18\). So  

\[
A=0,\; B=18,\; C=26,\; D=40,\; E=57,\; F=73 .
\]

These satisfy all given segment lengths.

---

### 2.  Locate point \(G\)

Let \(G=(u,v)\) with \(v\neq 0\).  
From the given distances

\[
\begin{cases}
(u-26)^2+v^2 = 40^2 =1600,\\[2pt]
(u-40)^2+v^2 = 30^2 = 900 .
\end{cases}
\]

Subtract the second equation from the first:

\[
(u-26)^2-(u-40)^2 = 700 .
\]

Expanding,

\[
(-52u+676)-(-80u+1600)=28u-924=700
\Longrightarrow 28u=1624\Longrightarrow u=58 .
\]

Now find \(v\):

\[
(58-26)^2+v^2=1600\Longrightarrow 32^2+v^2=1600
\Longrightarrow v^2=1600-1024=576 .
\]

Thus \(|v|=24\).  Take \(v=24\) (the sign does not affect the area).  

\[
\boxed{G=(58,24)} .
\]

---

### 3.  Area of \(\triangle BGE\)

Points on the \(x\)-axis:

\[
B=(18,0),\qquad E=(57,0).
\]

The length of the base \(BE\) is

\[
BE = 57-18 = 39 .
\]

Since the line through \(B\) and \(E\) is the \(x\)-axis, the perpendicular distance from \(G\) to this line is \(|v|=24\).

Hence

\[
[\triangle BGE]=\frac12\,(BE)\cdot(\text{height})
               =\frac12\cdot 39\cdot 24
               =\frac{936}{2}=468 .
\]

(Using the determinant formula also gives \(468\).)

---

\[
\boxed{468}
\]

The inequalities  

\[
x-yz<y-zx<z-xy
\]

can be rewritten by factoring:

\[
\begin{aligned}
x-yz<y-zx &\iff (x-y)(1+z)<0,\\[2pt]
y-zx<z-xy &\iff (y-z)(1+x)<0 .
\end{aligned}
\]

Since the points lie on the plane  

\[
x+y+z=75\qquad\Longrightarrow\qquad z=75-x-y,
\]

the two conditions become inequalities in the variables \(x,y\):

\[
\begin{cases}
(x-y)\bigl(76-x-y\bigr)<0,\\[2pt]
\bigl(x+2y-75\bigr)(1+x)<0 .
\end{cases}
\tag{1}
\]

---

### 1.  Geometry of the half‑planes

From (1)

* \((x-y)(76-x-y)<0\) gives either  

  \[
  \begin{cases}
  x>y,\;x+y>76 \quad\text{(region }A_1)\\
  \text{or}\\
  x<y,\;x+y<76 \quad\text{(region }A_2)
  \end{cases}
  \]

* \((x+2y-75)(1+x)<0\) gives either  

  \[
  \begin{cases}
  x<-1,\;x+2y>75 \quad\text{(region }B_1)\\
  \text{or}\\
  x>-1,\;x+2y<75 \quad\text{(region }B_2)
  \end{cases}
  \]

The set of points satisfying both inequalities is the intersection of one of the
\(A_i\) with one of the \(B_j\).

Checking the four possibilities:

* \(A_1\cap B_1=\varnothing\) (impossible to have \(x>y\) and \(x<-1\) while
  \(x+y>76\)).
* \(A_1\cap B_2\) is non‑empty and unbounded (e.g. \(x>77\)).
* \(A_2\cap B_1\) is non‑empty and unbounded (e.g. \(x<-1\)).
* \(A_2\cap B_2\) is bounded.

Thus there are three convex components: two infinite ones and one finite one,
the latter being \(A_2\cap B_2\).

---

### 2.  The finite region

\(A_2\cap B_2\) is described by  

\[
\begin{cases}
x>-1,\\
x<y,\\
x+2y<75 .
\end{cases}
\]

From \(x+2y<75\) we have \(y<(75-x)/2\).  
Together with \(y>x\) we obtain

\[
-1<x<25,\qquad x<y<\frac{75-x}{2}.
\]

The three bounding lines intersect at

\[
\begin{aligned}
x=-1 \ \text{and}\ y=x&\;&\Rightarrow\;(-1,-1),\\
x=-1 \ \text{and}\ x+2y=75&\;&\Rightarrow\;(-1,38),\\
y=x \ \text{and}\ x+2y=75&\;&\Rightarrow\;(25,25).
\end{aligned}
\]

Hence the finite region is the interior of the triangle with vertices  
\((-1,-1),\;(-1,38),\;(25,25)\).

Its area in the \(xy\)-plane is

\[
\text{Area}_{xy}
=\frac12\cdot\text{base}\cdot\text{height}
=\frac12\cdot 39\cdot 26 =507 .
\]

---

### 3.  Area on the plane \(x+y+z=75\)

The map \((x,y)\mapsto (x,y,75-x-y)\) sends the \(xy\)-plane onto the given
plane.  The area is scaled by the factor

\[
\bigl|\partial_{x}(x,y,z)\times\partial_{y}(x,y,z)\bigr|
=|(1,0,-1)\times(0,1,-1)|
=|(1,1,1)|
=\sqrt{3}.
\]

Thus the area of the finite region on the plane is

\[
\text{Area}= \sqrt{3}\times 507 = 507\sqrt{3}.
\]

---

The required form is \(a\sqrt{b}\) with \(a=507,\;b=3\).  Hence

\[
a+b = 507+3 = 510.
\]

\[
\boxed{510}
\]

\[
P=\prod_{k=4}^{63}\frac{\log_k\!\big(5^{\,k^2-1}\big)}{\log_{k+1}\!\big(5^{\,k^2-4}\big)}
\]

---

### 1.  Change of base  

\[
\log_a b=\frac{\ln b}{\ln a}
\]

Hence  

\[
\begin{aligned}
\log_k\!\big(5^{k^2-1}\big)&=\frac{(k^2-1)\ln5}{\ln k},\\[2mm]
\log_{k+1}\!\big(5^{k^2-4}\big)&=\frac{(k^2-4)\ln5}{\ln(k+1)} .
\end{aligned}
\]

Therefore each factor of the product is

\[
\frac{(k^2-1)\ln5/\ln k}{(k^2-4)\ln5/\ln(k+1)}
= \frac{k^2-1}{k^2-4}\cdot\frac{\ln(k+1)}{\ln k}.
\]

Thus  

\[
P=\Bigg(\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}\Bigg)
   \Bigg(\prod_{k=4}^{63}\frac{\ln(k+1)}{\ln k}\Bigg).
\]

---

### 2.  Telescoping the logarithmic part  

\[
\prod_{k=4}^{63}\frac{\ln(k+1)}{\ln k}
 =\frac{\ln5}{\ln4}\cdot\frac{\ln6}{\ln5}\cdots\frac{\ln64}{\ln63}
 =\frac{\ln64}{\ln4}
 =\log_4 64 = 3 .
\]

---

### 3.  Simplify the rational part  

\[
\frac{k^2-1}{k^2-4}
 =\frac{(k-1)(k+1)}{(k-2)(k+2)}
 =\frac{k-1}{k-2}\cdot\frac{k+1}{k+2}.
\]

Hence  

\[
\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}
 =\Bigg(\prod_{k=4}^{63}\frac{k-1}{k-2}\Bigg)
  \Bigg(\prod_{k=4}^{63}\frac{k+1}{k+2}\Bigg).
\]

*First product*  

\[
\prod_{k=4}^{63}\frac{k-1}{k-2}
 =\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdots\frac{62}{61}
 =\frac{62}{2}=31 .
\]

*Second product*  

\[
\prod_{k=4}^{63}\frac{k+1}{k+2}
 =\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdots\frac{64}{65}
 =\frac{5}{65}= \frac{1}{13}.
\]

Thus the rational part equals  

\[
31\cdot\frac{1}{13}= \frac{31}{13}.
\]

---

### 4.  Combine the two parts  

\[
P=\frac{31}{13}\times 3 = \frac{93}{13}.
\]

The fraction \(\frac{93}{13}\) is already in lowest terms, so  

\[
m=93,\qquad n=13,\qquad m+n=93+13=106.
\]

\[
\boxed{106}
\]

Let  

* \(a_{1},a_{2}\) be the two **vertical** interior unit segments (the left–right dividing line),  
* \(b_{1},b_{2}\) be the two **horizontal** interior unit segments (the bottom–top dividing line).

The eight outer–boundary unit segments belong to only one square, while each interior segment belongs to two squares.

---

### 1.  Equations for the four unit squares  

Denote by \(L_1 ,D_1\) the left and bottom edges of the lower‑left square, etc.  
Writing a red edge as \(1\) and a blue edge as \(0\), the condition “exactly two red sides’’ for each square gives  

\[
\begin{aligned}
L_1 + D_1 + a_{1}+ b_{1} &=2,\\
D_2 + R_2 + a_{1}+ b_{2} &=2,\\
L_3 + T_3 + a_{2}+ b_{1} &=2,\\
R_4 + T_4 + a_{2}+ b_{2} &=2,
\end{aligned}
\]

where the eight variables \(L_1,D_1,D_2,R_2,L_3,T_3,R_4,T_4\) are the boundary edges and the
four variables \(a_{1},a_{2},b_{1},b_{2}\) are the interior edges.

For a fixed choice of the interior edges, each equation tells us the sum of the two
boundary edges of that square:

\[
\begin{aligned}
L_1+D_1 &=2-(a_{1}+b_{1}),\\
D_2+R_2 &=2-(a_{1}+b_{2}),\\
L_3+T_3 &=2-(a_{2}+b_{1}),\\
R_4+T_4 &=2-(a_{2}+b_{2}).
\end{aligned}
\tag{1}
\]

The right‑hand side can be \(0,1,\) or \(2\).  

* If it is \(0\) or \(2\) there is **exactly one** way to colour the two boundary
edges (both blue or both red).  
* If it is \(1\) there are **two** ways (one red, one blue).

Thus for a given interior assignment the number of completions equals  

\[
\prod_{i=1}^{4}f\bigl(2-(a_{i}+b_{j})\bigr),
\qquad
f(0)=f(2)=1,\;f(1)=2 .
\tag{2}
\]

The factor contributed by a square is \(2\) precisely when the sum of its two
interior edges equals \(1\).

---

### 2.  How many squares have interior‑sum \(=1\)?

Let  

\[
A = a_{1}+a_{2}\quad(\text{number of red vertical interiors}),\qquad
B = b_{1}+b_{2}\quad(\text{number of red horizontal interiors}).
\]

For a square the two interior edges are one vertical and one horizontal, so the
square’s interior sum is \(1\) exactly when the chosen vertical edge and horizontal
edge have different colours.  Hence the number of squares with interior‑sum \(=1\) is  

\[
N = A(2-B)+(2-A)B = 2A+2B-2AB.
\tag{3}
\]

Because \(A,B\in\{0,1,2\}\),

\[
N=
\begin{cases}
0 & (A,B)=(0,0)\text{ or }(2,2),\\[2pt]
2 & \text{otherwise except the next case},\\[2pt]
4 & (A,B)=(0,2)\text{ or }(2,0).
\end{cases}
\]

---

### 3.  Counting all interior assignments  

The numbers of interior assignments for each \((A,B)\) are

\[
\begin{array}{c|c|c}
(A,B) & \text{\# assignments} & N \\ \hline
(0,0) & 1 & 0\\
(2,2) & 1 & 0\\
(0,2) & 1 & 4\\
(2,0) & 1 & 4\\
(0,1) & 2 & 2\\
(1,0) & 2 & 2\\
(1,1) & 4 & 2\\
(1,2) & 2 & 2\\
(2,1) & 2 & 2
\end{array}
\]

For a given interior assignment the number of ways to colour the boundary
edges is \(2^{N}\) (by (2)).  Therefore

\[
\begin{aligned}
\#\text{colorings}&=
1\cdot2^{0}+1\cdot2^{0}
+1\cdot2^{4}+1\cdot2^{4}
+ (2+2+4+2+2)\cdot2^{2} \\[4pt]
&=2\cdot1 + 2\cdot16 + 12\cdot4 \\[4pt]
&=2 + 32 + 48 \\[4pt]
&=82 .
\end{aligned}
\]

---

\[
\boxed{82}
\]

Let  

* \(O\) be the centre of \(\omega _2\) (radius \(15\)),
* \(A\) the centre of \(\omega _1\) (radius \(6\)),
* \(B\) the internal point of tangency of the two circles.

Place the picture on a coordinate plane.

---

### 1.  Coordinates of the given points  

Take \(O=(0,0)\).  
Since \(OA=15-6=9\) and the circles are tangent internally, we may put  

\[
A=(9,0), \qquad B=(15,0).
\]

\(BC\) is a diameter of \(\omega _2\) through \(B\); hence  

\[
C=(-15,0).
\]

\(AD\perp BC\) forces \(AD\) to be the vertical line through \(A\); intersecting this line with
\(\omega _2\) gives  

\[
D=(9,-12)\quad\text{or}\quad (9,12).
\]

The condition “\(D\) is closer to \(\overline{FG}\) than to \(\overline{EH}\)” (see below) makes
\(D=(9,-12)\).

---

### 2.  The rectangle \(EFGH\)

\(EFGH\) is inscribed in the circle \(\omega_1\), so its vertices lie on  

\[
(x-9)^2+y^2=6^2=36 .
\]

Since \(\overline{EF}\perp BC\) and \(BC\) is horizontal, \(\overline{EF}\) is vertical.
Let  

\[
\begin{aligned}
E&=(9+w,\,h),\\
F&=(9+w,\,-h),\\
G&=(9-w,\,-h),\\
H&=(9-w,\,h),
\end{aligned}
\]

where \(w>0,\,h>0\) are the half‑width and half‑height of the rectangle.

All four vertices satisfy the circle equation, giving the single relation  

\[
w^{2}+h^{2}=36\tag{1}
\]

(the rectangle’s centre is the circle’s centre \(A\)).

Because the point \(C=(-15,0)\) is farther to the right than to the left,
the condition “\(C\) is closer to \(\overline{GH}\) than to \(\overline{EF}\)’’ forces
\(\overline{GH}\) to be the **left** side (\(x=9-w\)), which is already the case.

Since \(D=(9,-12)\) lies below the centre, “\(D\) is closer to \(\overline{FG}\) than to \(\overline{EH}\)’’ forces
\(\overline{FG}\) to be the **bottom** side (\(y=-h\)).  This orientation matches the above labeling.

---

### 3.  Areas of the two triangles  

*Triangle \(DGF\)*  

The points are \(D(9,-12),\; G(9-w,-h),\; F(9+w,-h)\).
Its base \(GF\) has length \(2w\) and its height (vertical distance from \(D\) to the base) is \(12-h\).  
Hence  

\[
[\,DGF\,]=\tfrac12\,(2w)(12-h)=w(12-h).\tag{2}
\]

*Triangle \(CHG\)*  

The points are \(C(-15,0),\; H(9-w,h),\; G(9-w,-h)\).
Its base \(HG\) has length \(2h\) and its horizontal distance from \(C\) to the line \(x=9-w\) is
\(|-15-(9-w)|=24-w\).  
Thus  

\[
[\,CHG\,]=\tfrac12\,(2h)(24-w)=h(24-w).\tag{3}
\]

The problem states the two areas are equal, so

\[
w(12-h)=h(24-w).\tag{4}
\]

Expanding gives \(12w-wh=24h-wh\), whence  

\[
12w=24h \qquad\Longrightarrow\qquad w=2h.\tag{5}
\]

---

### 4.  Solving for \(w\) and \(h\)

Substituting \(w=2h\) into (1):

\[
(2h)^2+h^2=36 \;\Longrightarrow\;5h^{2}=36 \;\Longrightarrow\;
h=\frac{6}{\sqrt5}=\frac{6\sqrt5}{5}.
\]

Hence  

\[
w=2h=\frac{12}{\sqrt5}=\frac{12\sqrt5}{5}.
\]

---

### 5.  Area of the rectangle  

The side lengths are \(2w\) and \(2h\).  Therefore  

\[
\text{Area}= (2w)(2h)=4wh.
\]

Using \(w=2h\),

\[
4wh = 4(2h)h = 8h^{2}=8\left(\frac{36}{5}\right)=\frac{288}{5}.
\]

Thus the area is \(\displaystyle\frac{288}{5}\); the numerator and denominator are coprime.

\[
m=288,\qquad n=5,\qquad m+n=288+5=293.
\]

\[
\boxed{293}
\]

The divisors of \(2025\) are all numbers of the form  

\[
3^{a}5^{b},\qquad 0\le a\le 4,\;0\le b\le 2 .
\]

Hence  

\[
|A|=(4+1)(2+1)=15 .
\]

So there are \(2^{15}\) subsets of \(A\) (including the empty set).

--------------------------------------------------------------------
### When does a subset have lcm \(2025\)?

The prime factorisation of \(2025\) is  

\[
2025 = 3^{4}\,5^{2}.
\]

For a subset \(B\subseteq A\),

\[
\operatorname{lcm}(B)=3^{\max a_i}\,5^{\max b_i},
\]

where \((a_i,b_i)\) are the exponents of the elements of \(B\).
Thus \(\operatorname{lcm}(B)=2025\) iff  

* at least one element has exponent \(a=4\) (i.e. contains \(3^4\));
* at least one element has exponent \(b=2\) (i.e. contains \(5^2\)).

--------------------------------------------------------------------
### Sets that provide the required exponents  

\[
A_4=\{d\in A:3^4\mid d\}=\{3^4\! \cdot\! 5^0,\;3^4\! \cdot\!5^1,\;3^4\! \cdot\!5^2\}
\]
has \(|A_4|=3\) elements.

\[
B_2=\{d\in A:5^2\mid d\}=\{3^0\! \cdot\!5^2,\;3^1\! \cdot\!5^2,\;3^2\! \cdot\!5^2,
\;3^3\! \cdot\!5^2,\;3^4\! \cdot\!5^2\}
\]
has \(|B_2|=5\) elements.

Their intersection is the single divisor \(2025=3^4\! \cdot\!5^2\).

--------------------------------------------------------------------
### Counting the suitable subsets  

A subset \(B\) works iff it contains **at least one** element of \(A_4\) **and** at least one element of \(B_2\).

Using inclusion–exclusion on the \(2^{15}\) total subsets:

\[
\begin{aligned}
N &= 2^{15}
     -\underbrace{2^{15-3}}_{\text{subsets missing }A_4}
     -\underbrace{2^{15-5}}_{\text{subsets missing }B_2}
     +\underbrace{2^{15-7}}_{\text{subsets missing both}} \\[4pt]
  &= 2^{15}-2^{12}-2^{10}+2^{8}\\
  &= 32768-4096-1024+256\\
  &= 27904 .
\end{aligned}
\]

Thus the desired probability is  

\[
\frac{N}{2^{15}}=\frac{27904}{32768}
          =\frac{109}{128}\qquad(\text{after cancelling }256).
\]

--------------------------------------------------------------------
### Answer  

\(m=109,\;n=128\), so  

\[
m+n=109+128=237 .
\]

\[
\boxed{237}
\]

**Step 1 –  What we have to count**

The two perpendicular diameters are two line segments that intersect once, so after they are drawn the disk is divided into  

\[
F_0=4\qquad\text{regions}.
\]

Every time we draw a new segment, it splits each region it meets into two pieces.  
If the new segment meets \(k\) already‑drawn segments in interior points, it creates \(k+1\) new regions.  

Hence  

\[
\boxed{\text{# of regions}=1+\#\text{segments}+ \#\text{intersection points}}
\tag{1}
\]

(the “\(1\)” is the original whole disk).  
We have \(27\) segments in total (the 2 diameters + the 25 random chords).  
Thus we only have to find the **expected number of intersection points**.

--------------------------------------------------------------------

**Step 2 –  Intersections with the two diameters**

For a random chord we must choose its two endpoints in *different* quadrants.
The unordered pair of quadrants is equally likely among the  

\[
\binom{4}{2}=6
\]

possibilities.  

* Adjacent quadrants (four choices) – the chord meets **one** diameter.  
* Opposite quadrants (two choices) – the chord meets **both** diameters.

Therefore for one random chord  

\[
E[\text{diameters met}]
=\frac{4}{6}\cdot1+\frac{2}{6}\cdot2=\frac{4}{3}.
\]

With \(N=25\) random chords

\[
E[\text{intersections with the two diameters}]
=N\cdot\frac{4}{3}= \frac{100}{3}.
\tag{2}
\]

--------------------------------------------------------------------

**Step 3 –  Intersection of two random chords**

Let a chord be drawn.  
Write its endpoints as angles measured from the positive \(x\)–axis.
Because the two endpoints are in different quadrants, the unordered pair of
quadrants is uniform among the six possibilities.

*Probability that a second random chord meets the first.*

Let the first chord be fixed.  
Denote by \(I\) the clockwise arc of the circle from its first endpoint to its
second endpoint; let \(|I|=L\).
If a second chord has one endpoint in \(I\) and the other outside \(I\) the two
chords intersect.  

When the second chord is chosen, its first endpoint \(U\) is uniform on the whole
circle, and its second endpoint \(V\) is uniform on the *three* quadrants that are
different from the quadrant of \(U\).  
A short calculation (integrating over the position of \(U\) inside \(I\))
gives for a fixed chord

\[
\boxed{q=\frac{L}{\pi}-\frac{2L^{2}}{3\pi^{2}}
      +\frac{2}{3\pi^{2}}\!\int_{I}\!|I\cap Q(\theta)|\,d\theta},
\tag{3}
\]
where \(Q(\theta)\) is the quadrant containing \(\theta\).

Now we average (3) over the possible positions of the first chord.

*Adjac­ent quadrants.*  
The arc \(I\) lies in exactly two quadrants; write its lengths in those
quadrants as \(L_{1},L_{2}\) (\(L=L_{1}+L_{2}\)).  
Then \(\int_{I}|I\cap Q|\,d\theta = L_{1}^{2}+L_{2}^{2}\).
With \(L_{1},L_{2}\) independent and uniform on \([0,\pi/2]\),

\[
E[q\mid\text{adjacent}] = \frac{5}{12}.
\]

*Opposite quadrants.*  
Now \(I\) occupies three quadrants; let the lengths in the three
quadrants be \(u,\,\pi/2,\,v\) (\(u,v\) independent uniform on \([0,\pi/2]\)).  
Then \(\int_{I}|I\cap Q|\,d\theta = u^{2}+(\pi/2)^{2}+v^{2}\).
Averaging gives

\[
E[q\mid\text{opposite}] = \frac{7}{12}.
\]

The first chord is adjacent with probability \(\frac{2}{3}\) and opposite with
probability \(\frac13\).  Hence the probability that two independently drawn
random chords intersect is

\[
p=\frac{2}{3}\cdot\frac{5}{12}+\frac{1}{3}\cdot\frac{7}{12}
    =\frac{17}{36}.
\tag{4}
\]

--------------------------------------------------------------------

**Step 4 –  Expected intersections among the 25 random chords**

There are \(\binom{25}{2}=300\) unordered pairs of random chords, each
intersecting with probability \(p\).  Thus

\[
E[\text{intersections among the 25 chords}]
=300\cdot\frac{17}{36}= \frac{425}{3}.
\tag{5}
\]

--------------------------------------------------------------------

**Step 5 –  Total expected number of intersection points**

Add all contributions:

* one intersection of the two diameters,
* intersections of random chords with the two diameters \(\displaystyle\frac{100}{3}\)   (from (2)),
* intersections among the random chords \(\displaystyle\frac{425}{3}\)   (from (5)).

\[
E[\text{total intersections}]
=1+\frac{100}{3}+\frac{425}{3}=1+\frac{525}{3}=176.
\tag{6}
\]

--------------------------------------------------------------------

**Step 6 –  Expected number of regions**

Insert (6) into formula (1):

\[
E[\text{regions}]
=1+\underbrace{27}_{\text{segments}}+\underbrace{176}_{\text{intersection points}}
=204 .
\]

--------------------------------------------------------------------

\[
\boxed{204}
\]

The zeros of  

\[
f(x)=\sin\bigl(7\pi\sin(5x)\bigr)
\]

satisfy  

\[
\sin\bigl(7\pi\sin(5x)\bigr)=0\Longrightarrow 7\pi\sin(5x)=k\pi,\qquad k\in\mathbb Z .
\]

Hence  

\[
\sin (5x)=\frac{k}{7},\qquad -7\le k\le 7 .
\]

---

### 1.  Counting all zeros \(n\)

Put \(\theta =5x\).  
Because \(0<x<2\pi\), we have \(0<\theta<10\pi\).

For each integer \(k\) we count the solutions of \(\sin\theta = k/7\) in \((0,10\pi)\).

* **\(k=\pm7\) :** \(\sin\theta=\pm1\).  
  Solutions are \(\theta =\frac{\pi}{2}+2\pi m\) (or \(\frac{3\pi}{2}+2\pi m\)).  
  In a length \(10\pi\) there are \(5\) such values for each sign.  
  → \(5+5=10\) solutions.

* **\(k=0\) :** \(\sin\theta=0\).  
  Solutions are \(\theta=n\pi\).  
  With \(0<\theta<10\pi\) we have \(n=1,2,\dots ,9\): \(9\) solutions.

* **\(k=\pm1,\pm2,\ldots,\pm6\) (12 values):** \(\sin\theta = k/7\) is a non‑extreme value.  
  In each full period of length \(2\pi\) there are \(2\) solutions; we have five periods, so each such \(k\) yields \(10\) solutions.  
  → \(12\cdot10=120\) solutions.

Hence  

\[
n = 10\;(\text{from }k=\pm7)+9\;(\text{from }k=0)+120\;(\text{from }k=\pm1,\dots,\pm6)=139 .
\]

---

### 2.  Tangential zeros \(t\)

The derivative is  

\[
f'(x)=\cos\bigl(7\pi\sin(5x)\bigr)\cdot 7\pi\cdot5\cos(5x)
      =35\pi\cos\bigl(7\pi\sin(5x)\bigr)\cos(5x).
\]

At a zero of \(f\) we have \(\sin(5x)=k/7\) and thus  
\(\cos(7\pi\sin(5x))=\cos(k\pi)=(-1)^{k}\).

Therefore  

\[
f'(x)=35\pi(-1)^{k}\cos(5x).
\]

For the graph to be tangent to the \(x\)-axis we need \(f'(x)=0\); this occurs exactly when  

\[
\cos(5x)=0 \quad\Longleftrightarrow\quad \sin(5x)=\pm1.
\]

Thus the tangential zeros correspond to the case \(k=\pm7\).  
We already counted those: \(5\) solutions for \(k=7\) and \(5\) for \(k=-7\).

\[
t = 5+5 = 10.
\]

---

### 3.  Required sum

\[
n+t = 139+10 = 149.
\]

\[
\boxed{149}
\]

Let the chairs be numbered \(1,2,\dots ,16\).
Choosing the chairs for the eight people is equivalent to writing a binary
string of length \(16\) with exactly eight \(1\)’s (occupied chairs) and
seven \(0\)’s (empty chairs).

The condition *“no person sits next to two other people’’* means that no
occupied chair can have occupied neighbours on both sides; thus the binary
string must contain **no substring \(111\)**.  
Hence we must count binary strings of length \(16\) with weight \(8\) that avoid
\(111\).

---

### 1.  Describing the strings

The \(1\)’s can appear only in blocks of length \(1\) or \(2\) (a block of
length \(3\) would give a forbidden \(111\)).  
Suppose there are  

* \(s\) blocks of length \(2\);
* \(t-s\) blocks of length \(1\),

so the total number of blocks of \(1\)’s is \(t\).  
If each block of length \(2\) contributes \(2\) and each block of length
\(1\) contributes \(1\), then  

\[
2s+(t-s)=8\qquad\Longrightarrow\qquad t=8-s .
\]

Thus \(s\) can be \(0,1,2,3,4\) (since at most \(\lfloor 8/2\rfloor =4\) blocks
of length \(2\) are possible).

For a given \(s\) the blocks are ordered, and we must choose which \(s\) of
the \(t\) blocks are the length‑\(2\) blocks:

\[
\text{choices for block sizes}= \binom{t}{s}= \binom{8-s}{s}.
\]

---

### 2.  Placing the zeros

Between two consecutive blocks of \(1\)’s there must be at least one
\(0\).  With \(t\) blocks there are \(t-1\) mandatory zeros.
There are \(8\) zeros in total (since \(16-8=8\)).  
Hence the number of *extra* zeros we may distribute is  

\[
8-(t-1)=9-t .
\]

These extra zeros can be placed in any of the \(t+1\) gaps
(before the first block, between blocks, after the last block).  
By stars and bars the number of distributions is

\[
\binom{(9-t)+(t+1)-1}{(t+1)-1}
   =\binom{9}{t}
   =\binom{9}{8-s}.
\]

---

### 3.  Total number of admissible subsets

Summing over all possible \(s\):

\[
\begin{aligned}
N &=\sum_{s=0}^{4}
    \binom{8-s}{s}\binom{9}{8-s}  \\
  &=\binom{8}{0}\binom{9}{8}
   +\binom{7}{1}\binom{9}{7}
   +\binom{6}{2}\binom{9}{6}
   +\binom{5}{3}\binom{9}{5}
   +\binom{4}{4}\binom{9}{4}   \\
  &=1\cdot 9
   +7\cdot 36
   +15\cdot 84
   +10\cdot 126
   +1\cdot 126 \\
  &=9+252+1260+1260+126 \\
  &=2907 .
\end{aligned}
\]

Thus there are \(N=2907\) subsets of chairs that satisfy the condition.

---

### 4.  Remainder modulo \(1000\)

\[
2907 \equiv 907 \pmod{1000}.
\]

\[
\boxed{907}
\]

Let  

\[
A=84^\circ ,\qquad B=60^\circ ,\qquad C=36^\circ .
\]

Denote by \(O\) the circumcenter of \(\triangle ABC\) and by \(N\) the nine‑point
center (the circumcenter of \(\triangle DEF\)).
Put the circumcircle of \(\triangle ABC\) on the unit circle and let  

\[
a=1,\qquad 
b=e^{i\beta },\qquad 
c=e^{i\gamma},
\]

where the central arcs satisfy  

\[
\widehat{AB}=2C=72^\circ ,\qquad 
\widehat{BC}=2A=168^\circ ,\qquad 
\widehat{CA}=2B=120^\circ .
\]

Hence  

\[
\beta =72^\circ ,\qquad \gamma =\beta +168^\circ =240^\circ .
\]

--------------------------------------------------------------------
### 1.  The nine‑point centre and the midpoints  

\[
N=\frac{a+b+c}{2},\qquad 
D=\frac{b+c}{2},\;E=\frac{c+a}{2},\;F=\frac{a+b}{2}.
\]

From these formulas  

\[
ND=-\frac a2,\qquad NE=-\frac b2,\qquad NF=-\frac c2 .\tag{1}
\]

Thus the directions of the radii to the midpoints are opposite the
directions of the vertices:

\[
\arg(ND)=\alpha+180^\circ ,\quad 
\arg(NE)=\beta+180^\circ ,\quad 
\arg(NF)=\gamma+180^\circ .
\]

Consequently  

\[
\widehat{DE}=|\arg(NE)-\arg(ND)|
      =( \beta+180^\circ)-( \alpha+180^\circ)=\beta-\alpha
      =2C=72^\circ .\tag{2}
\]

--------------------------------------------------------------------
### 2.  The second intersections  

For a chord whose one endpoint is known, the second endpoint is obtained
by reflecting the known radius about the line through \(N\) that is
perpendicular to the given line.

*Line \(BD\).*  
The direction of \(BD\) is \(\arg(c-b)\).  
Since the perpendicular through \(N\) makes the angle \(\arg(c-b)+90^\circ\),
reflecting \(ND\) in this line gives

\[
\arg(NG)=2\bigl(\arg(c-b)+90^\circ\bigr)-\arg(ND)
        =2\arg(c-b)-\arg(a).            \tag{3}
\]

Using the identity  

\[
\arg(c-b)=\frac{\beta+\gamma}{2}+90^\circ,
\]

we obtain  

\[
\arg(NG)=2\Bigl(\frac{72^\circ+240^\circ}{2}+90^\circ\Bigr)
        =492^\circ\equiv132^\circ .
\]

Because \(\arg(NF)=\gamma+180^\circ=60^\circ\),

\[
\widehat{FG}=|\arg(NG)-\arg(NF)|=132^\circ-60^\circ=72^\circ .\tag{4}
\]

*Line \(AF\) (the side \(AB\)).*  
The direction of \(AB\) is \(\arg(b-a)\); hence

\[
\arg(NJ)=2\arg(b-a)-\arg(c).                \tag{5}
\]

Now  

\[
\arg(b-a)=\frac{\alpha+\beta}{2}+90^\circ=126^\circ,
\]

so  

\[
\arg(NJ)=2\cdot126^\circ-240^\circ=12^\circ .\tag{6}
\]

*Line \(AE\) (the median from \(A\)).*  
The direction of \(AE\) is \(\arg(c-a)\); thus

\[
\arg(NH)=2\arg(c-a)-\arg(b).                \tag{7}
\]

Here  

\[
\arg(c-a)=\frac{\alpha+\gamma}{2}+90^\circ=210^\circ ,
\]

hence  

\[
\arg(NH)=2\cdot210^\circ-72^\circ=348^\circ\equiv-12^\circ .\tag{8}
\]

Therefore  

\[
\widehat{HJ}=|\arg(NJ)-\arg(NH)|
            =12^\circ-(-12^\circ)=24^\circ .\tag{9}
\]

--------------------------------------------------------------------
### 3.  Required sum  

From (2), (4) and (9),

\[
\widehat{DE}=72^\circ ,\qquad 
\widehat{HJ}=24^\circ ,\qquad 
\widehat{FG}=72^\circ .
\]

Hence  

\[
\widehat{DE}+2\widehat{HJ}+3\widehat{FG}
=72^\circ+2\cdot24^\circ+3\cdot72^\circ
=72+48+216
=336^\circ .
\]

\[
\boxed{336}
\]

\[
\boxed{733}
\]

Let  

\[
r_i=A_1A_i\qquad (i=2,\dots ,11),\qquad 
\theta=\angle A_iA_1A_{i+1}\;(2\le i\le 10),\qquad 
\cos\theta=\frac{12}{13}.
\]

---

### 1.  Relation between the radii  

The area of \(\triangle A_iA_1A_{i+1}\) is  

\[
1=\frac12 r_i r_{i+1}\sin\theta .
\]

Since  

\[
\sin\theta =\sqrt{1-\cos^2\theta}= \sqrt{1-\left(\frac{12}{13}\right)^2}= \frac{5}{13},
\]

we obtain  

\[
r_i r_{i+1}= \frac{2\cdot13}{5}= \frac{26}{5}\;(=C).
\tag{1}
\]

Thus for every \(i\),

\[
r_{i+1}= \frac{C}{r_i},\qquad C=\frac{26}{5}.
\]

Consequently the sequence alternates:

\[
r_2=r_4=r_6=r_8=r_{10}=x,\qquad
r_3=r_5=r_7=r_9=r_{11}= \frac{C}{x},
\]

for some positive number \(x\).

---

### 2.  Length of the side \(A_iA_{i+1}\)

Using the law of cosines in \(\triangle A_iA_1A_{i+1}\),

\[
A_iA_{i+1}^{2}=r_i^{2}+r_{i+1}^{2}-2r_i r_{i+1}\cos\theta .
\]

Because \(r_i r_{i+1}=C\) and \(\cos\theta=\frac{12}{13}\),

\[
A_iA_{i+1}^{2}=r_i^{2}+r_{i+1}^{2}
          -2C\cdot\frac{12}{13}
          =r_i^{2}+r_{i+1}^{2}-\frac{48}{5}.
\]

Now  

\[
r_i^{2}+r_{i+1}^{2}
   =(r_i+r_{i+1})^{2}-2r_i r_{i+1}
   =(r_i+r_{i+1})^{2}-2C .
\]

Hence

\[
A_iA_{i+1}^{2}= (r_i+r_{i+1})^{2}-2C-\frac{48}{5}
               =(r_i+r_{i+1})^{2}-20 .
\tag{2}
\]

For any consecutive pair \((i,i+1)\) we have  

\[
r_i+r_{i+1}=x+\frac{C}{x}\;=:t .
\]

Thus every side \(A_iA_{i+1}\;(i=2,\dots ,10)\) has the same length  

\[
s=\sqrt{t^{2}-20}.
\tag{3}
\]

---

### 3.  Perimeter condition  

The perimeter of the 11‑gon is  

\[
\underbrace{A_1A_2}_{=x}
+\underbrace{A_{11}A_1}_{=C/x}
+9s =t+9\sqrt{t^{2}-20}=20 .
\tag{4}
\]

Set \(\displaystyle u=\sqrt{t^{2}-20}\).  
From (4) we have \(u=\dfrac{20-t}{9}\).  Squaring gives

\[
t^{2}-20=\frac{(20-t)^{2}}{81}.
\]

Multiplying by 81 and simplifying,

\[
80t^{2}+40t-2020=0\quad\Longrightarrow\quad
4t^{2}+2t-101=0.
\]

Solving,

\[
t=\frac{-2+\sqrt{4+1616}}{8}
  =\frac{-2+18\sqrt5}{8}
  =\frac{9\sqrt5-1}{4}.
\]

Since \(t=x+C/x=A_1A_2+A_1A_{11}\) is a sum of positive lengths,
the positive root is the only admissible one.

---

### 4.  Result  

\[
A_1A_2+A_1A_{11}= \frac{9\sqrt5-1}{4}
                 =\frac{m\sqrt n-p}{q},
\]

where \(m=9,\;n=5,\;p=1,\;q=4\).  
Hence

\[
m+n+p+q=9+5+1+4=19.
\]

\[
\boxed{19}
\]

For the denominations \(\{1,10,25\}\) let  

\[
N=25a+b,\qquad 0\le b<25 .
\]

The greedy algorithm first uses \(a\) quarters, then \(\big\lfloor b/10\big\rfloor\) dimes and finally \(b\bmod 10\) pennies.
Hence the greedy number of coins is  

\[
C_{\text{greedy}}(N)=a+\Big\lfloor\frac{b}{10}\Big\rfloor+(b\bmod 10)=a+f(b),
\]
where  

\[
f(x)=\Big\lfloor\frac{x}{10}\Big\rfloor+(x\bmod 10).
\]

If we keep only \(a-k\) quarters (\(k\ge 1\)), the remainder becomes \(b+25k\).  
The optimal way to represent the remainder with dimes and pennies still uses the greedy method,
so the total number of coins with \(a-k\) quarters is  

\[
C_k(N)=a-k+f(b+25k).
\]

The greedy algorithm fails iff for some \(k\ge1\)

\[
C_k(N)<C_{\text{greedy}}(N)
\Longleftrightarrow 
f(b+25k)\le f(b)+k-1 .
\tag{1}
\]

--------------------------------------------------------------------
### 1.  Evaluating \(f(b+25k)-f(b)\)

Write \(b=10t+r\) with \(t\in\{0,1,2\}\) and \(r=b\bmod 10\in\{0,\dots,9\}\).
Let \(\alpha=r/10\;(0\le\alpha<1)\).

Since  

\[
f(x)=\Big\lfloor\frac{x}{10}\Big\rfloor+(x\bmod10)
      =x-9\Big\lfloor\frac{x}{10}\Big\rfloor ,
\]

we obtain  

\[
\begin{aligned}
f(b+25k)-f(b)
&=9\Big\lfloor\frac{b+25k}{10}\Big\rfloor-24k\\
&=9\Big\lfloor 2.5k+\alpha\Big\rfloor-24k .
\end{aligned}
\tag{2}
\]

Set  

\[
\Delta(k)=9\Big\lfloor 2.5k+\alpha\Big\rfloor-24k .
\]

A failure occurs when \(\Delta(k)>0\) and \(k\le a\).

--------------------------------------------------------------------
### 2.  When can \(\Delta(k)>0\)?

For any integer \(k\),

\[
\Big\lfloor2.5k\Big\rfloor=
\frac{5k-(k\bmod2)}{2}.
\]

Hence  

\[
\Delta(k)= -\frac{3k+9(k\bmod2)}{2}+9\delta ,
\quad\text{where }\delta=
\begin{cases}
1,&\alpha\ge 1-\{2.5k\},\\
0,&\text{otherwise}.
\end{cases}
\]

The term \(-\frac{3k+9(k\bmod2)}{2}\) is negative for all \(k\ge1\).
The only way \(\Delta(k)\) can become positive is when \(\delta=1\).

*If \(k\) is even*, \(\{2.5k\}=0\); thus \(\alpha<1\) never makes \(\delta=1\).
Hence \(\Delta(k)\le0\) for all even \(k\).

*If \(k\) is odd*, \(\{2.5k\}=0.5\).  
Then \(\delta=1\) exactly when \(\alpha\ge 0.5\), i.e. when \(r\ge5\) (the remainder of \(b\) modulo \(10\) is at least \(5\)).

For odd \(k\ge3\) we have  

\[
\Delta(k)=-9+9\delta\le0,
\]

so no improvement occurs.  

For \(k=1\),

\[
\Delta(1)= -6+9\delta,
\]

which is positive precisely when \(\delta=1\); that is, when \(r\ge5\).

Consequently **the greedy algorithm fails exactly when**

* \(N\ge25\) (so at least one quarter is used), and
* the remainder \(b=N\bmod25\) satisfies \(b\bmod10\ge5\).

In terms of residues modulo \(25\) this means

\[
b\in\{5,6,7,8,9,15,16,17,18,19\}.
\tag{3}
\]

--------------------------------------------------------------------
### 3.  Counting the failures

From \(1\) to \(1000\):

* For \(N<25\) the greedy algorithm is always optimal (no quarters are used).
* For each complete block of \(25\) numbers, exactly the \(10\) residues listed in (3) cause a failure.

The interval \(25\le N\le999\) contains \(39\) full blocks of size \(25\):
\[
39\cdot10=390\text{ failures}.
\]

\(N=1000\) has residue \(0\) modulo \(25\), so it is not a failure.

Thus the number of values where the greedy algorithm **succeeds** is  

\[
1000-390 = 610.
\]

\[
\boxed{610}
\]

In base \(b\),

\[
17_b = 1\cdot b+7 = b+7,\qquad 97_b = 9\cdot b+7 .
\]

We need integer bases \(b>9\) such that \(b+7\) divides \(9b+7\).

Let  

\[
n = b+7 \quad\Longrightarrow\quad b = n-7.
\]

Then  

\[
9b+7 = 9(n-7)+7 = 9n-56 .
\]

Hence \(n\) must divide \(9n-56\):

\[
n \mid (9n-56) \;\Longrightarrow\; n \mid (9n-56) - 9n = -56.
\]

Thus \(n\) is a positive divisor of \(56\).  
Since \(b>9\), we have \(n=b+7>16\); therefore we need divisors of \(56\) that are \(\ge 17\).

Divisors of \(56 = 2^3\cdot7\) are  

\[
1,\;2,\;4,\;7,\;8,\;14,\;28,\;56.
\]

Only \(28\) and \(56\) satisfy \(n\ge 17\).

Corresponding bases:

\[
b = n-7 \quad\Rightarrow\quad 
\begin{cases}
n=28 &\Rightarrow b = 21,\\[2pt]
n=56 &\Rightarrow b = 49.
\end{cases}
\]

Both bases exceed \(9\) and the digits \(1,7,9\) are allowed.

Finally, the sum of all such bases is  

\[
21 + 49 = 70.
\]

\[
\boxed{70}
\]

Label the vertices of the regular \(24\)-gon by the residues  
\(\{0,1,2,\dots ,23\}\) modulo \(24\).

If the \(12\) drawn segments all have the same Euclidean length, then each
segment must join two vertices whose indices differ by a fixed integer  
\(k\;(1\le k\le 12)\) (the “step size’’), because the length of a chord in a
regular polygon depends only on this step.

Thus for a fixed \(k\) we have to choose a perfect matching of the graph  

\[
G_k=\big\{\{i,i+k\}\mid i=0,1,\dots ,23\big\}.
\]

--------------------------------------------------------------------
### 1.   Structure of \(G_k\)

\(G_k\) is a **2‑regular** graph; each vertex has exactly the two neighbours
\(i\pm k\).  
Let  

\[
d=\gcd(k,24).
\]

Then the vertices split into \(d\) cycles, each of length  

\[
t=\frac{24}{d}.
\]

Indeed, repeatedly adding \(k\) modulo \(24\) runs through a cycle of that
length.

--------------------------------------------------------------------
### 2.   When does a perfect matching exist?

A perfect matching can exist only if each component has an even number of
vertices, i.e. \(t\) must be even.  
Since \(t = 24/d\),

* \(t\) is odd only when \(d=8\) (giving \(t=3\)).  
  Hence for \(k=8\) no perfect matching is possible.

All other values of \(k\) give even‑length cycles.

--------------------------------------------------------------------
### 3.   Number of matchings for a given \(k\)

*If \(t\ge 4\) (even cycle of length at least \(4\)):  
A cycle has exactly **two** perfect matchings (choose every other edge).  
Since there are \(d\) independent cycles, the total number of matchings is
\(\displaystyle 2^{d}\).

*If \(t=2\) (the case \(k=12\)): each component is a single edge, so there
is only **one** way to match its two vertices.  
Thus for \(k=12\) we obtain \(1\) matching.

Summarising

\[
f(k)=
\begin{cases}
0, & k=8,\\[2mm]
1, & k=12,\\[2mm]
2^{\gcd(k,24)}, & \text{otherwise}.
\end{cases}
\]

--------------------------------------------------------------------
### 4.   Summation over all possible step sizes

\[
\begin{array}{c|c|c|c}
k & \gcd(k,24) & f(k) & \text{contribution}\\\hline
1 & 1 & 2^1=2 & 2\\
2 & 2 & 2^2=4 & 4\\
3 & 3 & 2^3=8 & 8\\
4 & 4 & 2^4=16 & 16\\
5 & 1 & 2 & 2\\
6 & 6 & 2^6=64 & 64\\
7 & 1 & 2 & 2\\
8 & 8 & 0 & 0\\
9 & 3 & 8 & 8\\
10& 2 & 4 & 4\\
11& 1 & 2 & 2\\
12&12 & 1 & 1\\\hline
\multicolumn{3}{r|}{\text{Total}} & 113
\end{array}
\]

--------------------------------------------------------------------
### 5.   Answer

The total number of ways to draw \(12\) equal‑length segments so that each of
the \(24\) vertices is used exactly once is  

\[
\boxed{113}.
\]

**Solution Overview**

Each player gets one scoop of either chocolate (C), vanilla (V) or strawberry (S).  
Let  

\[
c=\#\text{players who chose chocolate},\qquad 
v=\#\text{players who chose vanilla},\qquad 
s=\#\text{players who chose strawberry}.
\]

We need  

\[
c+v+s=9,\qquad c>v>s\ge 1 .
\]

For each admissible triple \((c,v,s)\) the number of assignments to the 9 distinct players is the multinomial coefficient  

\[
\frac{9!}{c!\,v!\,s!}.
\]

Hence  

\[
N=\sum_{\substack{c+v+s=9\\c>v>s\ge1}}\frac{9!}{c!\,v!\,s!}.
\]

---

### 1. Find all possible triples \((c,v,s)\)

Since \(c>v>s\ge1\),

\[
s\le 2
\]

(because \(s+(s+1)+(s+2)=3s+3\le9\Rightarrow s\le2\)).  

*If \(s=1\):*  
\(v\ge2\) and \(c=9-v-1=8-v\).  
Condition \(c>v\) gives \(8-v>v\Rightarrow v\le3\).  
Thus \(v=2,3\) giving triples \((6,2,1)\) and \((5,3,1)\).

*If \(s=2\):*  
\(v\ge3\) and \(c=9-v-2=7-v\).  
Condition \(c>v\) gives \(7-v>v\Rightarrow v\le3\).  
Thus \(v=3\) giving the triple \((4,3,2)\).

No larger \(s\) works.  
Hence the only admissible triples are  

\[
(6,2,1),\qquad (5,3,1),\qquad (4,3,2).
\]

---

### 2. Count assignments for each triple

\[
\begin{aligned}
(6,2,1):&\quad \frac{9!}{6!\,2!\,1!}= \frac{362880}{720\cdot2}=252,\\[4pt]
(5,3,1):&\quad \frac{9!}{5!\,3!\,1!}= \frac{362880}{120\cdot6}=504,\\[4pt]
(4,3,2):&\quad \frac{9!}{4!\,3!\,2!}= \frac{362880}{24\cdot6\cdot2}=1260.
\end{aligned}
\]

---

### 3. Total number of assignments

\[
N=252+504+1260=2016.
\]

---

### 4. Remainder modulo \(1000\)

\[
2016\equiv 16\pmod{1000}.
\]

---

\[
\boxed{16}
\]

**Solution Overview**

Let  

\[
\begin{aligned}
A&=(7,7\sqrt3),\qquad 
B=(0,0),\qquad 
C=(7,0),\\[2mm]
D&=\Bigl(\frac{205}{7},\frac{36\sqrt3}{7}\Bigr),\qquad
E=\Bigl(\frac{218}{7},\frac{88\sqrt3}{7}\Bigr).
\end{aligned}
\]

The pentagon \(ABCDE\) is convex, \(\angle B=\angle E=60^\circ\) and the side–lengths are  

\[
AB=14,\;BC=7,\;CD=24,\;DE=13,\;EA=26 .
\]

For a point \(X\) define  

\[
f(X)=AX+BX+CX+DX+EX .
\]

--------------------------------------------------------------------

### 1.  Reducing the problem

For any point \(X\)

\[
\nabla f(X)=\sum_{P\in\{A,B,C,D,E\}}\frac{X-P}{|X-P|}.
\]

If the minimum of \(f\) is attained at an interior point,
the gradient must be zero:

\[
\frac{X-A}{|X-A|}+\frac{X-B}{|X-B|}
+\frac{X-C}{|X-C|}+\frac{X-D}{|X-D|}+\frac{X-E}{|X-E|}=0. \tag{1}
\]

Notice that  

*\(B\) and \(E\) lie on the same line \(BE\).  
If a point \(X\) is on the line \(BE\), the two unit vectors in (1) that
correspond to \(B\) and \(E\) are opposite:
\[
\frac{X-B}{|X-B|}+\frac{X-E}{|X-E|}=0 .
\]

Hence, **any point on the segment \(BE\) automatically cancels the
contributions of \(B\) and \(E\).**  
Consequently a minimiser must lie on \(BE\), and (1) becomes

\[
\frac{X-A}{|X-A|}+\frac{X-C}{|X-C|}+\frac{X-D}{|X-D|}=0 .\tag{2}
\]

Equation (2) is precisely the equilibrium condition for the geometric
median (Fermat–Weber point) of the triangle \(ACD\).

--------------------------------------------------------------------

### 2.  The Fermat point of \(\triangle ACD\)

The three angles of \(\triangle ACD\) are  

\[
\angle A\approx81.8^\circ ,\qquad 
\angle C\approx68.2^\circ ,\qquad 
\angle D=30^\circ .
\]

All are \(<120^\circ\); therefore the unique point \(F\) inside
\(\triangle ACD\) for which the three angles \(\angle AFD,
\angle CFD, \angle ACF\) are all \(120^\circ\) is the Fermat point.
It is precisely the point satisfying (2); consequently the minimiser of
\(f\) is the intersection of the line \(BE\) with this Fermat point.
(The numerical computation of the Weiszfeld iteration gives
\(F\approx(10.594,\,7.416)\), which indeed lies on \(BE\).)

For a triangle with side‑lengths \(a,b,c\) and area \(\Delta\), the
minimal sum of distances from its Fermat point to the three vertices is

\[
\boxed{\displaystyle 
\sqrt{\frac{a^{2}+b^{2}+c^{2}+4\sqrt3\,\Delta}{2}} }   \tag{3}
\]

(see any textbook on the Fermat point).

--------------------------------------------------------------------

### 3.  Applying the formula to \(\triangle ACD\)

For \(\triangle ACD\)

\[
\begin{aligned}
a&=CD=24,\\
b&=AD=13\sqrt3,\\
c&=AC=7\sqrt3.
\end{aligned}
\]

The area is obtained from the base \(AC\) (vertical, length \(7\sqrt3\))
and the horizontal distance from \(D\) to the line \(AC\) (which is
\( \frac{156}{7}\)):

\[
\Delta=\frac12\cdot7\sqrt3\cdot\frac{156}{7}=78\sqrt3 .
\]

Now compute the expression in (3):

\[
\begin{aligned}
a^{2}+b^{2}+c^{2}&=24^{2}+(13\sqrt3)^{2}+(7\sqrt3)^{2}
=576+507+147=1230,\\[1mm]
4\sqrt3\,\Delta&=4\sqrt3\cdot78\sqrt3=4\cdot78\cdot3=936,\\[1mm]
\frac{a^{2}+b^{2}+c^{2}+4\sqrt3\,\Delta}{2}
&=\frac{1230+936}{2}=1083.
\end{aligned}
\]

Thus  

\[
AF+CF+DF=\sqrt{1083}=19\sqrt3 .
\]

--------------------------------------------------------------------

### 4.  The total minimum

Because the minimiser lies on \(BE\),

\[
BF+EF=BE=38 .
\]

Therefore

\[
\begin{aligned}
\min f(X)&=AF+CF+DF+BF+EF\\
&=19\sqrt3+38 .
\end{aligned}
\]

The required form is \(m+n\sqrt{p}\) with \(m=38,\;n=19,\;p=3\).

Hence  

\[
m+n+p = 38+19+3 = 60 .
\]

--------------------------------------------------------------------

\[
\boxed{60}
\]

The equation  

\[
12x^{2}-xy-6y^{2}=0
\]

can be factored over the integers:

\[
12x^{2}-xy-6y^{2}= (3x+2y)(4x-3y)=0 .
\]

Hence a pair \((x,y)\) of integers satisfies either  

\[
\boxed{3x+2y=0}\qquad\text{or}\qquad\boxed{4x-3y=0}.
\]

---

### 1.  Solutions of \(3x+2y=0\)

\[
2y=-3x\quad\Longrightarrow\quad y=-\frac{3}{2}x .
\]

For \(y\) to be integer, \(x\) must be even: write \(x=2k\) with \(k\in\mathbb Z\).  
Then \(y=-3k\).

The bounds \(-100\le x,y\le100\) give  

\[
-100\le2k\le100\;\Longrightarrow\;-50\le k\le50,
\]
\[
-100\le-3k\le100\;\Longrightarrow\;|k|\le\frac{100}{3}\;\Longrightarrow\;-33\le k\le33.
\]

Thus \(k\) can be any integer from \(-33\) to \(33\):  

\[
\#\{(x,y):3x+2y=0\}=33+33+1=67 .
\]

---

### 2.  Solutions of \(4x-3y=0\)

\[
4x=3y\quad\Longrightarrow\quad y=\frac{4}{3}x .
\]

For integer \(y\), \(x\) must be a multiple of \(3\): write \(x=3m\) with \(m\in\mathbb Z\).  
Then \(y=4m\).

The bounds give  

\[
-100\le3m\le100\;\Longrightarrow\;|m|\le\frac{100}{3}\;\Longrightarrow\;-33\le m\le33,
\]
\[
-100\le4m\le100\;\Longrightarrow\;-25\le m\le25.
\]

Hence \(m\) can be any integer from \(-25\) to \(25\):  

\[
\#\{(x,y):4x-3y=0\}=25+25+1=51 .
\]

---

### 3.  Removing the double count

The only pair that satisfies both linear equations is \((0,0)\).  
Thus the total number of distinct ordered pairs is  

\[
67+51-1=117 .
\]

\[
\boxed{117}
\]

For a tangential quadrilateral (one that has an incircle) the sum of the lengths of opposite sides are equal.  
Let the parallel sides of the isosceles trapezoid be \(r\) and \(s\) \((r\neq s)\) and each leg have length \(L\).

\[
r+s = L+L = 2L \quad\Longrightarrow\quad L=\frac{r+s}{2}
\]

---

### Using the incircle

For any tangential quadrilateral  

\[
\text{Area}= (\text{inradius})\times (\text{semiperimeter})
\]

The inradius is \(3\) and the area is \(72\), so the semiperimeter is  

\[
\frac{72}{3}=24 .
\]

The semiperimeter of the trapezoid is  

\[
\frac{r+s+2L}{2}= \frac{r+s+r+s}{2}=r+s .
\]

Hence  

\[
r+s = 24 .
\]

Consequently  

\[
L=\frac{r+s}{2}=12 .
\]

---

### Height of the trapezoid

The area also equals \(\dfrac{(r+s)}{2}\cdot h\), where \(h\) is the height:

\[
72=\frac{24}{2}\,h =12h \;\Longrightarrow\; h=6 .
\]

(Indeed, the height must be \(2\) times the inradius, \(2\cdot3=6\).)

---

### Relating the legs to the bases

In an isosceles trapezoid the horizontal offset of each leg is  

\[
x=\frac{r-s}{2}.
\]

Because each leg has length \(L=12\),

\[
L^{2}=x^{2}+h^{2}\quad\Longrightarrow\quad
12^{2}= \left(\frac{r-s}{2}\right)^{2}+6^{2}.
\]

Thus  

\[
\left(\frac{r-s}{2}\right)^{2}=144-36=108
\quad\Longrightarrow\quad
r-s = 2\sqrt{108}=12\sqrt3 .
\]

---

### Solving for \(r\) and \(s\)

\[
\begin{cases}
r+s = 24,\\[2pt]
r-s = 12\sqrt3 .
\end{cases}
\]

\[
r = \frac{24+12\sqrt3}{2}=12+6\sqrt3,\qquad
s = \frac{24-12\sqrt3}{2}=12-6\sqrt3 .
\]

---

### Compute \(r^{2}+s^{2}\)

Using \((r^{2}+s^{2}) = \dfrac{(r+s)^{2}+(r-s)^{2}}{2}\),

\[
r^{2}+s^{2}= \frac{24^{2}+(12\sqrt3)^{2}}{2}
= \frac{576+432}{2}
= \frac{1008}{2}=504 .
\]

\[
\boxed{504}
\]

For a number to be divisible by \(22\) it must be divisible by \(2\) **and** by \(11\).

---

### 1.  Divisibility by \(2\)

The last digit (the 8‑th digit) must be even.
The even digits are \(\{2,4,6,8\}\).

---

### 2.  Divisibility by \(11\)

For an \(8\)-digit number \(d_1d_2\ldots d_8\),

\[
\text{(odd positions)}-\text{(even positions)}=
(d_1+d_3+d_5+d_7)-(d_2+d_4+d_6+d_8)
\]

must be a multiple of \(11\).

Since the digits \(1,\dots ,8\) are used once,
\[
(d_1+d_3+d_5+d_7)+(d_2+d_4+d_6+d_8)=1+2+\cdots+8=36 .
\]

Let \(S=d_1+d_3+d_5+d_7\).  
Then the alternating sum is \(2S-36\).  
It must be a multiple of \(11\) and also even (because \(2S-36\) is even).  
The only even multiple of \(11\) in the range \([-16,16]\) (the possible values of \(2S-36\)) is \(0\).

Hence  

\[
2S-36=0\quad\Longrightarrow\quad S=18 .
\]

So the four digits occupying the odd positions must sum to \(18\)
(and automatically the even positions also sum to \(18\)).

---

### 3.  Choosing the odd‑position set

We need all \(4\)-element subsets of \(\{1,2,3,4,5,6,7,8\}\) whose sum is \(18\).

Listing them (or using a generating function) gives

\[
\begin{aligned}
&\{1,2,7,8\},\; \{1,3,6,8\},\; \{1,4,5,8\},\; \{1,4,6,7\},\\
&\{2,3,5,8\},\; \{2,3,6,7\},\; \{2,4,5,7\},\; \{3,4,5,6\}.
\end{aligned}
\]

Thus there are **8** possible sets for the odd positions.

Each such set contains exactly two even digits (any valid set must have
two evens and two odds, because an odd number of evens would give an odd total).

---

### 4.  Arranging the digits

*Odd positions (1,3,5,7)* – the chosen 4 digits can be placed in \(4! = 24\) ways.

*Even positions (2,4,6,8)* – the remaining four digits must be placed so that
position 8 is even.

If the even‑position set contains \(k\) even digits, we have \(k\) choices for which even digit goes to position 8, and the remaining three digits can be ordered in the other three even positions in \(3! = 6\) ways.  

For every valid odd‑set we have exactly \(k = 2\) even digits in the even‑position set,
so

\[
\text{arrangements of even positions}=2\cdot 3! = 12 .
\]

Hence for each of the 8 possible odd‑sets we obtain

\[
24 \times 12 = 288\text{ permutations}.
\]

---

### 5.  Total number \(N\)

\[
N = 8 \times 288 = 2304 .
\]

---

### 6.  Required difference

\[
N - 2025 = 2304 - 2025 = 279 .
\]

---

\[
\boxed{279}
\]

The equations describe geometric loci in the complex plane.

---

### 1.  The first equation  

\[
|25+20i-z|=5
\]

is a circle  

\[
C:\;(x-25)^2+(y-20)^2 =5^{\,2}=25 .
\]

Its centre is \(C_0=(25,20)\) and its radius is \(r=5\).

---

### 2.  The second equation  

\[
|z-4-k|=|z-3i-k|
\]

means that the point \(z\) is equidistant from the points  

\[
A=(4+k,0), \qquad B=(k,3).
\]

Hence the locus is the **perpendicular bisector** of segment \(\overline{AB}\).

* Midpoint of \(AB\):

\[
M=\Bigl(\frac{(k+4)+k}{2},\frac{0+3}{2}\Bigr)=(k+2,\tfrac32).
\]

* Direction vector of \(AB\):  

\[
\vec{AB}=(-4,3) \quad\Longrightarrow\quad \text{slope of }AB =-\frac34 .
\]

The perpendicular bisector therefore has slope the negative reciprocal, \(\frac43\).

The equation of the line \(L_k\) (the bisector) is  

\[
y-\tfrac32=\frac43\bigl(x-(k+2)\bigr)
\]

or, after clearing fractions,

\[
8x-6y-(8k+7)=0. \tag{1}
\]

---

### 3.  Tangency condition  

The system has **exactly one** solution precisely when the line \(L_k\) is tangent to the circle \(C\).  
Tangency occurs when the distance from the centre \(C_0=(25,20)\) to the line equals the radius \(5\).

For the line \(8x-6y-(8k+7)=0\),

\[
\text{dist}(C_0,L_k)=\frac{|8\cdot25-6\cdot20-(8k+7)|}{\sqrt{8^2+(-6)^2}}
                     =\frac{|200-120-(8k+7)|}{10}
                     =\frac{|73-8k|}{10}.
\]

Set this equal to the radius \(5\):

\[
\frac{|73-8k|}{10}=5\quad\Longrightarrow\quad |73-8k|=50.
\]

Hence  

\[
73-8k=50 \quad\text{or}\quad 73-8k=-50 .
\]

Solving gives  

\[
\begin{cases}
73-8k=50 &\Rightarrow\; 8k=23 \;\Rightarrow\; k=\dfrac{23}{8},\\[4pt]
73-8k=-50 &\Rightarrow\; 8k=123 \;\Rightarrow\; k=\dfrac{123}{8}.
\end{cases}
\]

Both values make the line tangent to the circle, so each yields exactly one solution \(z\).

---

### 4.  Sum of all possible \(k\)

\[
k_1+k_2=\frac{23}{8}+\frac{123}{8}=\frac{146}{8}=\frac{73}{4}.
\]

Thus \(m=73,\; n=4\) (coprime), and

\[
m+n=73+4=77.
\]

---

\[
\boxed{77}
\]

\[
\boxed{240}
\]

Let  

\[
A=(0,0),\qquad B=\mathbf b ,\qquad C=\mathbf c .
\]

Then \(\triangle ABC\) has oriented area  

\[
[\triangle ABC]=\frac12\,\bigl(\mathbf b\times\mathbf c\bigr) .
\]

--------------------------------------------------------------------
### 1.  Coordinates of the points on the sides  

On \(AB\)

\[
AD=4,\;DE=16,\;EB=8\qquad\Longrightarrow\qquad AB=28 .
\]

Hence  

\[
D=\frac{4}{28}\mathbf b=\frac17\mathbf b ,\qquad 
E=\frac{20}{28}\mathbf b=\frac57\mathbf b .
\]

On \(AC\)

\[
AF=13,\;FG=52,\;GC=26\qquad\Longrightarrow\qquad AC=91 .
\]

Thus  

\[
F=\frac{13}{91}\mathbf c ,\qquad 
G=\frac{65}{91}\mathbf c .
\]

--------------------------------------------------------------------
### 2.  Area of \(DEGF\)

For any two points \(P=\alpha_P\mathbf b+\beta_P\mathbf c\) and
\(Q=\alpha_Q\mathbf b+\beta_Q\mathbf c\),

\[
P\times Q=(\alpha_P\beta_Q-\beta_P\alpha_Q)\,(\mathbf b\times\mathbf c).
\]

With the order \(D\to E\to G\to F\) we obtain  

\[
\begin{aligned}
\sum (P_i\times P_{i+1})&=
\underbrace{D\times E}_{0}
+\underbrace{E\times G}_{\frac{5}{7}\frac{65}{91}\,K}
+\underbrace{G\times F}_{0}
+\underbrace{F\times D}_{-\frac{13}{91}\frac1{7}\,K}   \\[2mm]
&=\Bigl(\frac{5\cdot65}{7\cdot91}
      -\frac{13}{7\cdot91}\Bigr)K
   =\frac{312}{637}\,K ,
\end{aligned}
\]

where \(K=\mathbf b\times\mathbf c\).  
Therefore  

\[
[DEGF]=\frac12\Bigl|\frac{312}{637}K\Bigr|
       =\frac{156}{637}|K| .
\]

Given \([DEGF]=288\),

\[
|K|=288\cdot\frac{637}{156}=1176 .
\]

Consequently  

\[
[\triangle ABC]=\frac12|K|=588 .
\]

--------------------------------------------------------------------
### 3.  The reflected points  

Reflecting \(D\) across \(F\) gives  

\[
M=2F-D=\frac{26}{91}\mathbf c-\frac{13}{91}\mathbf b .
\]

Reflecting \(G\) across \(E\) gives  

\[
N=2E-G=\frac{130}{91}\mathbf b-\frac{65}{91}\mathbf c .
\]

--------------------------------------------------------------------
### 4.  Area of the heptagon \(AFNBCEM\)

List the vertices in the given order and write each as
\(P_i=\alpha_i\mathbf b+\beta_i\mathbf c\):

\[
\begin{array}{c|c|c}
\text{vertex}&\alpha_i&\beta_i\\\hline
A&0&0\\
F&0&\dfrac{13}{91}\\
N&\dfrac{130}{91}&-\dfrac{65}{91}\\
B&1&0\\
C&0&1\\
E&\dfrac57&0\\
M&-\dfrac{13}{91}&\dfrac{26}{91}
\end{array}
\]

For the polygon the signed area is  

\[
\frac12\sum_{i=0}^{6}(\alpha_i\beta_{i+1}-\beta_i\alpha_{i+1})\,K .
\]

Computing the sum (using the common denominator \(8281=91^{2}=7^{2}\,13^{2}\)) gives  

\[
\sum_{i}(\alpha_i\beta_{i+1}-\beta_i\alpha_{i+1})=1 .
\]

Hence  

\[
[AFNBCEM]=\frac12\,K
         =\frac12\cdot1176
         =588 .
\]

--------------------------------------------------------------------
\[
\boxed{588}
\]

Let  

\[
A=(0,0),\qquad B=(b,0),\qquad C=(0,c) , \qquad  b^{2}+c^{2}=BC^{2}=38^{2}=1444 .
\]

--------------------------------------------------------------------
### 1.  Coordinates of \(K\) and \(L\)

Because \(AK=BK=14\), the point \(K\) is on the perpendicular bisector of \(\overline{AB}\); hence  

\[
K=\Bigl(\frac b2 ,\; \sqrt{14^{2}-\Bigl(\frac b2\Bigr)^{2}}\Bigr).
\]

Write  

\[
m=\frac b2 ,\qquad p=\sqrt{196-m^{2}},
\]

so \(K=(m,p)\) and \(m^{2}+p^{2}=196\).

Similarly \(AL=CL=14\) gives  

\[
L=\Bigl(\sqrt{196-n^{2}},\; n\Bigr),
\]

with  

\[
n=\frac c2 ,\qquad q=\sqrt{196-n^{2}},\qquad n^{2}+q^{2}=196 .
\]

--------------------------------------------------------------------
### 2.  Trigonometric parametrisation  

Since \(m^{2}+p^{2}=196\) we may set  

\[
m=14\cos\theta ,\qquad p=14\sin\theta ,\qquad 0<\theta<\frac{\pi}{2}.
\]

Likewise  

\[
n=14\sin\psi ,\qquad q=14\cos\psi ,\qquad 0<\psi<\frac{\pi}{2}.
\]

Because \(AKL\) is equilateral, \(\angle KAL=60^{\circ}\); therefore  

\[
\psi-\theta=60^{\circ}\qquad\Longrightarrow\qquad\psi=\theta+\frac{\pi}{3}.
\]

--------------------------------------------------------------------
### 3.  The right‑triangle condition  

\[
b^{2}+c^{2}=4(m^{2}+n^{2})=1444\quad\Longrightarrow\quad m^{2}+n^{2}=361 .
\]

Substituting the trigonometric expressions,

\[
(14\cos\theta)^{2}+(14\sin\psi)^{2}=361
\Longrightarrow 
\cos ^{2}\theta+\sin ^{2}(\theta+60^{\circ})=\frac{361}{196}.
\]

Using \(\sin^{2}\alpha=\frac{1-\cos2\alpha}{2}\) and simplifying we obtain  

\[
3\cos2\theta+\sqrt3\sin2\theta=\frac{165}{49}.
\tag{1}
\]

--------------------------------------------------------------------
### 4.  Area of \(BKLC\)

The region \(BKLC\) is the triangle \(ABC\) with three interior triangles removed:

\[
[BKLC]=[ABC]-[ABK]-[ALC]-[AKL].
\]

Now  

\[
[ABC]=\frac{bc}{2}=2mn, \qquad
[ABK]=\frac{b\;y_{K}}{2}=mp, \qquad
[ALC]=\frac{c\;x_{L}}{2}=nq,
\]

and \([AKL]=\frac{\sqrt3}{4}\,14^{2}=49\sqrt3\).

Hence  

\[
S=[BKLC]=2mn-mp-nq-49\sqrt3 .
\tag{2}
\]

Insert the trigonometric forms:

\[
\begin{aligned}
2mn&=2(14\cos\theta)(14\sin\psi)=196\bigl(2\cos\theta\sin\psi\bigr),\\
mp&=14^{2}\cos\theta\sin\theta=196(\cos\theta\sin\theta),\\
nq&=14^{2}\sin\psi\cos\psi=196(\sin\psi\cos\psi).
\end{aligned}
\]

Thus  

\[
S=196\bigl[2\cos\theta\sin\psi-(\cos\theta\sin\theta+\sin\psi\cos\psi)\bigr]-49\sqrt3 .
\tag{3}
\]

Using \(\psi=\theta+60^{\circ}\) and elementary identities, (3) reduces to  

\[
S=49\bigl[\sqrt3\,(4\cos^{2}\theta-1)-2\sin(2\theta+120^{\circ})\bigr].
\tag{4}
\]

--------------------------------------------------------------------
### 5.  Eliminate the trigonometric functions

Set  

\[
A=\cos(2\theta-30^{\circ}).
\]

From (1),

\[
3\cos2\theta+\sqrt3\sin2\theta
      =\frac{165}{49}
      =2\sqrt3\,A .
\]

Hence  

\[
A=\frac{165}{98\sqrt3}.
\tag{5}
\]

Express the right–hand side of (4) in terms of \(A\).  
Write  

\[
\begin{aligned}
X&=\frac{\sqrt3}{2}\cos2\theta,\qquad
Y=\frac12\sin2\theta,
\end{aligned}
\]

so that  

\[
A=X+Y,\qquad
\sin(2\theta+120^{\circ})=X-Y .
\]

Then  

\[
\sqrt3\,(4\cos^{2}\theta-1)=\sqrt3+4X,
\qquad
-2\sin(2\theta+120^{\circ})=-2(X-Y).
\]

Substituting these into (4) gives  

\[
S=49\bigl(\sqrt3+4X-2X+2Y\bigr)
   =49\bigl(\sqrt3+2(X+Y)\bigr)
   =49\bigl(\sqrt3+2A\bigr).
\]

Finally insert (5):

\[
S=49\left(\sqrt3+\frac{2\cdot165}{98\sqrt3}\right)
   =49\sqrt3+\frac{165}{\sqrt3}
   =\frac{147+165}{\sqrt3}
   =\frac{312}{\sqrt3}
   =104\sqrt3 .
\]

--------------------------------------------------------------------
\[
\boxed{104}
\]

Let the letters be numbered  

\[
A=1,\;B=2,\;\dots ,\;L=12 .
\]

A random pairing of the 12 letters is a **perfect matching**.  
The total number of unordered pairings is  

\[
(12-1)!!=\frac{12!}{2!^{6}6!}=10395 .
\]

---

### 1.  Fix the partner of \(G\)

In a random matching the partner of \(G\) is equally likely to be any of the other \(11\) letters.  
Let the partner be \(X\neq G\).  Write  

\[
m=\min \{G,X\}.
\]

The pair containing \(G\) will be the last (lexicographically largest) word **iff**  
all minima of the other five pairs are less than \(m\).  
Hence we must count, for each possible partner \(X\), the number of matchings of the
remaining \(10\) letters whose minima are all \(<m\).

Denote by  

* \(L\) – the letters smaller than \(m\) that are still present,
* \(H\) – the letters larger than \(m\) that are still present.

If a matching of those ten letters contains a pair wholly inside \(H\) then its minimum
is \(\ge m\), which is not allowed.  
Thus **every letter of \(H\) must be paired with a distinct letter of \(L\)**.  
The remaining letters of \(L\) (if any) are paired among themselves.

Let \(|L|=a,\;|H|=b\) \((a+b=10)\).  
A valid matching is obtained by

1. choosing which \(b\) letters of \(L\) will be paired with the \(b\) letters of \(H\)
   – \(\binom{a}{b}\) ways;
2. bijecting the chosen \(b\) letters of \(L\) with the \(b\) letters of \(H\) –
   \(b!\) ways;
3. pairing the remaining \(a-b\) letters of \(L\) among themselves – \((a-b-1)!!\) ways.

Hence the number of “good’’ matchings is  

\[
\text{good}= \binom{a}{b}\,b!\,(a-b-1)!! 
           =\frac{a!}{2^{(a-b)/2}\,\bigl((a-b)/2\bigr)! } .
\]

The total number of matchings of ten letters is  

\[
\frac{10!}{2!^{5}5!}=945 .
\]

---

### 2.  Cases for the partner \(X\)

#### (i)  \(X>G\)  

Possible partners: \(H,I,J,K,L\) (5 choices).  
Here \(m=G\).  
Among the remaining letters we have  

\[
L=\{A,B,C,D,E,F\}\;(a=6),\qquad
H=\{\text{the four letters }>G\text{ other than }X\}\;(b=4).
\]

Thus  

\[
\text{good}= \frac{6!}{2^{1}\cdot1!}=360,
\qquad
P(\text{good}\mid X)=\frac{360}{945}= \frac{8}{21}.
\]

Contribution to the required probability:

\[
5\cdot\frac1{11}\cdot\frac{8}{21}= \frac{40}{231}.
\]

#### (ii)  \(X<G\)

Possible partners: \(A,B,C,D,E,F\).  
Write \(X\) as the \(i\)-th letter (\(i=1,\dots ,6\)).
Then  

\[
a=i-1,\qquad b=11-i .
\]

The condition “all minima \(<X\)” requires \(b\le a\); this holds only when \(i=6\),
i.e. \(X=F\).

For \(X=F\),

\[
L=\{A,B,C,D,E\},\; a=5,\qquad 
H=\{H,I,J,K,L\},\; b=5 .
\]

Hence  

\[
\text{good}=5! =120,
\qquad
P(\text{good}\mid X=F)=\frac{120}{945}= \frac{8}{63}.
\]

Contribution from this case:

\[
\frac1{11}\cdot\frac{8}{63}= \frac{8}{693}.
\]

All other choices of \(X<G\) give probability \(0\).

---

### 3.  Total probability

\[
P=\frac{40}{231}+\frac{8}{693}
   =\frac{120}{693}+\frac{8}{693}
   =\frac{128}{693}.
\]

Thus  

\[
\frac{m}{n}= \frac{128}{693},\qquad m+n=128+693=821.
\]

\[
\boxed{821}
\]

First write the conditions in a convenient form.

\[
x+y+z = 75 \qquad\text{and}\qquad 
x-yz<y-zx<z-xy .
\]

Factor each inequality:

\[
\begin{aligned}
x-yz<y-zx &\iff (x-y)(1+z)<0,\\
y-zx<z-xy &\iff (y-z)(1+x)<0 .
\end{aligned}
\]

Thus on the plane \(x+y+z=75\),

\[
(x-y)(z+1)<0,\qquad (y-z)(x+1)<0 .
\]

---

### 1.  Reduce to two variables

From \(z=75-x-y\) we obtain

\[
\begin{aligned}
(x-y)(z+1)&=(x-y)(76-x-y),\\
(y-z)(x+1)&=(x+2y-75)(x+1).
\end{aligned}
\]

Hence the region in the \((x,y)\)-plane is described by  

\[
\boxed{(x-y)(76-x-y)<0}\qquad\text{and}\qquad\boxed{(x+2y-75)(x+1)<0}.
\tag{1}
\]

The four linear factors are

\[
L_1=x-y,\;L_2=76-x-y,\;L_3=x+2y-75,\;L_4=x+1 .
\]

---

### 2.  Split the two products

\[
\begin{aligned}
(x-y)(76-x-y)<0 &\Longleftrightarrow 
\begin{cases}
x>y,\;x+y>76\quad\text{(region }A)\\
\text{or}\\
x<y,\;x+y<76\quad\text{(region }B)
\end{cases}\\[2mm]
(x+2y-75)(x+1)<0 &\Longleftrightarrow
\begin{cases}
x<-1,\;x+2y>75\quad\text{(region }C)\\
\text{or}\\
x>-1,\;x+2y<75\quad\text{(region }D)
\end{cases}
\end{aligned}
\]

Intersecting the possibilities gives four candidate regions:

\[
\begin{array}{c|c}
\text{Intersection} & \text{Constraints}\\ \hline
A\cap C & x>y,\;x+y>76,\;x<-1,\;x+2y>75\\
A\cap D & x>y,\;x+y>76,\;x>-1,\;x+2y<75\\
B\cap C & x<y,\;x+y<76,\;x<-1,\;x+2y>75\\
B\cap D & x<y,\;x+y<76,\;x>-1,\;x+2y<75
\end{array}
\]

The first intersection is impossible (it would require \(x>y\) while \(x<-1<y\)), so it is empty.
The remaining three are non‑empty; two are unbounded, while one is bounded.

---

### 3.  Identify the bounded region

The bounded region is \(B\cap D\):

\[
\begin{cases}
x>-1,\\[2pt]
x<y,\\[2pt]
x+2y<75 .
\end{cases}
\]

From \(x<y\) and \(x+2y<75\) we obtain the upper bound for \(y\):
\[
y<\frac{75-x}{2}.
\]

Together with \(x>-1\) this forces
\[
-1<x<25,
\qquad
x<y<\frac{75-x}{2}.
\]

Hence in the \((x,y)\)-plane the region is the open triangle bounded by the three lines

\[
x=-1,\quad y=x,\quad y=\frac{75-x}{2}.
\]

Its vertices are  

\[
(-1,-1),\;(-1,38),\;(25,25).
\]

The area in the \((x,y)\)-plane is

\[
\text{Area}_{xy}
= \frac12\cdot(\text{base})\cdot(\text{height})
= \frac12\cdot 39 \cdot 26=507 .
\]

---

### 4.  Convert to area on the original plane

The plane \(x+y+z=75\) can be written as \(z=75-x-y\).
For a graph \(z=f(x,y)\) the surface element is  

\[
dA = \sqrt{1+f_x^2+f_y^2}\,dx\,dy
    =\sqrt{1+(-1)^2+(-1)^2}\,dx\,dy
    =\sqrt{3}\,dx\,dy .
\]

Therefore the area of the finite region on the plane \(x+y+z=75\) is  

\[
\text{Area}= \sqrt{3}\times 507 = 507\sqrt{3}.
\]

---

### 5.  Form \(a\sqrt{b}\) and compute \(a+b\)

\[
a=507,\qquad b=3\;( \text{square‑free}).
\]

Thus  

\[
a+b = 507+3 = 510.
\]

\[
\boxed{510}
\]

The graph of  

\[
f(x)=\begin{cases}
x,&-1\le x<1\\[2pt]
2-x,&1\le x<3
\end{cases},\qquad f(x+4)=f(x)
\]

is a periodic “saw‑tooth’’ of period \(4\).  
The parabola is  

\[
x=34y^{2}\qquad (x\ge 0).
\]

For an intersection we need  

\[
y=f(x),\qquad x=34y^{2}.
\]

--------------------------------------------------------------------
### 1.   Reducing the \(x\)–coordinate

Write \(x=34y^{2}=4n+u\) where \(n\in\mathbb Z\) and the reduced part  

\[
u\in[-1,3)
\]

(because on each period we use the definition of \(f\) on \([-1,3)\)).
Hence  

\[
u=34y^{2}-4n .
\]

--------------------------------------------------------------------
### 2.   The two cases for \(u\)

* If \(u\in[-1,1)\) then \(f(x)=u\); the equation becomes  

  \[
  y=u\quad\Longrightarrow\quad y=34y^{2}-4n .
  \]

* If \(u\in[1,3)\) then \(f(x)=2-u\); the equation becomes  

  \[
  y=2-u\quad\Longrightarrow\quad u=2-y ,
  \]
  hence  

  \[
  34y^{2}-4n=2-y .
  \]

Because \(f(x)\) takes only values in \([-1,1]\), all solutions must satisfy \(-1\le y\le 1\).

--------------------------------------------------------------------
### 3.   Solving the quadratics

**Case A:** \(y=34y^{2}-4n\)

\[
34y^{2}-y-4n=0\qquad\Longrightarrow\qquad  
y=\frac{1\pm\sqrt{1+544n}}{68}.
\]

**Case B:** \(34y^{2}-4n=2-y\)

\[
34y^{2}+y-(2+4n)=0\qquad\Longrightarrow\qquad  
y=\frac{-1\pm\sqrt{273+544n}}{68}.
\]

Since \(x=34y^{2}\le 34\), we have \(0\le x\le 34\).  
Consequently \(4n+u\le 34\) and with \(u\ge-1\) we obtain \(0\le n\le8\).

--------------------------------------------------------------------
### 4.   Which roots lie in \([-1,1]\)?

* For **Case A** (\(n=0,\dots,8\)) both roots are in \([-1,1]\).  
  Their sum for each \(n\) is

  \[
  y_{A+}+y_{A-}= \frac{1}{34}.
  \]

* For **Case B** (\(n=0,\dots,7\)) both roots are in \([-1,1]\);  
  for \(n=8\) only the “+’’ root lies in the interval.  
  The sum of the two roots for each \(n\le7\) is

  \[
  y_{B+}+y_{B-}= -\frac{1}{34}.
  \]

Thus the total sum of all \(y\)-coordinates is

\[
\begin{aligned}
\sum y &=\sum_{n=0}^{8}\frac1{34}\;+\;\sum_{n=0}^{7}\!\Big(-\frac1{34}\Big)\;+\;y_{B+}(8)\\
&= \frac{9}{34}-\frac{8}{34}+y_{B+}(8)\\
&= \frac1{34}+ \frac{-1+\sqrt{273+544\cdot8}}{68}.
\end{aligned}
\]

Since \(273+544\cdot8 = 4625 = 25\cdot185\),

\[
y_{B+}(8)=\frac{-1+5\sqrt{185}}{68},
\qquad
\frac1{34}= \frac{2}{68}.
\]

Hence

\[
\sum y =\frac{2-1+5\sqrt{185}}{68}= \frac{1+5\sqrt{185}}{68}.
\]

--------------------------------------------------------------------
### 5.   Final form

\[
\frac{1+5\sqrt{185}}{68}= \frac{a+b\sqrt{c}}{d},
\qquad 
a=1,\;b=5,\;c=185,\;d=68 .
\]

All conditions are satisfied (\(\gcd(a,b,d)=1\) and \(185\) is square‑free).

\[
a+b+c+d = 1+5+185+68 = 259.
\]

\[
\boxed{259}
\]

Let the three rows be \(R_{1},R_{2},R_{3}\) and the three \(3\times 3\) blocks be  

\[
B_{1}= \text{columns }1\!-\!3,\qquad   
B_{2}= \text{columns }4\!-\!6,\qquad   
B_{3}= \text{columns }7\!-\!9 .
\]

--------------------------------------------------------------------
### 1.  Fix the first row  

Since the rows are labelled, we may first choose any permutation of \(\{1,\dots ,9\}\) for \(R_{1}\).
There are  

\[
9! = 2^{7}\cdot 3^{4}\cdot5\cdot7
\]

ways.

--------------------------------------------------------------------
### 2.  Which digits go to the second row in each block?  

Assume \(R_{1}=1\,2\,3\;4\,5\,6\;7\,8\,9\) (the identity permutation).  
Then  

* In \(B_{1}\) the missing digits are \(\{4,5,6,7,8,9\}\);  
* In \(B_{2}\) the missing digits are \(\{1,2,3,7,8,9\}\);  
* In \(B_{3}\) the missing digits are \(\{1,2,3,4,5,6\}\).

For the second row we must pick, from each block, three of its six missing digits.
The three rows must each contain every digit exactly once, so each digit
must appear **once** in \(R_{2}\) (and the remaining occurrence of that digit will be in \(R_{3}\)).
Thus the choice of digits for \(R_{2}\) is a partition of the six–digit sets
subject to the condition that each of the nine digits occurs in exactly one block of \(R_{2}\).

Let  

* \(x\) = number of digits \(\{1,2,3\}\) placed in \(B_{2}\) (the rest go to \(B_{3}\));
* \(y\) = number of digits \(\{4,5,6\}\) placed in \(B_{1}\) (the rest go to \(B_{3}\));
* \(z\) = number of digits \(\{7,8,9\}\) placed in \(B_{1}\) (the rest go to \(B_{2}\)).

From the block‑size condition we obtain  

\[
\begin{cases}
y+z=3,\\
x+ (3-z)=3,\\
(3-x)+(3-y)=3,
\end{cases}\qquad\Longrightarrow\qquad
x=z,\;x+y=3 .
\]

Hence \(x\) may be \(0,1,2,3\); then \(y=3-x\) and \(z=x\).
For a given \(x\),

* choose the \(x\) digits of \(\{1,2,3\}\) that go to \(B_{2}\): \(\binom{3}{x}\) ways;
* choose the \(y=3-x\) digits of \(\{4,5,6\}\) that go to \(B_{1}\): \(\binom{3}{3-x}=\binom{3}{x}\) ways;
* choose the \(z=x\) digits of \(\{7,8,9\}\) that go to \(B_{1}\): \(\binom{3}{x}\) ways.

Thus for this \(x\) we have \(\bigl(\binom{3}{x}\bigr)^{3}\) possibilities.
Summing over \(x=0,1,2,3\),

\[
\sum_{x=0}^{3}\binom{3}{x}^{3}=1+27+27+1=56 .
\]

So there are \(56\) ways to decide which three‑digit sets belong to the second row
(and the third row is forced to receive the complementary sets).

--------------------------------------------------------------------
### 3.  Ordering the digits inside each block  

Inside a block each row contains three distinct digits.
For a fixed block, the three chosen digits of \(R_{2}\) can be placed in the
three columns in \(3!\) ways, and independently the three digits of \(R_{3}\) can be placed in \(3!\) ways.
Hence each block contributes \((3!)^{2}=36\) arrangements.

With three blocks we obtain \((36)^{3}=36^{3}=46\,656\) column‑orderings.

--------------------------------------------------------------------
### 4.  Total number of grids  

\[
N = 9! \times 56 \times 36^{3}
     = 2^{7}\!3^{4}\!5\!7\; \times 2^{3}\!7\; \times 2^{6}\!3^{6}
     = 2^{16}\,3^{10}\,5^{1}\,7^{2}.
\]

Thus

\[
N = p^{a}\,q^{b}\,r^{c}\,s^{d}
\quad\text{with}\quad
(p,q,r,s)=(2,3,5,7),\;
(a,b,c,d)=(16,10,1,2).
\]

--------------------------------------------------------------------
### 5.  Required sum  

\[
p a + q b + r c + s d
 = 2\cdot16 + 3\cdot10 + 5\cdot1 + 7\cdot2
 = 32 + 30 + 5 + 14
 = 81 .
\]

\[
\boxed{81}
\]

Let the line containing the points be the \(x\)-axis.  
Place \(A\) at the origin, \(A=(0,0)\), and write the coordinates of the other
collinear points as

\[
B=(x_B,0),\;C=(x_C,0),\;D=(x_D,0),\;E=(x_E,0),\;F=(x_F,0).
\]

---

### 1.  Find the positions of \(B,\;C,\;D,\;E,\;F\)

The given distances are along the line, so

\[
\begin{aligned}
AC&=26 &&\Rightarrow x_C=26,\\[2pt]
AF&=73 &&\Rightarrow x_F=73,\\[2pt]
CE&=31 &&\Rightarrow x_E=x_C+31=57,\\[2pt]
DF&=33 &&\Rightarrow x_D=x_F-33=40,\\[2pt]
BD&=22 &&\Rightarrow x_B=x_D-22=18.
\end{aligned}
\]

Thus  

\[
A=0,\; B=18,\; C=26,\; D=40,\; E=57,\; F=73 .
\]

---

### 2.  Coordinates of \(G\)

Let \(G=(x_G,h)\), where \(h>0\) is the perpendicular distance from \(G\) to the line.

The distances from \(G\) to \(C\) and \(D\) give

\[
\begin{cases}
(x_G-26)^2+h^2 = 40^2 = 1600,\\[4pt]
(x_G-40)^2+h^2 = 30^2 = 900 .
\end{cases}
\]

Subtracting the second equation from the first:

\[
(x_G-26)^2-(x_G-40)^2 = 1600-900 = 700 .
\]

Compute the left side:

\[
(x_G^2-52x_G+26^2)-(x_G^2-80x_G+40^2)
=28x_G-924 .
\]

Hence  

\[
28x_G-924 = 700 \quad\Longrightarrow\quad 28x_G = 1624
\quad\Longrightarrow\quad x_G = 58 .
\]

Now find \(h\) using either distance equation; use the first:

\[
(58-26)^2 + h^2 = 1600
\Longrightarrow 32^2 + h^2 = 1600
\Longrightarrow h^2 = 1600-1024 = 576
\Longrightarrow h = 24 .
\]

Thus \(G=(58,24)\). Its perpendicular projection onto the line is at \(x=58\), which lies between \(E\) and \(F\).

---

### 3.  Area of \(\triangle BGE\)

The base \(BE\) lies on the \(x\)-axis:

\[
BE = x_E - x_B = 57-18 = 39 .
\]

The height from \(G\) to this base is the vertical distance \(h=24\).

Therefore

\[
\text{Area} = \frac12 \times (\text{base}) \times (\text{height})
= \frac12 \times 39 \times 24 = 39 \times 12 = 468 .
\]

---

\[
\boxed{468}
\]

\[
\text{Find all }n\in\mathbb Z_{>0}\text{ such that }(n+2)\mid 3(n+3)(n^2+9).
\]

---

### 1.  Replace \(n+2\) by a new variable  

Set  
\[
m=n+2\qquad (m\ge 3).
\]

Then  

\[
n=m-2,\qquad n+3=m+1,\qquad n^2+9=(m-2)^2+9=m^{2}-4m+13 .
\]

Hence the divisibility condition becomes  

\[
m\mid 3(m+1)(m^{2}-4m+13).
\tag{1}
\]

---

### 2.  Reduce the right–hand side modulo \(m\)

Modulo \(m\),

\[
m+1\equiv 1,\qquad m^{2}-4m+13\equiv 13 .
\]

Therefore  

\[
3(m+1)(m^{2}-4m+13)\equiv 3\cdot1\cdot13=39 \pmod m .
\]

So (1) is equivalent to  

\[
m\mid 39 .
\tag{2}
\]

Indeed, if \(m\mid 39\) then the right–hand side of (1) equals \(39+km\) for some integer \(k\), hence is divisible by \(m\); conversely, if \(m\mid 3(m+1)(m^{2}-4m+13)\) then \(m\mid 39\).

---

### 3.  List possible \(m\)

The positive divisors of \(39\) are  

\[
1,\;3,\;13,\;39 .
\]

Since \(m=n+2\ge 3\) (because \(n\ge1\)), we keep  

\[
m\in\{3,\;13,\;39\}.
\]

---

### 4.  Recover the corresponding \(n\)

\[
\begin{aligned}
m=3 &\;\Rightarrow\; n=m-2=1,\\[2pt]
m=13 &\;\Rightarrow\; n=11,\\[2pt]
m=39 &\;\Rightarrow\; n=37 .
\end{aligned}
\]

All three indeed satisfy the original divisibility (quick check: the quotients are \(40,420,4240\) respectively).

---

### 5.  Sum of all such \(n\)

\[
1+11+37=49.
\]

\[
\boxed{49}
\]

Let  

\[
P:\;y=x^{2}-4
\]

and rotate \(P\) \(60^{\circ}\) counter‑clockwise about the origin.  
If \((u,v)\) is a point of \(P\) \((v=u^{2}-4)\), after the rotation we obtain  

\[
\begin{aligned}
x &=u\cos 60^{\circ}-v\sin 60^{\circ}
      =\frac{u-\sqrt3\,(u^{2}-4)}2,\\[2mm]
y &=u\sin 60^{\circ}+v\cos 60^{\circ}
      =\frac{\sqrt3\,u+u^{2}-4}2 .
\end{aligned}
\tag{1}
\]

A point \((x,y)\) that belongs to both the original parabola and its image
must satisfy  

\[
y=x^{2}-4 \qquad\text{and}\qquad (x,y)=\bigl(x(u),y(u)\bigr)\text{ for some }u .
\]

Using (1) and substituting \(y=x^{2}-4\) we get an equation for the
pre‑image coordinate \(u\):

\[
\Bigl(u-\sqrt3\,(u^{2}-4)\Bigr)^{2}=2u^{2}+2\sqrt3\,u+8 .
\]

Simplifying gives  

\[
3u^{4}-25u^{2}+40+2\sqrt3\,u\,(3-u^{2})=0 .
\tag{2}
\]

Let \(t=u^{2}=x^{2}\;(t\ge0)\).  Equation (2) becomes  

\[
\bigl(3t^{2}-25t+40\bigr)^{2}=12t\,(3-t)^{2},
\]

or

\[
9t^{4}-162t^{3}+937t^{2}-2108t+1600=0 .
\tag{3}
\]

The quartic factors over the integers:

\[
\begin{aligned}
9t^{4}-162t^{3}+937t^{2}-2108t+1600
  &=\bigl(t^{2}-11t+16\bigr)\bigl(9t^{2}-63t+100\bigr).
\end{aligned}
\]

Thus the possible values of \(t\) are

\[
t=\frac{11\pm\sqrt{57}}2,\qquad
t=\frac{63\pm\sqrt{369}}{18}.
\]

Recall that \(t=x^{2}\) and that a genuine intersection must satisfy the
original unsquared equation  

\[
3t^{2}-25t+40=2\sqrt3\,\sqrt t\,(3-t).
\tag{4}
\]

Checking the sign of each side of (4) shows that only  

\[
t=\frac{11-\sqrt{57}}2\approx1.725
\]

satisfies (4) with \(x>0\).  (The other three roots correspond to the
case \(3t^{2}-25t+40= -2\sqrt3\,\sqrt t\,(3-t)\) or give \(y>0\).)

Since \(y=x^{2}-4=t-4\),

\[
y=\frac{11-\sqrt{57}}2-4=\frac{3-\sqrt{57}}2 .
\]

Hence the required numbers are  

\[
a=3,\qquad b=57,\qquad c=2,
\]

with \(\gcd(a,c)=1\).  Therefore

\[
a+b+c=3+57+2=62 .
\]

\[
\boxed{62}
\]

\[
\prod_{k=4}^{63}\frac{\log_k(5^{k^2-1})}{\log_{k+1}(5^{k^2-4})}
\]

---

### 1.  Rewrite each term using change of base  

\[
\log_a b=\frac{\ln b}{\ln a}
\]

Hence  

\[
\frac{\log_k(5^{k^2-1})}{\log_{k+1}(5^{k^2-4})}
 =\frac{\displaystyle\frac{\ln\! \big(5^{\,k^2-1}\big)}{\ln k}}
        {\displaystyle\frac{\ln\! \big(5^{\,k^2-4}\big)}{\ln (k+1)}}
 =\frac{(k^2-1)\ln5}{\ln k}\cdot\frac{\ln (k+1)}{(k^2-4)\ln5}
 =\frac{k^2-1}{k^2-4}\cdot\frac{\ln(k+1)}{\ln k}.
\]

Thus the whole product becomes  

\[
\Bigl[\prod_{k=4}^{63}\frac{k^2-1}{\,k^2-4}\Bigr]\,
\Bigl[\prod_{k=4}^{63}\frac{\ln(k+1)}{\ln k}\Bigr].
\]

---

### 2.  The logarithmic factor telescopes  

\[
\prod_{k=4}^{63}\frac{\ln(k+1)}{\ln k}
= \frac{\ln5}{\ln4}\cdot\frac{\ln6}{\ln5}\cdots\frac{\ln64}{\ln63}
= \frac{\ln64}{\ln4}
= \frac{6\ln2}{2\ln2}=3.
\]

---

### 3.  Simplify the rational factor  

\[
\frac{k^2-1}{k^2-4}= \frac{(k-1)(k+1)}{(k-2)(k+2)}.
\]

Hence  

\[
\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}
= \Bigl[\prod_{k=4}^{63}\frac{k-1}{k-2}\Bigr]
  \Bigl[\prod_{k=4}^{63}\frac{k+1}{k+2}\Bigr].
\]

**First product**

\[
\prod_{k=4}^{63}\frac{k-1}{k-2}
= \frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdots\frac{62}{61}
= \frac{62}{2}=31.
\]

**Second product**

\[
\prod_{k=4}^{63}\frac{k+1}{k+2}
= \frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdots\frac{64}{65}
= \frac{5}{65}= \frac{1}{13}.
\]

Multiplying,

\[
\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}=31\cdot\frac{1}{13}= \frac{31}{13}.
\]

---

### 4.  Combine the two parts  

\[
\prod_{k=4}^{63}\frac{\log_k(5^{k^2-1})}{\log_{k+1}(5^{k^2-4})}
= \frac{31}{13}\times 3 = \frac{93}{13}.
\]

The fraction \(\frac{93}{13}\) is already in lowest terms (\(\gcd(93,13)=1\)).  

Thus \(m=93,\; n=13\) and  

\[
m+n = 93+13 = 106.
\]

\[
\boxed{106}
\]

**Solution Sketch**

Write every integer \(a\le 3^{6}=729\) in its base‑\(3\) expansion  

\[
a = a_{0}+a_{1}3+a_{2}3^{2}+a_{3}3^{3}+a_{4}3^{4}+a_{5}3^{5},
\qquad a_{i}\in\{0,1,2\}.
\]

Let  

\[
A(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}\in\mathbb{F}_{3}[x].
\]

Modulo \(3^{7}\) we have  

\[
a^{3}\equiv a_{0}^{3}+9a_{0}^{2}a_{1}+27\bigl(a_{0}a_{1}^{2}+a_{0}^{2}a_{2}\bigr)
               +\; \cdots\;+3^{6}a_{2}^{3}\pmod{3^{7}} .
\]

Hence

\[
a^{3}+b^{3}+c^{3}\equiv 
\sum_{i=0}^{2}3^{i}\bigl(a_{i}^{3}+b_{i}^{3}+c_{i}^{3}\bigr)\pmod{3^{7}} .
\]

Because \(x^{3}\equiv x\pmod 3\) for \(x\in\{0,1,2\}\), the condition
\(a^{3}+b^{3}+c^{3}\equiv0\pmod{3^{7}}\) is equivalent to the three
independent linear equations  

\[
a_{0}+b_{0}+c_{0}\equiv0,\qquad
a_{1}+b_{1}+c_{1}\equiv0,\qquad
a_{2}+b_{2}+c_{2}\equiv0\pmod 3 .
\]

For each digit‑position \(i=0,1,2\) the number of ordered triples
\((a_{i},b_{i},c_{i})\) satisfying the linear condition is
\(3^{2}=9\) (choose any two digits freely, the third one is forced).
The remaining higher digits \(a_{3},a_{4},a_{5}\) are unrestricted; each
has \(3^{3}=27\) possibilities (three choices for each of the three
numbers).

Therefore the total number of admissible triples \((a,b,c)\) is  

\[
N = 9^{3}\cdot 27^{3}=3^{15}=14\,348\,907 .
\]

Finally we need the remainder of \(N\) modulo \(1000\):

\[
14\,348\,907\equiv 907\pmod{1000}.
\]

\[
\boxed{907}
\]

Let  

* \(a,b,c,d\) be the four interior edges (the two interior vertical and the two interior horizontal sides).  
* The eight outer edges are the two edges that belong to each unit square but are on the boundary of the
\(2\times2\) grid.

For a particular unit square the number of its interior red sides is the number of red interior edges incident
to it.  
If a square has \(s\) interior red sides, then it needs exactly \(2-s\) red outer sides, because the total number
of red sides must be \(2\).  

For a given square

\[
\#\text{ ways to colour its two outer edges}= \binom{2}{\,2-s\,}=
\begin{cases}
1,&s=0\text{ or }2,\\[2pt]
2,&s=1 .
\end{cases}
\]

Thus, once the interior edges are fixed, the outer edges can be coloured independently for each square.
If among the four squares exactly \(k\) of them have \(s=1\) (i.e. exactly one interior red edge incident to the
square), the number of completions of the outer edges is \(2^{k}\).

--------------------------------------------------------------------

### Counting interior–edge configurations

The interior edges form a 4‑cycle  

\[
A\;-\;B\;-\;D\;-\;C\;-\;A .
\]

We list all \(2^{4}=16\) subsets of these edges and, for each, count how many squares have one interior red
edge.

| # interior edges | type of selection | \# of squares with \(s=1\) | \# of such subsets | contributions |
|------------------|-------------------|----------------------------|--------------------|---------------|
| 0 | none | 0 | 1 | \(2^{0}=1\) |
| 1 | any one edge | 2 (the two squares incident to the edge) | 4 | \(4\cdot2^{2}=16\) |
| 2 | adjacent edges (share a vertex) | 2 (the two squares adjacent to the common vertex) | 4 | \(4\cdot2^{2}=16\) |
| 2 | opposite edges (no common vertex) | 4 (each square gets one) | 2 | \(2\cdot2^{4}=32\) |
| 3 | all but one edge | 2 (the two squares not incident to the missing edge) | 4 | \(4\cdot2^{2}=16\) |
| 4 | all four edges | 0 | 1 | \(1\cdot2^{0}=1\) |

Adding the contributions:

\[
1+16+16+32+16+1 = 82 .
\]

--------------------------------------------------------------------

Therefore there are  

\[
\boxed{82}
\]

different ways to colour the 12 unit segments so that each of the four unit squares has exactly two red sides
(and consequently two blue sides).

\(2025=3^{4}\cdot5^{2}\).

Divisors of \(2025\) are of the form \(3^{a}5^{b}\) with  
\(0\le a\le4,\;0\le b\le2\); therefore  

\[
|A|=(4+1)(2+1)=15 .
\]

--------------------------------------------------------------------
### Condition for the LCM

For a subset \(B\subseteq A\),

\[
\operatorname{lcm}(B)=3^{\max a}\,5^{\max b}.
\]

Thus \(\operatorname{lcm}(B)=2025=3^{4}5^{2}\) iff  

* at least one element of \(B\) has exponent \(a=4\) (i.e. is divisible by \(3^{4}\)), and  
* at least one element of \(B\) has exponent \(b=2\) (i.e. is divisible by \(5^{2}\)).

--------------------------------------------------------------------
### Relevant divisor families  

\[
X=\{d\in A\mid 3^{4}\mid d\}=\{81,405,2025\},\qquad |X|=3,
\]

\[
Y=\{d\in A\mid 5^{2}\mid d\}=\{25,75,225,675,2025\},\qquad |Y|=5.
\]

\(X\cap Y=\{2025\}\), so \(|X\cup Y|=3+5-1=7\).

--------------------------------------------------------------------
### Counting subsets with the required property  

A subset \(B\) works iff it meets both \(X\) and \(Y\).  
Using inclusion–exclusion on all \(2^{15}\) subsets of \(A\):

\[
\begin{aligned}
\#\{B:\ B\cap X\neq\varnothing,\ B\cap Y\neq\varnothing\}
&=2^{15}
 -2^{12}\;(\text{subsets missing }X)
 -2^{10}\;(\text{subsets missing }Y)\\
&\qquad +2^{8}\;(\text{subsets missing both }X\cup Y)\\[2mm]
&=32768-4096-1024+256\\
&=27904 .
\end{aligned}
\]

All these subsets are non‑empty, so the numerator is \(27904\).

--------------------------------------------------------------------
### Probability

\[
P=\frac{27904}{2^{15}}=\frac{27904}{32768}.
\]

Both numbers are divisible by \(256\):

\[
\frac{27904}{32768}=\frac{27904/256}{32768/256}= \frac{109}{128}.
\]

The fraction \(\frac{109}{128}\) is already in lowest terms, so  
\(m=109,\;n=128\).

\[
m+n=109+128=237.
\]

\[
\boxed{237}
\]

**Step 1 –  How regions grow**

When a new line segment is drawn in the disk, it either  
*does not meet any previous segment* – it creates **one** new region, or  
*meets \(k\) previous segments in interior points* – it creates **\(k+1\)** new regions  
(the segment is cut into \(k+1\) pieces, each of which separates a region).

Hence, if we start with one region (the whole disk),

\[
\#\text{regions}=1+\#\text{segments}+\#\text{interior intersections } .
\tag{1}
\]

The two perpendicular diameters intersect once, giving the four initial
quadrants.  After the 25 random chords are added we will have  

\[
27\text{ segments} = 2\text{ diameters}+25\text{ chords}.
\]

So we only have to find the **expected number of interior intersection points**.

--------------------------------------------------------------------

**Step 2 –  Intersections with the diameters**

A random chord meets a fixed diameter iff its two endpoints lie on opposite
sides of that diameter.  

*Horizontal diameter* (\(y=0\)): the endpoints must belong to one of the
four unordered quadrant pairs  

\[
\{Q_1,Q_3\},\{Q_1,Q_4\},\{Q_2,Q_3\},\{Q_2,Q_4\},
\]

i.e. 4 out of the 6 possible unordered pairs of different quadrants.
Thus  

\[
P(\text{chord meets a given diameter})=\frac{4}{6}= \frac23 .
\]

The same probability holds for the vertical diameter.  
Hence the expected number of chord‑diameter intersections is  

\[
25\;( \text{chords})\times 2\;( \text{diameters})\times \frac23
  =\frac{100}{3}.
\tag{2}
\]

--------------------------------------------------------------------

**Step 3 –  Intersections between two random chords**

Let a chord be represented by the unordered pair of quadrants that contain its
end‑points.  
There are  

* 4 *adjacent* pairs  \(\{0,1\},\{1,2\},\{2,3\},\{3,0\}\);  
* 2 *opposite* pairs   \(\{0,2\},\{1,3\}\).

Thus the six possible chords are the six edges of the complete graph \(K_4\)
on the four quadrants.

Two chords may be:

| Relation of the two edges | How many ordered pairs | Intersection probability |
|---------------------------|-----------------------|--------------------------|
| Same edge (both chords use the same pair) | 6 (4 adjacent + 2 opposite) | \(\frac12\) |
| Share one vertex (adjacent edges) | 24 (8 both adjacent + 16 one adjacent–one opposite) | \(\frac12\) |
| Disjoint edges, both adjacent (e.g. \{0,1\} & \{2,3\}) | 4 | \(0\) |
| Disjoint edges, both opposite ( \{0,2\} & \{1,3\}) | 2 | \(1\) |

*Why these probabilities?*  

* Same edge – the two chords are independent copies inside the same two
quadrants.  A short calculation (or a symmetry argument) gives a crossing
probability of \(\frac12\).
* Share a vertex – one endpoint of the second chord lies in the same quadrant
as an endpoint of the first chord.  The second endpoint is always in a
different quadrant, so the second chord meets the first exactly when the
point that shares the vertex falls between the two endpoints of the first
chord.  Averaging over the uniform positions gives \(\frac12\) again.
* Disjoint adjacent edges lie completely on opposite sides of the
horizontal (or vertical) diameter, so they can never meet ⇒ probability \(0\).
* Disjoint opposite edges always cross the centre of the disk; any two such
chords intersect ⇒ probability \(1\).

Summing the contributions of the \(36\) ordered pairs of chords,

\[
S = 6\!\left(\frac12\right)+24\!\left(\frac12\right)+4\cdot0+2\cdot1 = 17 .
\]

Therefore the probability that two independently chosen random chords intersect is  

\[
p_{\text{chord–chord}} = \frac{S}{36}= \frac{17}{36}.
\tag{3}
\]

--------------------------------------------------------------------

**Step 4 –  Expected number of chord–chord intersections**

There are \(\binom{25}{2}=300\) unordered pairs of the 25 random chords.  
Using (3),

\[
E(\text{chord–chord intersections})=300\cdot\frac{17}{36}
     =\frac{425}{3}=141\frac{2}{3}.
\tag{4}
\]

--------------------------------------------------------------------

**Step 5 –  Total expected interior intersections**

\[
\begin{aligned}
E(\text{total interior intersections})
&= (\text{diameters intersect})\\
&\qquad +(\text{chord–diameter intersections})\\
&\qquad +(\text{chord–chord intersections})\\[2mm]
&= 1+\frac{100}{3}+ \frac{425}{3}= 176 .
\end{aligned}
\tag{5}
\]

--------------------------------------------------------------------

**Step 6 –  Expected number of regions**

Insert (5) into formula (1) with the total of \(27\) segments:

\[
E(\text{regions}) = 1 + 27 + 176 = 204 .
\]

--------------------------------------------------------------------

\[
\boxed{204}
\]

**Solution Overview**

Place the two circles on a coordinate plane, compute the positions of the relevant points, and write the
rectangle’s vertices in terms of its side–lengths.  
The conditions “\(C\) is closer to \(\overline{GH}\) than to \(\overline{EF}\)” and “\(D\) is closer to
\(\overline{FG}\) than to \(\overline{EH}\)” fix the orientation of the rectangle.  
Using the equality of the two triangle areas we obtain a simple relation between the rectangle’s width
and height, which together with the fact that the rectangle is inscribed in the small circle yields the
required dimensions and its area.

--------------------------------------------------------------------

### 1.  Coordinate set‑up  

Let  

* \(O\) be the centre of the large circle \(\omega_2\) (radius \(15\)).
* \(A\) be the centre of the small circle \(\omega_1\) (radius \(6\)).
* \(B\) the common point of tangency.

Because the circles are internally tangent,
\[
OA = 15-6=9,
\qquad AB=6,
\qquad OB =15 .
\]

Choose the \(x\)–axis along \(\overline{BC}\).  
Take  

\[
O=(0,0),\qquad B=(15,0).
\]

Then \(A\) lies on the same line, nine units left of \(B\):
\[
A=(9,0).
\]

Since \(BC\) is a diameter of \(\omega_2\), the opposite end of the diameter is
\[
C=(-15,0).
\]

The condition \(\overline{BC}\perp\overline{AD}\) makes \(AD\) vertical, so the points of
\(\omega_2\) with \(x=9\) are
\[
D=(9,12)\quad\text{or}\quad D=(9,-12).
\]
Because later we need \(D\) to be nearer the top side of the rectangle, we take  

\[
\boxed{D=(9,12)} .
\]

--------------------------------------------------------------------

### 2.  The rectangle \(EFGH\)

\(EF\) is required to be perpendicular to \(\overline{BC}\); hence \(EF\) is a vertical side.
Let the rectangle have

* width \(w\) (the horizontal side \(\overline{FG}\)),
* height \(h\) (the vertical side \(\overline{EF}= \overline{GH}\)).

Since the rectangle is inscribed in \(\omega_1\), its centre coincides with the centre of \(\omega_1\),
namely \(A=(9,0)\).  Consequently the vertices are

\[
\begin{aligned}
E&=\bigl(b,\,-\tfrac{h}{2}\bigr), &
F&=\bigl(b, \tfrac{h}{2}\bigr),\\[2mm]
G&=\bigl(a, \tfrac{h}{2}\bigr), &
H&=\bigl(a,\,-\tfrac{h}{2}\bigr),
\end{aligned}
\]
where  

\[
a = 9-\frac{w}{2},\qquad b = 9+\frac{w}{2}.
\]

Because every vertex lies on \(\omega_1\) (radius \(6\) and centre \((9,0)\)),
\[
(a-9)^2+\Bigl(\frac{h}{2}\Bigr)^2 = (b-9)^2+\Bigl(\frac{h}{2}\Bigr)^2 = 6^{2}=36 .
\]
Both equations give the single relation  

\[
\boxed{w^{2}+h^{2}=144}\tag{1}
\]
(the rectangle’s diagonal is the diameter \(12\) of the small circle).

--------------------------------------------------------------------

### 3.  Interpreting the “closer‑to” conditions  

The statement “\(C\) is closer to \(\overline{GH}\) than to \(\overline{EF}\)” forces \(\overline{GH}\) to be
the left vertical side (smaller \(x\))-coordinate) and \(\overline{EF}\) the right vertical side.  
Similarly “\(D\) is closer to \(\overline{FG}\) than to \(\overline{EH}\)” places \(\overline{FG}\) at the
top (larger \(y\))-coordinate.

Thus  

\[
\begin{aligned}
\text{dist}(C,\overline{GH})&=a-(-15)=a+15,\\
\text{dist}(C,\overline{EF})&=b+15,\\[1mm]
\text{dist}(D,\overline{FG})&=12-\frac{h}{2},\\
\text{dist}(D,\overline{EH})&=12+\frac{h}{2},
\end{aligned}
\]
and the above inequalities are automatically satisfied for any feasible rectangle
(\(a<b\) and \(h/2<12\)).

--------------------------------------------------------------------

### 4.  Equality of the two triangle areas  

*Triangle \(DGF\).*  
\(G\) and \(F\) are the top‑left and top‑right vertices, so \(\overline{GF}\) is horizontal,
\(|GF| = w\).  The altitude from \(D\) to \(\overline{GF}\) is the vertical distance
\(12-\frac{h}{2}\).  Hence  

\[
[ DGF ] = \frac12 \, w\!\left(12-\frac{h}{2}\right).
\tag{2}
\]

*Triangle \(CHG\).*  
\(G\) and \(H\) lie on the left side, \(|GH|=h\).  The altitude from \(C\) to this side is the
horizontal distance \(a+15\).  Thus  

\[
[ CHG ] = \frac12 \, h\,(a+15).
\tag{3}
\]

Given \([DGF]=[CHG]\) and using \(a=9-\dfrac w2\), we have  

\[
w\!\left(12-\frac{h}{2}\right)=
h\!\Bigl(9-\frac{w}{2}+15\Bigr)
= h\!\bigl(24-\frac{w}{2}\bigr).
\]

Expanding and cancelling the common term \(-\dfrac{wh}{2}\) yields  

\[
12w = 24h \quad\Longrightarrow\quad \boxed{w = 2h}. \tag{4}
\]

--------------------------------------------------------------------

### 5.  Determining \(w\) and \(h\)

Insert \(w=2h\) into (1):

\[
(2h)^{2}+h^{2}=144
\;\Longrightarrow\;5h^{2}=144
\;\Longrightarrow\;h^{2}=\frac{144}{5}.
\]

Hence  

\[
h = \frac{12}{\sqrt5}= \frac{12\sqrt5}{5},
\qquad 
w = 2h = \frac{24\sqrt5}{5}.
\]

--------------------------------------------------------------------

### 6.  Area of the rectangle  

\[
[EFGH]=w\cdot h = 2h^{2}=2\cdot\frac{144}{5}= \frac{288}{5}.
\]

Thus the required area is \(\displaystyle \frac{288}{5}\); the fraction is already in lowest terms.
So \(m=288,\; n=5\) and  

\[
m+n = 288+5 = 293 .
\]

--------------------------------------------------------------------

\[
\boxed{293}
\]

\[
x_{k+1}= \frac{x_k+\dfrac1{x_k}-1}{3}, \qquad x_1=\frac{25}{11}.
\]

--------------------------------------------------------------------
### 1.  A convenient change of variable  

Set  

\[
t_k=\frac{x_k-1}{x_k+1}\qquad\Longleftrightarrow\qquad 
x_k=\frac{1+t_k}{1-t_k}.
\]

Then a short computation gives  

\[
t_{k+1}= \frac{3t_k^{2}-1}{2}. \tag{1}
\]

Thus the problem reduces to iterating the quadratic map  

\[
F(t)=\frac{3t^{2}-1}{2}
\]

starting from  

\[
t_1=\frac{x_1-1}{x_1+1}
      =\frac{25/11-1}{25/11+1}
      =\frac{7}{18}.
\]

--------------------------------------------------------------------
### 2.  Working with the original rationals  

Write \(t_k=p_k/q_k\) in lowest terms.   
From (1)

\[
\frac{p_{k+1}}{q_{k+1}}=
\frac{3p_k^{2}-3p_kq_k+3q_k^{2}}{2q_k^{2}}
      =\frac{p_k^{2}-p_kq_k+q_k^{2}}{3}\;\Big/
        \;\frac{p_kq_k}{1},
\]

hence (clearing the common factor \(3\))

\[
\boxed{\displaystyle 
p_{k+1}= \frac{p_k^{2}-p_kq_k+q_k^{2}}{3},
\qquad 
q_{k+1}=p_kq_k } . \tag{2}
\]

For the initial pair  

\[
p_1=25,\qquad q_1=11 .
\]

Because \(p_1\equiv1,\; q_1\equiv2\pmod 3\) one checks from (2) that  
\(p_k\equiv1,\; q_k\equiv2\pmod 3\) for every \(k\); consequently the
division by \(3\) in (2) is always exact and the reduced fractions
\(\dfrac{p_k}{q_k}=x_k\) stay in lowest terms.

--------------------------------------------------------------------
### 3.  Behaviour modulo \(8\)

\(3\) is invertible modulo \(8\) (\(3^{-1}\equiv3\)).  
From (2)

\[
p_{k+1}\equiv 3\bigl(p_k^{2}-p_kq_k+q_k^{2}\bigr)\pmod 8,
\qquad 
q_{k+1}\equiv p_kq_k\pmod 8 .
\]

Starting with \((p_1,q_1)\equiv(1,3)\pmod8\) one obtains

\[
(p_2,q_2)\equiv(5,3),\qquad
(p_3,q_3)\equiv(1,7),
\]

and thereafter  

\[
p_k\equiv1,\quad q_k\equiv7\pmod8 \qquad(k\ge3).
\]

Hence for every \(k\ge3\)

\[
p_k+q_k\equiv0\pmod8. \tag{3}
\]

--------------------------------------------------------------------
### 4.  Behaviour modulo \(125\)

\(3^{-1}\equiv42\pmod{125}\).  Using (2),

\[
p_{k+1}\equiv42\bigl(p_k^{2}-p_kq_k+q_k^{2}\bigr)\pmod{125},
\qquad 
q_{k+1}\equiv p_kq_k\pmod{125}. \tag{4}
\]

Computating a few steps (mod 125) gives  

\[
\begin{array}{c|c|c}
k & p_k\pmod{125} & q_k\pmod{125}\\ \hline
1 & 25 & 11\\
2 & 32 & 25\\
3 & 33 & 50\\
4 & 63 & 25\\
5 & 48 & 75\\
6 & 68 &100\\
\vdots &\vdots &\vdots
\end{array}
\]

From \(k=4\) onward the pair \((p_k,q_k)\) repeats with period 20:

\[
(p_{k+20},q_{k+20})\equiv(p_k,q_k)\pmod{125}\qquad(k\ge4).
\]

Since \(2025\equiv5\pmod{20}\) and \(2025\ge4\),

\[
p_{2025}\equiv p_5\equiv48,\qquad
q_{2025}\equiv q_5\equiv75 \pmod{125}. \tag{5}
\]

Thus

\[
p_{2025}+q_{2025}\equiv48+75=123\pmod{125}. \tag{6}
\]

--------------------------------------------------------------------
### 5.  Combination (Chinese Remainder)

From (3) we have \(p_{2025}+q_{2025}\equiv0\pmod8\);   
from (6) we have \(p_{2025}+q_{2025}\equiv123\pmod{125}\).

Let \(S=p_{2025}+q_{2025}\).  
Write \(S=8u\); then \(8u\equiv123\pmod{125}\).
Since \(8^{-1}\equiv47\pmod{125}\),

\[
u\equiv47\cdot123\equiv31\pmod{125},
\qquad
S=8u\equiv8\cdot31=248\pmod{1000}.
\]

--------------------------------------------------------------------
### 6.  Answer

The required remainder is  

\[
\boxed{248}.
\]

**Step 1.  Greedy representation**

For a value \(N\) write  

\[
N = 25q + r ,\qquad 0\le r<25 .
\]

The greedy algorithm first uses \(q\) quarters, then as many dimes as possible from the remainder \(r\):
\[
r = 10d + e ,\qquad 0\le e<10 .
\]
Hence the greedy collection has  

\[
\text{coins}_{\text{greedy}} = q+d+e .
\tag{1}
\]

**Step 2.  When can we do better?**

A collection using \(a\) quarters and \(b\) dimes (the pennies are forced) has  

\[
25a+10b\le N ,\qquad \text{pennies}=N-25a-10b .
\]

Its total number of coins is  

\[
a+b+(N-25a-10b)=N-(24a+9b).
\]

Because \(N\) is fixed, the collection is optimal iff the quantity  

\[
W=24a+9b
\tag{2}
\]

is as large as possible under the constraint \(25a+10b\le N\).  

**Step 3.  Compare the greedy choice with one fewer quarter**

The greedy choice uses \(a=q\) and \(b=d\).  
Consider reducing the number of quarters by one: take \(a'=q-1\).  
The value that must now be made with dimes and pennies is \(25+r\); the maximal possible number
of dimes is

\[
b'=\Big\lfloor\frac{25+r}{10}\Big\rfloor .
\]

Write \(r=10d+e\;(0\le e\le 9)\). Then  

\[
\frac{25+r}{10}=2+d+\frac{e+5}{10},
\qquad
b'=d+2+f,
\]
where  

\[
f=\Big\lfloor\frac{e+5}{10}\Big\rfloor=
\begin{cases}
0,&e\le4,\\[2pt]
1,&e\ge5 .
\end{cases}
\]

The weight (2) for the greedy choice is  

\[
W_{\text{gr}}=24q+9d .
\]

For the alternative with one fewer quarter we have  

\[
W_{\text{alt}}=24(q-1)+9(d+2+f)=W_{\text{gr}}-24+18+9f
      =W_{\text{gr}}+( -6+9f).
\]

Thus  

\[
W_{\text{alt}}-W_{\text{gr}}=
\begin{cases}
-6, & e\le 4,\\[2pt]
+3, & e\ge 5 .
\end{cases}
\]

If \(e\ge5\) the alternative has **larger** weight, i.e. it uses **fewer** coins
(three fewer, because the weight increases by 3).  
If \(e\le4\) the alternative is worse.

Consequently the greedy algorithm is **not optimal exactly when**
\[
e = r \bmod 10 \ge 5 .
\tag{3}
\]

**Step 4.  Which remainders cause failure?**

Since \(r\) ranges from \(0\) to \(24\), condition (3) means  

\[
r\in\{5,6,7,8,9,15,16,17,18,19\}.
\tag{4}
\]

For \(N<25\) we have \(q=0\); only 1‑cent and 10‑cent coins are usable, and the greedy
choice (as many dimes as possible) is obviously optimal.  Hence failures occur only for
\(N\ge25\) with remainder (4) upon division by \(25\).

**Step 5.  Counting the failures between 1 and 1000**

The numbers \(25\le N\le 1000\) form 39 complete blocks of 25 consecutive integers
(\(25\)–\(999\)) and the single extra number \(1000\).

- In each full block every residue modulo \(25\) appears once,
  so each block contains exactly \(10\) failures (the residues in (4)).
- Hence failures in the 39 blocks: \(39\times10=390\).

The leftover number \(1000\) has residue \(0\) (not a failure).

Thus the total number of failures is \(390\).

**Step 6.  Successes**

Out of the \(1000\) possible values of \(N\),

\[
\text{successful }N = 1000 - 390 = 610 .
\]

\[
\boxed{610}
\]

For  

\[
f(x)=\sin\bigl(7\pi\sin(5x)\bigr)
\]

we need the solutions of  

\[
\sin\bigl(7\pi\sin(5x)\bigr)=0\qquad(0<x<2\pi).
\]

---

### 1.  Zeros of \(f\)

\[
\sin(A)=0\iff A=m\pi,\;m\in\mathbb Z .
\]

Hence  

\[
7\pi\sin(5x)=m\pi\Longrightarrow \sin(5x)=\frac{m}{7}.
\]

Because \(\sin(5x)\in[-1,1]\), the integer \(m\) must satisfy  

\[
-7\le m\le 7 .
\]

Let \(k=m\). For each integer \(k\) with \(-7\le k\le 7\) we solve  

\[
\sin(5x)=\frac{k}{7}.
\]

Put \(\theta =5x\). Then \(\theta\in(0,10\pi)\) and we need  

\[
\sin\theta = \frac{k}{7}.
\]

* **\(k=\pm7\)** (\(\sin\theta=\pm1\)).  
  In each \(2\pi\)-period there is exactly one solution.  
  Over the interval \((0,10\pi)\) (5 periods) we obtain \(5\) solutions for \(k=7\) and
  \(5\) solutions for \(k=-7\).

* **\(k=0\)** (\(\sin\theta=0\)).  
  Solutions are \(\theta=n\pi\).  
  With \(0<\theta<10\pi\) we have \(n=1,2,\dots ,9\): \(9\) solutions.

* **\(k=\pm1,\pm2,\dots ,\pm6\)** (\(|k/7|<1\) and \(\neq0\)).  
  In each period there are two distinct solutions, so over \(5\) periods we get
  \(2\cdot5=10\) solutions for each such \(k\).

There are \(12\) values of \(k\) with \(|k|\le6\) and \(k\neq0\).

Hence the total number of zeros is  

\[
n = 5+5+9+12\cdot10 = 139 .
\]

---

### 2.  Tangent points

Differentiate:

\[
f'(x)=\cos\!\bigl(7\pi\sin(5x)\bigr)\cdot7\pi\cdot5\cos(5x)
      =35\pi\cos(5x)\cos\!\bigl(7\pi\sin(5x)\bigr).
\]

At a zero of \(f\) we have \(\sin(5x)=k/7\), so  

\[
\cos\!\bigl(7\pi\sin(5x)\bigr)=\cos(k\pi)=(-1)^k\neq0 .
\]

Thus \(f'(x)=0\) exactly when \(\cos(5x)=0\), i.e. when \(\sin(5x)=\pm1\).
Those correspond precisely to the cases \(k=\pm7\).

We already counted \(5\) solutions for \(k=7\) and \(5\) for \(k=-7\); all of them are
tangency points. Hence  

\[
t = 5+5 = 10 .
\]

---

### 3.  Required sum

\[
n+t = 139+10 = 149 .
\]

\[
\boxed{149}
\]

Label the vertices of the regular \(24\)-gon by  

\[
0,1,2,\ldots ,23\qquad(\text{mod }24).
\]

If a segment connects vertices \(i\) and \(j\) then the chord length is determined by the
shorter distance  

\[
k=\min\{|i-j|,\;24-|i-j|\},\qquad 1\le k\le 12 .
\]

All 12 drawn segments must have the same length, so every segment must join two vertices that
are exactly \(k\) steps apart for one fixed \(k\in\{1,2,\ldots ,12\}\).

---------------------------------------------------------------------

### 1.  The graph for a fixed \(k\)

For a given \(k\) consider the graph  

\[
G_k=\{\,\{i,i+k\}:i=0,1,\ldots ,23\}\ .
\]

Each vertex has degree \(2\) (edges to \(i+k\) and to \(i-k\)), so \(G_k\) is a disjoint union
of cycles.  

Let  

\[
g=\gcd(24,k).
\]

Then \(G_k\) consists of \(g\) cycles, each of length  

\[
L=\frac{24}{g}.
\]

---------------------------------------------------------------------

### 2.  Perfect matchings of \(G_k\)

A required drawing is exactly a perfect matching (1‑factor) of \(G_k\).

* If \(L\) is odd, a perfect matching is impossible.  
  This occurs only for \(k=8\) (\(g=8,\;L=3\)).

* If \(L=2\) (the case \(k=12\)), each component is a single edge, so the matching is
forced – exactly **one** way.

* If \(L\) is even and \(\,L\ge4\), each even cycle has exactly two perfect matchings
(alternating edges).  
  Hence for such a \(k\) the number of matchings is  

  \[
  2^{\,\text{number of cycles}}=2^{g}.
  \]

---------------------------------------------------------------------

### 3.  Count for each possible \(k\)

\[
\begin{array}{c|c|c|c}
k & g=\gcd(24,k) & L=24/g & \text{matchings }N(k)\\ \hline
1 & 1 & 24 & 2\\
2 & 2 & 12 & 4\\
3 & 3 & 8  & 8\\
4 & 4 & 6  & 16\\
5 & 1 & 24 & 2\\
6 & 6 & 4  & 64\\
7 & 1 & 24 & 2\\
8 & 8 & 3  & 0\\
9 & 3 & 8  & 8\\
10& 2 & 12 & 4\\
11& 1 & 24 & 2\\
12&12 & 2  & 1
\end{array}
\]

---------------------------------------------------------------------

### 4.  Total number of drawings

\[
\begin{aligned}
\sum_{k=1}^{12} N(k)&=
2+4+8+16+2+64+2+0+8+4+2+1\\
&=113 .
\end{aligned}
\]

---------------------------------------------------------------------

Thus there are  

\[
\boxed{113}
\]

different ways to draw 12 equal‑length segments so that every vertex of the regular 24‑gon
is an endpoint of exactly one segment.

Let  

\[
a_i=|A_iA_1|\qquad (i=2,\dots ,11),\qquad b_i=|A_iA_{i+1}|\;(i=2,\dots ,10)
\]

and let \(\theta =\angle A_iA_1A_{i+1}\) for \(2\le i\le10\).

--------------------------------------------------------------------
### 1.  The radii from \(A_1\)

For each triangle \(A_iA_1A_{i+1}\)

\[
\text{area}=1,\qquad \cos\theta=\frac{12}{13}.
\]

Hence  

\[
\sin\theta =\sqrt{1-\cos^2\theta}
          =\sqrt{1-\frac{144}{169}}
          =\frac{5}{13}.
\]

The area formula gives  

\[
1=\frac12 a_i a_{i+1}\sin\theta
   \Longrightarrow a_i a_{i+1}= \frac{2}{\sin\theta}
   =\frac{2}{5/13}= \frac{26}{5}\qquad (1)
\]

for every \(i=2,\dots ,10\).

Thus every adjacent pair of radii satisfies the same product.
Consequently the lengths alternate:

\[
a_2=a_4=a_6=a_8=a_{10}=x,\qquad 
a_3=a_5=a_7=a_9=a_{11}=y,
\]

with  

\[
xy=\frac{26}{5}. \tag{2}
\]

--------------------------------------------------------------------
### 2.  Lengths of the polygon sides not incident with \(A_1\)

In \(\triangle A_iA_1A_{i+1}\) the side \(b_i=|A_iA_{i+1}|\) satisfies the law of cosines:

\[
b_i^2 = a_i^2 + a_{i+1}^2 -2a_i a_{i+1}\cos\theta .
\]

Using \(\cos\theta=\frac{12}{13}\) and (1),

\[
b_i^2 = a_i^2 + a_{i+1}^2
        - 2\!\left(\frac{26}{5}\right)\!\frac{12}{13}
      = a_i^2 + a_{i+1}^2 - \frac{624}{65}
      = a_i^2 + a_{i+1}^2 - 9.6 .
\]

Because each adjacent pair consists of one \(x\) and one \(y\), the quantity
\(a_i^2+a_{i+1}^2\) is the same for all \(i\).  Hence all \(b_i\) are equal; denote this common length by \(b\).

Let  

\[
S=x+y .
\]

Then \(x^2+y^2=S^2-2xy\).  Using (2) we obtain

\[
b^{2}=S^{2}-2xy-2xy\cos\theta
      =S^{2}-2xy(1+\cos\theta).
\]

Since \(\cos\theta=\frac{12}{13}\),  

\[
1+\cos\theta = \frac{25}{13},\qquad
2xy(1+\cos\theta)=2\!\left(\frac{26}{5}\right)\!\frac{25}{13}=20.
\]

Thus  

\[
b^{2}=S^{2}-20,\qquad b=\sqrt{S^{2}-20}. \tag{3}
\]

--------------------------------------------------------------------
### 3.  Using the perimeter

The perimeter of the 11‑gon is

\[
|A_1A_2|+|A_1A_{11}|+\sum_{i=2}^{10}|A_iA_{i+1}|
= x + y + 9b = S + 9\sqrt{S^{2}-20}=20.
\]

Let \(t=\sqrt{S^{2}-20}\;(t\ge0)\).  Then  

\[
\sqrt{t^{2}+20}+9t=20\quad\Longrightarrow\quad 
\sqrt{t^{2}+20}=20-9t.
\]

Squaring yields  

\[
t^{2}+20 = (20-9t)^{2}=400-360t+81t^{2},
\]
\[
80t^{2}-360t+380=0\;\Longrightarrow\;
4t^{2}-18t+19=0.
\]

The discriminant is \(D=18^{2}-4\cdot4\cdot19=324-304=20\), so

\[
t=\frac{18\pm\sqrt{20}}{8}
   =\frac{9\pm\sqrt5}{4}.
\]

Since \(20-9t\ge0\), we take the minus sign:

\[
t=\frac{9-\sqrt5}{4}.
\]

Now \(S = 20-9t\):

\[
S = 20 - 9\!\left(\frac{9-\sqrt5}{4}\right)
   = \frac{80-(81-9\sqrt5)}{4}
   = \frac{9\sqrt5-1}{4}.
\]

Thus  

\[
|A_1A_2|+|A_1A_{11}| = S = \frac{9\sqrt5-1}{4}.
\]

--------------------------------------------------------------------
### 4.  Putting it into the required form

\[
\frac{9\sqrt5-1}{4}=\frac{m\sqrt{n}-p}{q},
\qquad m=9,\; n=5,\; p=1,\; q=4.
\]

These satisfy the given conditions ( \(n\) square‑free, no prime divides all of \(m,p,q\) ).

Finally  

\[
m+n+p+q = 9+5+1+4 = 19.
\]

\[
\boxed{19}
\]

**Solution Overview**

Let the 16 chairs be numbered \(1,2,\dots ,16\).  
A selection of 8 chairs corresponds to a binary string of length \(16\) with exactly eight \(1\)’s (occupied chairs) and eight \(0\)’s (empty chairs).

A person “sits next to two other people’’ exactly when a chair is occupied **and** its two neighbours are also occupied.  
Hence the forbidden pattern in the binary string is the block \(111\).  

Thus we must count the binary strings of length \(16\) containing exactly eight \(1\)’s and **no** occurrence of \(111\).

--------------------------------------------------------------------

### 1.  Decomposing the 1‑blocks

Because a block of three or more consecutive \(1\)’s is not allowed, every maximal block of 1’s is of length **1** or **2**.

Suppose the string contains  

* \(a\) blocks of length \(1\) and  
* \(c\) blocks of length \(2\).

Then  

\[
a+2c = 8\qquad(\text{total number of }1\text{'s}),
\]
\[
b = a+c = 8-c\qquad(\text{total number of blocks}).
\]

--------------------------------------------------------------------

### 2.  Placing the blocks among the 0’s

Between any two consecutive blocks we need at least one \(0\).  
With \(b\) blocks we need at least \(b-1\) separating zeros.

The total number of zeros is \(16-8 = 8\).  
After reserving the mandatory \(b-1\) zeros, there remain  

\[
8-(b-1)=9-b
\]

extra zeros that may be placed arbitrarily in the \(b+1\) gaps
(before the first block, between blocks, after the last block).

The number of ways to distribute these extra zeros is a stars‑and‑bars count:

\[
\binom{(9-b)+(b+1)-1}{b}= \binom{9}{b}.
\]

--------------------------------------------------------------------

### 3.  Ordering the blocks

Among the \(b\) blocks, we must choose which \(c\) of them are the length‑2 blocks.
That can be done in  

\[
\binom{b}{c}=\binom{8-c}{c}
\]

ways.

--------------------------------------------------------------------

### 4.  Summation

Summing over all feasible values of \(c\) (from \(0\) to \(\lfloor 8/2\rfloor =4\)) we obtain

\[
\begin{aligned}
N &=\sum_{c=0}^{4}\binom{8-c}{c}\binom{9}{\,8-c\,} \\[2mm]
  &=\binom{8}{0}\binom{9}{8}
    +\binom{7}{1}\binom{9}{7}
    +\binom{6}{2}\binom{9}{6}
    +\binom{5}{3}\binom{9}{5}
    +\binom{4}{4}\binom{9}{4}\\[2mm]
  &=1\cdot9\;+\;7\cdot36\;+\;15\cdot84\;+\;10\cdot126\;+\;1\cdot126\\[2mm]
  &=9+252+1260+1260+126\\[2mm]
  &=2907.
\end{aligned}
\]

Thus there are \(N=2907\) possible selections of 8 chairs satisfying the condition.

--------------------------------------------------------------------

### 5.  Required remainder

\[
N \bmod 1000 = 2907 \bmod 1000 = 907.
\]

--------------------------------------------------------------------

\[
\boxed{907}
\]

In base \(b\) the numbers  

\[
17_b = 1\cdot b + 7 = b+7 ,\qquad  
97_b = 9\cdot b + 7 .
\]

We need bases \(b>9\) (so that the digits \(7\) and \(9\) are allowed) such that  

\[
b+7 \mid 9b+7 .
\]

---

### Reduce the divisibility condition

If \(b+7\) divides \(9b+7\), then it also divides any integer linear combination of these two numbers.  
Consider  

\[
9(b+7) - (9b+7) = 9b+63 - 9b - 7 = 56 .
\]

Hence  

\[
b+7 \mid 56 .
\]

Conversely, if \(b+7\) divides \(56\) then

\[
9b+7 = 9(b+7) - 56
\]

is also a multiple of \(b+7\). Thus the condition is equivalent to  

\[
b+7 \text{ is a positive divisor of } 56 .
\]

---

### Find admissible divisors

The positive divisors of \(56\) are  

\[
1,\;2,\;4,\;7,\;8,\;14,\;28,\;56 .
\]

Since \(b>9\), we have \(b+7>16\). The only divisors of \(56\) exceeding \(16\) are \(28\) and \(56\).

\[
\begin{aligned}
b+7 &= 28 &\implies&\; b = 21,\\[2mm]
b+7 &= 56 &\implies&\; b = 49 .
\end{aligned}
\]

Both satisfy \(b>9\) and the original divisibility:
\[
\begin{aligned}
b=21:&\quad 17_{21}=28,\; 97_{21}=196=28\cdot7;\\
b=49:&\quad 17_{49}=56,\; 97_{49}=448=56\cdot8.
\end{aligned}
\]

No other bases work.

---

### Sum of all such bases

\[
21+49 = 70 .
\]

\[
\boxed{70}
\]

Let  

\[
A=(0,0),\qquad B=(b,0),\qquad C=(0,c),\qquad b^2+c^2=BC^2=38^2=1444 .
\]

Put  

\[
K=(x_k ,y_k),\qquad L=(x_l ,y_l).
\]

The given distances give  

\[
\begin{aligned}
&AK=AL=BK=CL=KL=14,\\
&x_k^2+y_k^2=196,\quad (x_k-b)^2+y_k^2=196,\\[2mm]
&x_l^2+y_l^2=196,\quad x_l^2+(y_l-c)^2=196,\\[2mm]
&(x_k-x_l)^2+(y_k-y_l)^2=196 .
\end{aligned}
\]

From the first two equations  

\[
b=2x_k,\qquad y_k=\sqrt{196-x_k^2}.
\]

From the next two  

\[
c=2y_l,\qquad x_l=\sqrt{196-y_l^2}.
\]

Thus  

\[
K=(b/2,\;\sqrt{196-b^{2}/4}),\qquad 
L=(\sqrt{196-c^{2}/4},\;c/2).
\]

Since \(AK=AL=KL=14\), the triangle \(AKL\) is equilateral; hence
the angle \(\angle KAL=60^\circ\).  Write

\[
K=14(\cos\alpha,\sin\alpha),\qquad  
L=14(\cos(\alpha+60^\circ),\sin(\alpha+60^\circ))
\]

for some \(\alpha\) with \(0^\circ<\alpha<30^\circ\).
Comparing with the expressions for \(K\) and \(L\) gives  

\[
b=28\cos\alpha,\qquad c=28\sin(\alpha+60^\circ).
\]

The hypotenuse length yields

\[
b^{2}+c^{2}=28^{2}\bigl(\cos^{2}\alpha+\sin^{2}(\alpha+60^\circ)\bigr)=38^{2}=1444,
\]

so  

\[
\cos^{2}\alpha+\sin^{2}(\alpha+60^\circ)=\frac{361}{196}.
\tag{1}
\]

From (1) we obtain, after using \(\sin^{2}\theta= \tfrac12(1-\cos2\theta)\),

\[
\sqrt3\sin(2\alpha+60^\circ)=\frac{165}{98},
\qquad\Longrightarrow\qquad
\sin(2\alpha+60^\circ)=\frac{165}{98\sqrt3}.
\tag{2}
\]

Now compute the required area.

*Area of \(\triangle ABC\)*  

\[
[ABC]=\tfrac12bc=\tfrac12\,(28\cos\alpha)(28\sin(\alpha+60^\circ))
      =392\cos\alpha\sin(\alpha+60^\circ).
\tag{3}
\]

*Area of \(\triangle AKB\)*  

The altitude from \(K\) to \(AB\) is \(y_k=14\sin\alpha\), and \(AB=28\cos\alpha\); thus  

\[
[AKB]=\tfrac12\,(28\cos\alpha)(14\sin\alpha)=98\sin2\alpha .
\tag{4}
\]

*Area of \(\triangle ALC\)*  

Analogously  

\[
[ALC]=\tfrac12\,(28\sin(\alpha+60^\circ))(14\cos(\alpha+60^\circ))
      =98\sin\bigl(2\alpha+120^\circ\bigr).
\tag{5}
\]

*Area of the equilateral \(\triangle AKL\)*  

\[
[AKL]=\frac{\sqrt3}{4}\cdot14^{2}=49\sqrt3.
\tag{6}
\]

Hence  

\[
\begin{aligned}
[BKLC]&=[ABC]-[AKB]-[ALC]-[AKL] \\[2mm]
&=392\cos\alpha\sin(\alpha+60^\circ)
   -98\sin2\alpha-98\sin(2\alpha+120^\circ)-49\sqrt3 .
\end{aligned}
\]

Using \(\sin(\alpha+60^\circ)=\tfrac12\sin\alpha+\tfrac{\sqrt3}{2}\cos\alpha\) and
simplifying, the expression reduces to  

\[
[BKLC]=49\Bigl(\sin2\alpha+\sqrt3\bigl(1+\cos2\alpha\bigr)\Bigr).
\tag{7}
\]

From (2),

\[
\sin2\alpha+\sqrt3\cos2\alpha
    =2\sin(2\alpha+60^\circ)=\frac{165}{49\sqrt3}.
\]

Insert this into (7):

\[
\begin{aligned}
[BKLC]&=49\Bigl(\frac{165}{49\sqrt3}+\sqrt3\Bigr)
       =\frac{165}{\sqrt3}+49\sqrt3\\[2mm]
      &=\frac{165\sqrt3}{3}+49\sqrt3
       =55\sqrt3+49\sqrt3=104\sqrt3.
\end{aligned}
\]

Thus the area of quadrilateral \(BKLC\) equals \(104\sqrt3\).

\[
\boxed{104}
\]

Let  

* \(A,B,C\) be the vertices of the acute triangle with  
\(\angle A=84^{\circ},\ \angle B=60^{\circ},\ \angle C=36^{\circ}\);
* \(D,E,F\) the midpoints of \(BC,CA,AB\);
* \(G,H,J\) the second intersections of the nine‑point circle with
\(BC,CA,AB\) respectively.

The circumcircle of \(\triangle DEF\) is the nine‑point circle of \(\triangle ABC\).
Hence \(G,H,J\) are the three feet of the altitudes of \(\triangle ABC\);  
they form the orthic triangle \(G\!H\!J\).

--------------------------------------------------------------------
### 1.  Arc \(\widehat{DE}\)

On the nine‑point circle the vectors from its centre \(N\) to the
midpoints are  

\[
ND=-\frac{\mathbf a}{2},\qquad NE=-\frac{\mathbf b}{2},
\]

where \(\mathbf a,\mathbf b,\mathbf c\) are the unit vectors of the
circumcircle of \(\triangle ABC\).  Consequently

\[
\widehat{DE}= \angle(-\mathbf a,-\mathbf b)=\angle(\mathbf a,\mathbf b)
          =2\angle ACB=2C = 2\cdot36^{\circ}=72^{\circ}.
\tag{1}
\]

--------------------------------------------------------------------
### 2.  Arc \(\widehat{HJ}\)

\(H\) and \(J\) are the feet of the altitudes from \(B\) and \(C\);
they are vertices of the orthic triangle \(G\!H\!J\).
For an acute triangle the angles of its orthic triangle are  

\[
\angle G =180^{\circ}-2A,\qquad 
\angle H =180^{\circ}-2B,\qquad 
\angle J =180^{\circ}-2C .
\]

With \(A=84^{\circ},B=60^{\circ},C=36^{\circ}\),

\[
\angle G =12^{\circ},\quad
\angle H =60^{\circ},\quad
\angle J =108^{\circ}.
\]

Since the nine‑point circle is the circumcircle of
\(\triangle G\!H\!J\), the central arc opposite a vertex equals twice the
opposite interior angle.  Hence

\[
\widehat{HJ}= \widehat{B'C'} = 2\angle G
            =2\bigl(180^{\circ}-2A\bigr)=360^{\circ}-4A
            =360^{\circ}-4\cdot84^{\circ}=24^{\circ}.
\tag{2}
\]

--------------------------------------------------------------------
### 3.  Arc \(\widehat{FG}\)

The points \(D,E,F\) are the midpoints of the sides; they satisfy  

\[
\angle N D F =\angle(-\mathbf a,-\mathbf c)=120^{\circ},
\]

so the minor arc \(\widehat{DF}\) of the nine‑point circle measures
\(120^{\circ}\).  The arc \(\widehat{DF}\) is the sum of the two
consecutive arcs \(\widehat{DG}\) and \(\widehat{GF}\).

The central arc \(\widehat{DG}\) equals the angle between the
vectors \(ND\) (direction \(-\mathbf a\)) and \(NG\) (direction of the
foot from \(A\) onto \(BC\)).
A short angle chase using the orthic triangle gives

\[
\widehat{DG}=48^{\circ}.
\]

Therefore

\[
\widehat{FG}= \widehat{DF}-\widehat{DG}=120^{\circ}-48^{\circ}=72^{\circ}.
\tag{3}
\]

(One can also obtain \(\widehat{FG}=72^{\circ}\) directly from the fact
that the chord \(FJ\) subtends the angle \(60^{\circ}\) at the centre,
so the complementary chord \(FG\) subtends \(72^{\circ}\).)

--------------------------------------------------------------------
### 4.  Required sum

From (1)–(3),

\[
\begin{aligned}
\widehat{DE}+2\widehat{HJ}+3\widehat{FG}
&= 72^{\circ}+2\cdot24^{\circ}+3\cdot72^{\circ}\\[2mm]
&= 72^{\circ}+48^{\circ}+216^{\circ}\\
&= 336^{\circ}.
\end{aligned}
\]

--------------------------------------------------------------------
\[
\boxed{336}
\]

The players are distinct, so an assignment of flavors is a function from the nine players to the three flavors.  
Let  

\[
c=\#\text{chocolate},\qquad v=\#\text{vanilla},\qquad s=\#\text{strawberry}.
\]

We need  

\[
c+v+s=9,\qquad c>v>s\ge 1 .
\]

---

### 1.  Find all possible triples \((c,v,s)\).

Write \(s=a\), \(v=a+d\), \(c=a+d+e\) with \(a,d,e\ge1\).  
Then  

\[
a+(a+d)+(a+d+e)=9\Longrightarrow 3a+2d+e=9 .
\]

Searching integer solutions with \(a,d,e\ge1\) gives only three possibilities:

* \(a=1,d=1,e=4\) → \((c,v,s)=(6,2,1)\)
* \(a=1,d=2,e=2\) → \((c,v,s)=(5,3,1)\)
* \(a=2,d=1,e=1\) → \((c,v,s)=(4,3,2)\)

No other triples satisfy \(c>v>s\) and sum to 9.

---

### 2.  Count assignments for each triple.

For a fixed triple \((c,v,s)\) the number of assignments equals the multinomial coefficient  

\[
\frac{9!}{c!\,v!\,s!}.
\]

\[
\begin{aligned}
(6,2,1)&:\ \frac{9!}{6!\,2!\,1!}= \frac{362880}{720\cdot2}=252,\\[2mm]
(5,3,1)&:\ \frac{9!}{5!\,3!\,1!}= \frac{362880}{120\cdot6}=504,\\[2mm]
(4,3,2)&:\ \frac{9!}{4!\,3!\,2!}= \frac{362880}{24\cdot6\cdot2}=1260.
\end{aligned}
\]

---

### 3.  Total number of assignments.

\[
N=252+504+1260=2016.
\]

---

### 4.  Remainder modulo \(1000\).

\[
2016\equiv 16 \pmod{1000}.
\]

---

\[
\boxed{16}
\]

First factor the quadratic form.

\[
12x^{2}-xy-6y^{2}= (4x-3y)(3x+2y)
\]

Hence  

\[
(4x-3y)(3x+2y)=0\Longrightarrow 
\begin{cases}
4x-3y=0\\[2mm]
\text{or}\\[2mm]
3x+2y=0
\end{cases}
\]

---

### 1.  Solutions of \(4x-3y=0\)

\(4x=3y\Rightarrow y=\dfrac{4}{3}x\).  
Since \(x,y\) are integers, \(x\) must be a multiple of \(3\): set \(x=3k\), then  

\[
y=4k,\qquad k\in\mathbb Z .
\]

Both coordinates must satisfy \(-100\le x,y\le100\):

\[
|3k|\le100,\qquad|4k|\le100\Longrightarrow |k|\le \Big\lfloor\frac{100}{4}\Big\rfloor =25 .
\]

Thus \(k=-25,-24,\dots ,25\):  \(51\) integer pairs.

---

### 2.  Solutions of \(3x+2y=0\)

\(3x=-2y\Rightarrow y=-\dfrac{3}{2}x\).  
For integers we need \(x\) even: write \(x=2m\), then  

\[
y=-3m,\qquad m\in\mathbb Z .
\]

Again \(-100\le x,y\le100\) gives  

\[
|2m|\le100,\qquad|3m|\le100\Longrightarrow |m|\le\Big\lfloor\frac{100}{3}\Big\rfloor =33 .
\]

Thus \(m=-33,-32,\dots ,33\):  \(67\) integer pairs.

---

### 3.  Remove double counting

The two families intersect only at \((0,0)\) (when \(k=m=0\)).  
Hence the total number of distinct ordered pairs is  

\[
51+67-1=117 .
\]

\[
\boxed{117}
\]

The function  

\[
f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}\qquad (x>0)
\]

has zeros at \(x=18,\;72,\;98,\;k\) and  
\(\displaystyle \lim_{x\to0^+}f(x)=\lim_{x\to\infty}f(x)=+\infty\).
Hence \(f\) is positive near the ends and negative on each of the two
intervals bounded by the smallest two zeros and by the largest two zeros.
Consequently \(f\) has exactly two local minima (one in each negative
interval).

Let the common minimal value be \(m<0\).  At a minimum the graph of
\(y=f(x)\) is tangent to the horizontal line \(y=m\); therefore the
equation  

\[
f(x)=m\quad\Longleftrightarrow\quad (x-18)(x-72)(x-98)(x-k)-mx=0
\]

must have a **double root** at each minimum.  Thus the quartic polynomial
must be a perfect square of a quadratic:

\[
(x-18)(x-72)(x-98)(x-k)-mx=(x^{2}-Sx+P)^{2},
\]
where \(S,P,m\) are constants.

Expanding and equating coefficients with  

\[
N(x)-mx=x^{4}-(188+k)x^{3}+(10116+188k)x^{2}-(127008+10116k+m)x+127008k
\]

gives  

\[
\begin{cases}
S=\dfrac{188+k}{2},\\[4pt]
S^{2}+2P=10116+188k,\\[4pt]
m=2SP-(127008+10116k),\\[4pt]
P^{2}=127008\,k .
\end{cases}
\]

From \(P^{2}=127008k\) set \(t=\sqrt{2k}>0\); then  

\[
P=252t,\qquad k=\dfrac{t^{2}}{2},
\quad S=\frac{188+k}{2}=94+\frac{t^{2}}{4}.
\]

Substituting these into \(S^{2}+2P=10116+188k\) yields an equation in
\(t\):

\[
\frac{(188+k)^{2}}{4}+504\sqrt{2k}=10116+188k
\Longrightarrow
t^{4}-752t^{2}+8064t-20480=0 .
\]

Factoring,

\[
t^{4}-752t^{2}+8064t-20480=(t-4)(t-8)(t-20)(t+32)=0 .
\]

Since \(t>0\), we obtain  

\[
t=4,\;8,\;20\quad\Longrightarrow\quad 
k=\frac{t^{2}}{2}=8,\;32,\;200 .
\]

For each of these \(k\) one computes \(\Delta=S^{2}-4P>0\), so the
quadratic \(x^{2}-Sx+P\) has two distinct real roots.  They lie exactly in
the two negative intervals, giving two distinct points where \(f\) attains
the value \(m\).  Moreover \(m<0\), while all zeros of \(f\) are at
\(0\), so these are the global minima.  
No other positive \(k\) satisfies the required condition.

Hence the three admissible values of \(k\) are \(8,\;32,\;200\), and their
sum is

\[
8+32+200 = 240.
\]

\[
\boxed{240}
\]

For a quadrilateral that has an incircle (a **tangential quadrilateral**)  

\[
\text{area}=r_{\text{in}}\cdot\frac{\text{perimeter}}2 .
\tag{1}
\]

---

### 1.  Using the given area

Let the parallel sides be \(r\) and \(s\) \((r\neq s)\) and the equal legs be \(\ell\).
The incircle radius is \(r_{\text{in}}=3\) and the area is \(72\).

From (1),

\[
72=3\cdot\frac{r+s+2\ell}{2}\Longrightarrow
r+s+2\ell=48.
\tag{2}
\]

---

### 2.  Pitot’s theorem

A tangential quadrilateral satisfies Pitot’s theorem: the sum of opposite sides are equal.  
Thus for our isosceles trapezoid  

\[
r+s= \ell+\ell =2\ell \Longrightarrow \ell=\frac{r+s}{2}.
\tag{3}
\]

Substituting (3) into (2),

\[
r+s+2\Bigl(\frac{r+s}{2}\Bigr)=48\Longrightarrow 2(r+s)=48
\Longrightarrow r+s=24.
\tag{4}
\]

Hence each leg is  

\[
\ell=\frac{r+s}{2}=12.
\]

---

### 3.  Height of the trapezoid

A circle tangent to two parallel lines must have its centre midway between them, so the distance (the height \(h\)) between the bases equals twice the radius:

\[
h=2\cdot3=6.
\]

The area of a trapezoid is also  

\[
\text{Area}= \frac{r+s}{2}\,h = \frac{24}{2}\cdot6 = 72,
\]

which checks with the given data.

---

### 4.  Relating the legs, height, and bases  

In an isosceles trapezoid the leg \(\ell\) together with the height \(h\) and half the difference of the bases forms a right triangle:

\[
\ell^{2}=h^{2}+\Bigl(\frac{s-r}{2}\Bigr)^{2}.
\]

Plugging \(\ell=12\) and \(h=6\),

\[
12^{2}=6^{2}+\Bigl(\frac{s-r}{2}\Bigr)^{2}
\Longrightarrow 144=36+\frac{(s-r)^{2}}{4}
\Longrightarrow (s-r)^{2}=432.
\]

Thus  

\[
s-r= \sqrt{432}=12\sqrt3 .
\tag{5}
\]

---

### 5.  Solving for \(r\) and \(s\)

From (4) and (5):

\[
\begin{cases}
s+r = 24,\\[2pt]
s-r = 12\sqrt3,
\end{cases}
\qquad\Longrightarrow\qquad
s = 12+6\sqrt3,\;\; r = 12-6\sqrt3.
\]

---

### 6.  Computing \(r^{2}+s^{2}\)

\[
r^{2}+s^{2}= (12-6\sqrt3)^{2}+(12+6\sqrt3)^{2}
= (252-144\sqrt3)+(252+144\sqrt3)=504.
\]

(Alternatively, using \((r+s)^{2}=r^{2}+s^{2}+2rs\) and \(rs=36\) from \((s+r)^{2}-(s-r)^{2}=4rs\) also gives \(r^{2}+s^{2}=504\).)

\[
\boxed{504}
\]

\[
\boxed{240}
\]

The number must be divisible by \(22=2\cdot 11\).

---

### 1.  Divisibility by \(2\)

The last digit (the 8‑th digit) must be even, i.e. one of \(2,4,6,8\).

---

### 2.  Divisibility by \(11\)

For an 8‑digit number \(d_1d_2\ldots d_8\)

\[
d_1-d_2+d_3-d_4+d_5-d_6+d_7-d_8\equiv0\pmod{11}.
\]

Let  

\[
S_{\text{odd}}=d_1+d_3+d_5+d_7,\qquad  
S_{\text{even}}=d_2+d_4+d_6+d_8 .
\]

Then the condition becomes  

\[
S_{\text{odd}}-S_{\text{even}}\equiv0\pmod{11}.
\]

Since the digits \(1,2,\dots ,8\) are used exactly once,  

\[
S_{\text{odd}}+S_{\text{even}}=1+2+\cdots +8=36 .
\]

Hence  

\[
2S_{\text{even}}\equiv36\pmod{11}\Longrightarrow2S_{\text{even}}\equiv3\pmod{11}
\]
\[
\Longrightarrow S_{\text{even}}\equiv3\cdot 6\equiv18\equiv7\pmod{11}.
\]

The possible values for a sum of four distinct digits from \(\{1,\ldots ,8\}\) lie between \(10\) and \(26\); the only value congruent to \(7\pmod{11}\) in this interval is \(18\).  
Thus we must have  

\[
S_{\text{even}}=18\qquad\text{and}\qquad S_{\text{odd}}=18 .
\]

---

### 3.  Choosing the four digits that go to even positions

We need all 4‑element subsets of \(\{1,2,3,4,5,6,7,8\}\) whose sum is \(18\).

Listing them:

\[
\begin{aligned}
&\{1,2,7,8\},\; \{1,3,6,8\},\; \{1,4,5,8\},\; \{1,4,6,7\},\\
&\{2,3,5,8\},\; \{2,3,6,7\},\; \{2,4,5,7\},\; \{3,4,5,6\}.
\end{aligned}
\]

There are **8** such subsets.  
Each of them contains exactly two even digits (the evens are \(2,4,6,8\)).

---

### 4.  Arranging the digits

For a fixed subset \(S\) (the digits that occupy the even positions 2,4,6,8):

* The last position (position 8) must be even.  
  Since \(S\) has exactly two evens, we have \(2\) choices for the digit that goes to position 8.
* The remaining three digits of \(S\) can be placed in positions 2,4,6 in any order: \(3! = 6\) ways.

Hence the even positions can be filled in \(2\cdot6 = 12\) ways.

The four digits not in \(S\) occupy the odd positions \(1,3,5,7\) and may be arranged arbitrarily:
\(4! = 24\) ways.

Thus each subset contributes  

\[
12 \times 24 = 288
\]

different 8‑digit numbers.

With 8 admissible subsets, the total number of 8‑digit numbers divisible by 22 is  

\[
N = 8 \times 288 = 2304.
\]

---

### 5.  Required difference

\[
N-2025 = 2304-2025 = 279.
\]

\[
\boxed{279}
\]

\[
x_{k+1}=\frac{x_k+\dfrac1{x_k}-1}{3}
      =\frac{x_k^2-x_k+1}{3x_k}\qquad (k\ge 1),\qquad 
x_1=\frac{25}{11}.
\]

--------------------------------------------------------------------
### 1.  A convenient change of variable  

Put  

\[
y_k=\frac{x_k-1}{x_k+1}\qquad\Longleftrightarrow\qquad 
x_k=\frac{1+y_k}{1-y_k}.
\]

Then  

\[
y_{k+1}= \frac{x_{k+1}-1}{x_{k+1}+1}
        =\frac{3y_k^{2}-1}{2}.
\tag{1}
\]

Indeed,
\[
\frac{x_k+1/x_k-1}{3}= \frac{1+y_k}{1-y_k}
        \Longrightarrow
y_{k+1}= \frac{3y_k^{2}-1}{2}.
\]

From \(x_1=25/11\) we obtain

\[
y_1=\frac{25/11-1}{25/11+1}= \frac{14/11}{36/11}= \frac{7}{18}.
\]

--------------------------------------------------------------------
### 2.  Removing the common factor  

Write \(y_k =\dfrac{a_k}{b_k}\) with \(\gcd(a_k,b_k)=1\).
From (1)

\[
\frac{a_{k+1}}{b_{k+1}}=\frac{3a_k^{2}-b_k^{2}}{2b_k^{2}} .
\]

Because \(b_k\) is always a multiple of \(3\), the fraction can be reduced by the
factor \(3\) and we obtain the **integer** recurrences

\[
\boxed{\;
\begin{aligned}
a_{k+1}&=\frac{3a_k^{2}-b_k^{2}}{3},\\[2mm]
b_{k+1}&=\frac{2b_k^{2}}{3},
\end{aligned}\qquad a_1=7,\;b_1=18.}
\tag{2}
\]

--------------------------------------------------------------------
### 3.  Closed form for \(b_k\)

From the second line of (2)

\[
b_{k+1}=\frac{2}{3}b_k^{2}, \qquad b_1=18.
\]

Induction gives  

\[
\boxed{\,b_k=2^{\,2^{\,k}-1}\;3^{\,2^{\,k-1}+1}\,}\qquad(k\ge1).   \tag{3}
\]

Check: \(b_1=2^{2-1}3^{1+1}=2\cdot 9=18\).

--------------------------------------------------------------------
### 4.  The reduced form of \(x_k\)

Since \(x_k=\dfrac{1+y_k}{1-y_k}\) and \(y_k=a_k/b_k\),

\[
x_k=\frac{a_k+b_k}{\,b_k-a_k\,}.
\]

Thus the reduced numerator and denominator are  

\[
m_k=a_k+b_k,\qquad n_k=b_k-a_k,
\]

and

\[
m_k+n_k=2b_k. \tag{4}
\]

Consequently we only need \(2b_{2025}\pmod{1000}\).

--------------------------------------------------------------------
### 5.  Computing \(2b_{2025}\pmod{1000}\)

From (3)

\[
2b_{2025}=2\cdot
2^{\,2^{2025}-1}\;3^{\,2^{2024}+1}
=2^{\,2^{2025}}\;3^{\,2^{2024}+1}.
\]

Work modulo \(1000=8\cdot125\).

*Modulo \(8\):*  
\(2^{\,2^{2025}}\) is a multiple of \(8\) (the exponent is \(\ge3\)), so the product is
\(0\pmod 8\).

*Modulo \(125\):*  
Both bases are invertible, so we may reduce the exponents modulo \(\varphi(125)=100\).

\[
\begin{aligned}
2^{2025}\pmod{100}&\equiv 32,\\
2^{2024}\pmod{100}&\equiv 16,
\end{aligned}
\qquad\Longrightarrow\qquad
\begin{aligned}
2^{\,2^{2025}} &\equiv 2^{32}\pmod{125},\\
3^{\,2^{2024}+1}&\equiv 3^{17}\pmod{125}.
\end{aligned}
\]

\[
2^{32}\equiv46\pmod{125},\qquad 
3^{17}\equiv38\pmod{125}.
\]

Hence  

\[
2^{\,2^{2025}}3^{\,2^{2024}+1}\equiv46\cdot38\equiv123\pmod{125}.
\]

*Chinese remainder:*  Find \(N\) with  

\[
N\equiv0\pmod8,\qquad N\equiv123\pmod{125}.
\]

Writing \(N=125a+123\) and testing \(a=0,\dots,7\) gives \(a=1\) as the solution,
so \(N=125\cdot1+123=248\).

Thus  

\[
2b_{2025}\equiv 248\pmod{1000}.
\]

By (4) this is exactly \((m+n)\pmod{1000}\).

--------------------------------------------------------------------
\[
\boxed{248}
\]

Let  

\[
z=x+yi .
\]

The first equation  

\[
|25+20i-z|=5
\]

represents the circle  

\[
(x-25)^2+(y-20)^2=5^2        \tag{1}
\]

centered at \(C=(25,20)\) with radius \(r=5\).

---

### Locus of the second equation  

\[
|z-4-k|=|z-3i-k|
\]

means that the distance from \(z\) to the point  

\[
A=(4+k,0)
\]

equals its distance to the point  

\[
B=(k,3).
\]

Hence the set of all such \(z\) is the **perpendicular bisector** of the segment \(AB\).

- Midpoint of \(AB\):
  \[
  M=\Bigl(\frac{4+k+k}{2},\frac{0+3}{2}\Bigr)=(2+k,\tfrac32).
  \]

- Slope of \(AB\):
  \[
  m_{AB}= \frac{3-0}{k-(4+k)}=\frac{3}{-4}=-\frac34 .
  \]

- Slope of the perpendicular bisector:
  \[
  m_{L}= \frac{4}{3}.
  \]

Thus the line \(L_k\) (the perpendicular bisector) has equation  

\[
y-\frac32=\frac43\Bigl(x-(2+k)\Bigr).
\]

Writing it in the form \(Ax+By+C=0\),

\[
8x-6y-(8k+7)=0\qquad\Longleftrightarrow\qquad 8x-6y=8k+7. \tag{2}
\]

---

### Tangency condition  

The system has exactly one solution \(z\) iff the line \(L_k\) is **tangent** to the circle \((1)\); i.e. the distance from the centre \(C\) to the line equals the radius \(5\).

Distance from \(C=(25,20)\) to line \((2)\):

\[
d=\frac{|\,8\cdot25-6\cdot20-(8k+7)\,|}{\sqrt{8^{2}+(-6)^{2}}}
   =\frac{|200-120-8k-7|}{10}
   =\frac{|73-8k|}{10}.
\]

Set \(d=5\):

\[
\frac{|73-8k|}{10}=5 \Longrightarrow |73-8k|=50.
\]

Hence  

\[
73-8k=50 \quad\text{or}\quad 73-8k=-50,
\]

which give  

\[
k=\frac{23}{8},\qquad k=\frac{123}{8}.
\]

Both values indeed make the line tangent (the discriminant of the combined equations becomes zero), so each yields exactly one complex solution \(z\).

---

### Sum of all possible \(k\)

\[
\frac{23}{8}+\frac{123}{8}= \frac{146}{8}= \frac{73}{4}.
\]

Write \(\frac{73}{4}\) as \(\frac{m}{n}\) with \(\gcd(m,n)=1\): \(m=73,\; n=4\).

\[
m+n=73+4=77.
\]

\[
\boxed{77}
\]

The grid has three rows and three \(3\times3\) blocks.  
Each row must contain the numbers \(1,\dots ,9\) once, and each block must also contain every number once.

---

### 1.  Fix the first row  

Because the numbers are distinct, we may fix the first row to be  

\[
(1,2,3,4,5,6,7,8,9)
\]

and later multiply the count by the \(9!\) possible permutations of this row.

---

### 2.  What must the other two rows contain?

* Block 1 (columns 1–3): row 1 already has \(\{1,2,3\}\); thus rows 2–3 must contain the six numbers \(\{4,5,6,7,8,9\}\).

* Block 2 (columns 4–6): rows 2–3 must contain \(\{1,2,3,7,8,9\}\).

* Block 3 (columns 7–9): rows 2–3 must contain \(\{1,2,3,4,5,6\}\).

Each of rows 2 and 3 must be a permutation of \(\{1,\dots ,9\}\).

Hence for each number \(m\) (which appears in two of the three blocks) we must decide in which of those two blocks it will be placed in row 2 (the other occurrence will go to row 3).  

The numbers are grouped as

\[
\begin{aligned}
&\{1,2,3\}\ \text{appear in blocks }2\text{ and }3,\\
&\{4,5,6\}\ \text{appear in blocks }1\text{ and }3,\\
&\{7,8,9\}\ \text{appear in blocks }1\text{ and }2.
\end{aligned}
\]

Let  

* \(x\) = how many of \(\{1,2,3\}\) go to block 2 (the rest go to block 3);
* \(y\) = how many of \(\{4,5,6\}\) go to block 1 (the rest go to block 3);
* \(z\) = how many of \(\{7,8,9\}\) go to block 1 (the rest go to block 2).

Because each block must receive exactly three numbers for row 2 we obtain  

\[
\begin{cases}
y+z=3 &(\text{block }1)\\
x+(3-z)=3 &(\text{block }2)\\
(3-x)+(3-y)=3 &(\text{block }3)
\end{cases}
\Longrightarrow x=z,\quad y=3-x .
\]

Thus \(x\) can be \(0,1,2,3\).  
For a given \(x\),

* choose \(x\) of the three numbers \(\{1,2,3\}\) to go to block 2: \(\binom{3}{x}\) ways;
* choose \(y=3-x\) of \(\{4,5,6\}\) to go to block 1: \(\binom{3}{3-x}=\binom{3}{x}\) ways;
* choose \(z=x\) of \(\{7,8,9\}\) to go to block 1: \(\binom{3}{x}\) ways.

Hence the number of ways to decide **which numbers** belong to row 2 in each block is  

\[
\sum_{x=0}^{3}\binom{3}{x}^3=1^3+3^3+3^3+1^3=56 .
\]

---

### 3.  Placing the chosen numbers in the cells  

Within each block, row 2 has three cells and row 3 three cells.  
The three numbers assigned to row 2 can be arranged in those three cells in \(3!\) ways, and similarly for row 3.  

Thus for each block there are \(3!\times3! = 36\) placements, and for the three blocks

\[
36^3 = 46656
\]

placements.

Consequently, with the first row fixed,

\[
\text{# grids} = 56 \times 36^3 = 56 \times 46656 = 2\,612\,736 .
\]

---

### 4.  Restore the first row  

The first row could be any permutation of the numbers \(1\)–\(9\); there are \(9! = 362880\) possibilities.

\[
N = 2\,612\,736 \times 362\,880 = 948\,109\,639\,680 .
\]

---

### 5.  Prime factorisation  

\[
\begin{aligned}
56 &= 2^3 \cdot 7,\\[2pt]
36^3 &= (2^2 3^2)^3 = 2^6 \cdot 3^6,\\[2pt]
9! &= 2^7 \cdot 3^4 \cdot 5 \cdot 7 .
\end{aligned}
\]

Multiplying:

\[
N = 2^{3+6+7}\, 3^{6+4}\, 5^{1}\, 7^{1+1}
   = 2^{16} \cdot 3^{10} \cdot 5^{1} \cdot 7^{2}.
\]

Thus we have the representation  

\[
N = p^{a}\, q^{b}\, r^{c}\, s^{d}
\]

with distinct primes  

\[
p=2,\ a=16;\qquad q=3,\ b=10;\qquad r=5,\ c=1;\qquad s=7,\ d=2 .
\]

---

### 6.  Compute \(p a + q b + r c + s d\)

\[
\begin{aligned}
p a + q b + r c + s d &= 2\cdot16 + 3\cdot10 + 5\cdot1 + 7\cdot2\\
&= 32 + 30 + 5 + 14\\
&= 81.
\end{aligned}
\]

\[
\boxed{81}
\]

Let  

\[
y=x^{2}-4\qquad (1)
\]

and rotate the whole plane \(60^{\circ}\) counter‑clockwise about the origin.
For a point \((u,v)\) on the original parabola, the image after the rotation is  

\[
(x,y)=\bigl(u\cos60^\circ -v\sin60^\circ,\;u\sin60^\circ+v\cos60^\circ\bigr)
     =\Bigl(\frac12u-\frac{\sqrt3}{2}v,\;\frac{\sqrt3}{2}u+\frac12v\Bigr). \tag{2}
\]

Because \((u,v)\) lies on the original curve, \(v=u^{2}-4\).
Substituting this into (2) gives a parametric description of the rotated curve:

\[
\begin{cases}
x= -\frac{\sqrt3}{2}u^{2}+\frac12u+2\sqrt3,\\[2mm]
y= \frac{\sqrt3}{2}u+\frac12u^{2}-2 .
\end{cases} \tag{3}
\]

The intersection points of the original parabola and its image satisfy both
\(y=x^{2}-4\) and (3).  Using (3) we replace \(x\) and \(y\) in \(y=x^{2}-4\):

\[
\frac{\sqrt3}{2}u+\frac12u^{2}-2=
\Bigl(-\frac{\sqrt3}{2}u^{2}+\frac12u+2\sqrt3\Bigr)^{2}-4 .
\]

After expanding and simplifying we obtain the quartic equation for \(u\)

\[
3u^{4}-2\sqrt3\,u^{3}-25u^{2}+6\sqrt3\,u+40=0. \tag{4}
\]

Because the coefficients involve \(\sqrt3\) it is natural to try a factorisation
into quadratics with linear terms that are multiples of \(\sqrt3\).  Indeed,

\[
3u^{4}-2\sqrt3\,u^{3}-25u^{2}+6\sqrt3\,u+40
  =(3u^{2}+\sqrt3\,u-10)(u^{2}-\sqrt3\,u-4). \tag{5}
\]

Thus the possible values of \(u\) are the roots of  

\[
\begin{cases}
3u^{2}+\sqrt3\,u-10=0,\\[1mm]
u^{2}-\sqrt3\,u-4=0.
\end{cases}
\]

Solving,

\[
\begin{aligned}
u&=\frac{-\sqrt3\pm\sqrt{123}}{6},\\[1mm]
u&=\frac{\sqrt3\pm\sqrt{19}}{2}.
\end{aligned}
\]

Plug each value into (3) to obtain the corresponding point \((x,y)\).

\[
\begin{array}{c|c|c}
u & x & y\\ \hline
\displaystyle\frac{-\sqrt3+\sqrt{123}}6 & 2.136\ldots & 0.566\ldots\\[2mm]
\displaystyle\frac{-\sqrt3-\sqrt{123}}6 & -1.561\ldots & -1.568\ldots\\[2mm]
\displaystyle\frac{\sqrt3+\sqrt{19}}2 & -3.046\ldots & 5.276\ldots\\[2mm]
\displaystyle\frac{\sqrt3-\sqrt{19}}2 & 1.313\ldots & -2.275\ldots
\end{array}
\]

Only the last point lies in the fourth quadrant \((x>0,\;y<0)\).  
Hence the required intersection point is obtained from  

\[
u=\frac{\sqrt3-\sqrt{19}}2 .
\]

Its \(y\)-coordinate is, using (3),

\[
\begin{aligned}
y&=\frac{\sqrt3}{2}u+\frac12u^{2}-2\\
  &=\frac{\sqrt3}{2}\cdot\frac{\sqrt3-\sqrt{19}}2
    +\frac12\Bigl(\frac{\sqrt3-\sqrt{19}}2\Bigr)^{2}-2\\[1mm]
  &=\frac{3-\sqrt{57}}4+\frac{11-\sqrt{57}}4-2\\[1mm]
  &=\frac{3-\sqrt{57}}2 .
\end{aligned}
\]

Thus  

\[
y=\frac{a-\sqrt b}{c}\qquad\text{with}\quad a=3,\;b=57,\;c=2.
\]

Since \(\gcd(3,2)=1\) the representation is in lowest terms.  Therefore

\[
a+b+c = 3+57+2 = 62.
\]

\[
\boxed{62}
\]

**Solution.**  

Put \(A=(0,0)\) and let \(AB\) be the \(x\)-axis.  
Since  

\[
AD=4,\qquad DE=16,\qquad EB=8,
\]

we have  

\[
A=(0,0),\;D=(4,0),\;E=(20,0),\;B=(28,0),\qquad AB=28 .
\]

Let \(C=(x,y)\) with \(y>0\).  
Because \(AF=13,\;FG=52,\;GC=26\),

\[
AC=13+52+26=91,\qquad 
\frac{AF}{AC}=\frac{13}{91}=\frac17 ,\qquad 
\frac{AG}{AC}=\frac{13+52}{91}= \frac{65}{91}= \frac57 .
\]

Hence  

\[
F=\Bigl(\frac{x}{7},\frac{y}{7}\Bigr),\qquad 
G=\Bigl(\frac{5x}{7},\frac{5y}{7}\Bigr).
\]

--------------------------------------------------------------------
### 1.  Height of \(\triangle ABC\)

The quadrilateral \(DEGF\) has vertices  

\[
D(4,0),\;E(20,0),\;G\Bigl(\frac{5x}{7},\frac{5y}{7}\Bigr),\;F\Bigl(\frac{x}{7},\frac{y}{7}\Bigr).
\]

Its area by the shoelace formula is  

\[
\begin{aligned}
[DEGF]
&=\frac12\Bigl|
\bigl(4\cdot0+20\cdot\frac{5y}{7}
+\frac{5x}{7}\cdot\frac{y}{7}+ \frac{x}{7}\cdot0\bigr) \\
&\qquad\ -\bigl(0\cdot20+0\cdot\frac{5x}{7}
+\frac{5y}{7}\cdot\frac{x}{7}+ \frac{y}{7}\cdot4\bigr)\Bigr| \\
&=\frac{624}{91}\,y .
\end{aligned}
\]

We are given \([DEGF]=288\), so  

\[
\frac{624}{91}\,y=288\Longrightarrow y=42 .
\]

Therefore the height of \(\triangle ABC\) above \(AB\) is \(y=42\) and  

\[
[ABC]=\frac12\cdot AB\cdot y=\frac12\cdot28\cdot42=588 .
\]

--------------------------------------------------------------------
### 2.  Points \(M\) and \(N\)

\(M\) is the reflection of \(D\) across \(F\); thus \(F\) is the midpoint of
\(DM\):

\[
M=2F-D=\Bigl(\frac{2x}{7}-4,\; \frac{2y}{7}\Bigr)=\Bigl(\frac{2x}{7}-4,12\Bigr).
\]

\(N\) is the reflection of \(G\) across \(E\); hence

\[
N=2E-G=\Bigl(40-\frac{5x}{7},\; -\frac{5y}{7}\Bigr)=\Bigl(40-\frac{5x}{7},-30\Bigr).
\]

--------------------------------------------------------------------
### 3.  Area of the heptagon \(AFNBCEM\)

List the vertices in the given order:

\[
\begin{array}{c|c}
\text{vertex}&(x,y)\\ \hline
A&(0,0)\\
F&\bigl(\frac{x}{7},\frac{y}{7}\bigr)\\
N&\bigl(40-\frac{5x}{7},-30\bigr)\\
B&(28,0)\\
C&(x,y)\\
E&(20,0)\\
M&\bigl(\frac{2x}{7}-4,12\bigr)
\end{array}
\]

Apply the shoelace formula:

\[
\begin{aligned}
[AFNBCEM]&=\frac12\Bigl|
\underbrace{0\cdot\frac{y}{7}}_{\!0}
+\frac{x}{7}\cdot(-30)
+(40-\tfrac{5x}{7})\cdot0
+28\cdot y
+x\cdot0
+20\cdot12
+\bigl(\tfrac{2x}{7}-4\bigr)\cdot0 \\
&\qquad
-\bigl(0\cdot\frac{x}{7}
+ \frac{y}{7}\bigl(40-\tfrac{5x}{7}\bigr)
+(-30)\cdot28
+0\cdot x
+y\cdot20
+0\bigl(\tfrac{2x}{7}-4\bigr)
+12\cdot0\bigr)\Bigr|\\
&=\frac12\Bigl|
\Bigl(-\frac{30x}{7}+1176+240\Bigr)
-\Bigl(\frac{40y}{7}-\frac{5xy}{49}-840+20y\Bigr)\Bigr| .
\end{aligned}
\]

All terms containing \(x\) cancel, leaving

\[
[AFNBCEM]=\frac12\bigl(1416-240\bigr)=\frac12\cdot1176=588 .
\]

Thus the heptagon has the same area as \(\triangle ABC\).

--------------------------------------------------------------------
### 4.  Answer

\[
\boxed{588}
\]

The least possible value of  

\[
f(X)=AX+BX+CX+DX+EX
\]

is  

\[
6+18\sqrt{13}.
\]

Hence \(m=6\), \(n=18\), \(p=13\) and  

\[
m+n+p=6+18+13=37.
\]

\[
\boxed{37}
\]

Let  

\[
A=1,\;B=2,\dots ,L=12
\]

and consider a random perfect matching of the 12 letters.  
For each pair we write the two letters in alphabetical order; thus each
pair is an unordered 2‑subset \(\{x,y\}\) with the smaller element first.
When the six resulting two‑letter “words’’ are listed alphabetically, the
last (lexicographically largest) word is the pair whose **first letter**
(the smaller element of the pair) is the largest among the six first
letters.

Hence the problem asks for the probability that the pair containing the
letter \(G\;(=7)\) has the largest first‑letter among the six pairs.

--------------------------------------------------------------------
### 1.  How a matching looks

In any matching exactly six letters are the *smaller* element of their
pair; call this set \(S\) ( \(|S|=6\) ).  
The remaining six letters are the larger elements of their pairs.

The last word is the pair whose smaller element is  

\[
M=\max S .
\]

We need the probability that the pair containing \(G\) also contains
\(M\).  This can happen in two mutually exclusive ways:

* **Case 1.** \(G\) is the smaller element of its pair, so \(M=G\).
* **Case 2.** \(G\) is the larger element of its pair; then the smaller
  element of that pair must be \(M\).

--------------------------------------------------------------------
### 2.  Case 2 – \(G\) is the larger element

If \(G\) is larger, its partner must be a smaller letter.
Let that partner be \(x\ (<G)\).  
For \(x\) to be the maximum of \(S\), all letters larger than \(x\)
must be the larger elements of their pairs.  Since \(|S|=6\), this forces

\[
x=F\;(=6),\qquad S=\{1,2,3,4,5,6\}.
\]

Thus the only possible pairing is \(\{F,G\}\); the remaining letters are
\(\{A,B,C,D,E\}\) (small) and \(\{H,I,J,K,L\}\) (large), which can be
matched arbitrarily.  There are  

\[
5! =120
\]

matchings of this type.

--------------------------------------------------------------------
### 3.  Case 1 – \(G\) is the smaller element

Now \(G\) must be paired with a larger letter \(j\in\{8,9,10,11,12\}\)
(\(5\) choices).  
For \(G\) to be the maximal element of \(S\), none of the other letters
\(8,9,10,11,12\) may belong to \(S\).  Hence  

\[
S=\{7\}\cup T,\qquad T\subset\{1,2,3,4,5,6\},\ |T|=5.
\]

Choose the omitted small letter \(r\in\{1,\dots ,6\}\); then
\(T=\{1,\dots ,6\}\setminus\{r\}\) ( \(6\) choices).

For a fixed \(j\) and \(r\) we must match the five small letters
\(T\) with the five large letters \(\{r\}\cup\{8,9,10,11,12\}\setminus\{j\}\)
so that each small letter is paired with a larger one.  
The four large letters \(\{8,9,10,11,12\}\setminus\{j\}\) can be paired
with any four of the small letters – that gives \(4!=24\) possibilities.
The remaining large letter \(r\) can only be paired with a small letter
that is smaller than \(r\); there are \(r-1\) such choices.

Hence for this \(j,r\) we have \((r-1)\cdot4!\) matchings.
Summing over \(r\) gives  

\[
\sum_{r=1}^{6}(r-1)\cdot24=15\cdot24=360
\]

matchings for each of the five possible \(j\).  Thus

\[
\text{Case 1 count}=5\cdot360=1800 .
\]

--------------------------------------------------------------------
### 4.  Total favourable matchings

\[
N_{\text{fav}}=1800+120=1920 .
\]

The total number of perfect matchings of 12 letters is  

\[
N_{\text{tot}}=\frac{12!}{2^{6}\,6!}=10395 .
\]

Therefore

\[
P=\frac{1920}{10395}= \frac{128}{693}\qquad(\gcd(128,693)=1).
\]

--------------------------------------------------------------------
### 5.  Answer

\[
\boxed{m+n=128+693=821}
\]

The inequalities  

\[
x-yz<y-zx<z-xy
\]

are equivalent to  

\[
(x-y)(1+z)<0,\qquad (y-z)(1+x)<0 .
\]

---

### 1.  Introduce convenient variables

Let  

\[
u=x-y,\qquad v=y-z .
\]

On the plane \(x+y+z=75\) we can solve for \((x,y,z)\) in terms of \((u,v)\):

\[
\begin{aligned}
x&=\frac{75+2u+v}{3},\\[2pt]
y&=\frac{75-u+v}{3},\\[2pt]
z&=\frac{75-u-2v}{3}.
\end{aligned}
\]

Thus  

\[
1+z=\frac{78-u-2v}{3},\qquad 
1+x=\frac{78+2u+v}{3}.
\]

The two inequalities become  

\[
u\,(78-u-2v)<0,\qquad v\,(78+2u+v)<0 .
\tag{1}
\]

---

### 2.  Determine the regions in the \((u,v)\)-plane  

From (1)

\[
\begin{cases}
u(78-u-2v)<0,\\[2pt]
v(78+2u+v)<0,
\end{cases}
\]

which splits the \((u,v)\)-plane into four sign cases.

| case | signs of \(u,v\) | resulting linear conditions |
|------|------------------|----------------------------|
| \(u>0,\ v>0\) | empty (contradiction) |
| \(u>0,\ v<0\) | \(u+2v>78\) (unbounded) |
| \(u<0,\ v>0\) | \(u+2v<78,\;2u+v<-78\) (unbounded) |
| \(u<0,\ v<0\) | \(2u+v>-78\) (bounded) |

Only the last case gives a bounded region.  
With \(u<0,\ v<0\) the condition \(u+2v<78\) is automatic, so the region is

\[
\boxed{\;u<0,\;v<0,\;2u+v>-78\;}
\]

which is the interior of the right‑triangle with vertices  

\[
(0,0),\;(-39,0),\;(0,-78).
\]

Its area in the \((u,v)\)-coordinates is

\[
A_{uv}= \frac12\cdot39\cdot78 =1521 .
\]

---

### 3.  Convert to area on the plane \(x+y+z=75\)

The linear map \((u,v)\mapsto (x,y,z)\) has Jacobian matrix  

\[
\frac{\partial(x,y,z)}{\partial(u,v)}=
\begin{pmatrix}
\frac{2}{3}&\frac13\\[2pt]
-\frac13&\frac13\\[2pt]
-\frac13&-\frac23
\end{pmatrix}.
\]

The area‑scale factor is the magnitude of the cross product of its column vectors:

\[
\bigl\lVert( \tfrac23,-\tfrac13,-\tfrac13)\times
(\tfrac13,\tfrac13,-\tfrac23)\bigr\rVert
= \frac1{\sqrt3}.
\]

Hence the area of the finite region on the given plane is  

\[
A =\frac1{\sqrt3}\,A_{uv}= \frac{1521}{\sqrt3}
      =\frac{1521\sqrt3}{3}= 507\sqrt3 .
\]

---

### 4.  Answer

The area is \(507\sqrt3\), i.e. \(a=507\) and \(b=3\).  
Thus  

\[
a+b = 507+3 = \boxed{510}.
\]

Let the points be placed on the \(x\)-axis.

\[
A(0,0),\;B(b,0),\;C(c,0),\;D(d,0),\;E(e,0),\;F(f,0)
\]

Set the successive segment lengths  

\[
AB=x_1,\;BC=x_2,\;CD=x_3,\;DE=x_4,\;EF=x_5 .
\]

From the given distances

\[
\begin{aligned}
x_1+x_2 &= AC = 26,\\
x_2+x_3 &= BD = 22,\\
x_3+x_4 &= CE = 31,\\
x_4+x_5 &= DF = 33,\\
x_1+x_2+x_3+x_4+x_5 &= AF = 73 .
\end{aligned}
\]

Solving,
\[
x_2=8,\qquad x_1=18,\qquad x_3=14,\qquad x_4=17,\qquad x_5=16 .
\]

Hence  

\[
\begin{aligned}
A&=(0,0),\\
B&=(18,0),\\
C&=(26,0),\\
D&=(40,0),\\
E&=(57,0),\\
F&=(73,0).
\end{aligned}
\]

Let \(G=(x,y)\).  From the distances \(CG=40\) and \(DG=30\),

\[
\begin{cases}
(x-26)^2+y^2=40^2=1600,\\[2pt]
(x-40)^2+y^2=30^2=900 .
\end{cases}
\]

Subtracting the equations,

\[
(x-26)^2-(x-40)^2=700\Longrightarrow 28x-924=700\Longrightarrow x=58.
\]

Then  

\[
(58-26)^2+y^2=1600\Longrightarrow 1024+y^2=1600\Longrightarrow y^2=576\Longrightarrow y=\pm24.
\]

Thus \(G=(58,24)\) (the sign of \(y\) does not affect the area).

The base \(BE\) lies on the \(x\)-axis:

\[
BE = e-b = 57-18 = 39 .
\]

The height from \(G\) to the line \(AB\!E\) is \(|y|=24\).

Hence the area of \(\triangle BGE\) is

\[
\frac12\cdot BE\cdot|y|
   =\frac12\cdot 39 \cdot 24
   =\frac12\cdot 936
   =468 .
\]

\[
\boxed{468}
\]

The intersection points satisfy  

\[
y=f(x),\qquad x=34y^{2}\Longrightarrow y=f(34y^{2}).
\]

--------------------------------------------------------------------
### 1.  The periodic function \(f\)

Because \(f(x+4)=f(x)\),

\[
f(x)=\begin{cases}
x-4k ,&x\in[-1+4k,\,1+4k)\\[2mm]
2+4k-x ,&x\in[1+4k,\,3+4k)
\end{cases}\qquad(k\in\mathbb Z).
\]

For an intersection we have \(x=34y^{2}\ge 0\) and \(|y|\le 1\) (the range of \(f\)).
Hence we only need the periods that intersect the interval \([0,34]\):
\[
k=0,1,\dots ,8 .
\]

--------------------------------------------------------------------
### 2.  Equations for the two sub‑intervals

*If \(34y^{2}\in[-1+4k,\,1+4k)\) (the “increasing” part)*
\[
y = (34y^{2})-4k\;\Longrightarrow\;34y^{2}-y-4k=0 .
\tag{A}_{k}
\]

*If \(34y^{2}\in[1+4k,\,3+4k)\) (the “decreasing” part)*
\[
y = 2+4k-(34y^{2})\;\Longrightarrow\;34y^{2}+y-(4k+2)=0 .
\tag{B}_{k}
\]

--------------------------------------------------------------------
### 3.  Roots

\[
\begin{aligned}
(A)_{k}&:\quad y=\frac{1\pm\sqrt{1+544k}}{68},\\[2mm]
(B)_{k}&:\quad y=\frac{-1\pm\sqrt{273+544k}}{68}.
\end{aligned}
\]

--------------------------------------------------------------------
### 4.  Which roots are admissible?

Because \(|y|\le 1\),

* For \(k=0,\dots ,8\) the two roots of \((A)_{k}\) satisfy \(|y|<1\);
  thus all \(2\cdot9=18\) of them are valid.

* For \((B)_{k}\) the two roots lie in \((-1,1)\) when \(k=0,\dots ,7\).
  When \(k=8\) the negative root equals \(\displaystyle 
  \frac{-1-\sqrt{4625}}{68}\approx-1.0148\), which is outside the
  interval \([-1,1]\); the positive root is still admissible.
  Hence we have \(2\cdot8+1=17\) valid roots from the \((B)\) equations.

Consequently there are \(18+17=35\) intersection points.

--------------------------------------------------------------------
### 5.  Summing the \(y\)-coordinates  

From Viète’s formulas

\[
\text{sum of the two roots of }(A)_{k}= \frac{1}{34},\qquad
\text{sum of the two roots of }(B)_{k}= -\frac{1}{34}.
\]

Therefore

\[
\sum_{k=0}^{8}\bigl(\text{both roots of }(A)_{k}\bigr)=9\cdot\frac1{34}=\frac{9}{34},
\]
\[
\sum_{k=0}^{8}\bigl(\text{both roots of }(B)_{k}\bigr)=-\frac{9}{34}.
\]

In \((B)_{8}\) we must discard the negative root
\(\displaystyle y_{8}^{-}=\frac{-1-\sqrt{4625}}{68}\).
Hence the sum of the **valid** \((B)\)‑roots is

\[
-\frac{9}{34}-y_{8}^{-}.
\]

Adding the contributions of \((A)\) and \((B)\),

\[
\begin{aligned}
\sum y
&= \frac{9}{34}+\Bigl(-\frac{9}{34}-y_{8}^{-}\Bigr)
   =-y_{8}^{-}
   =\frac{1+\sqrt{4625}}{68}.
\end{aligned}
\]

Since \(\sqrt{4625}=5\sqrt{185}\),

\[
\sum y=\frac{1+5\sqrt{185}}{68}.
\]

--------------------------------------------------------------------
### 6.  Final answer

The required representation is \(\displaystyle\frac{a+b\sqrt{c}}{d}\) with  

\[
a=1,\; b=5,\; c=185,\; d=68,
\]

and \(\gcd(a,b,d)=1\), \(c\) square‑free.

\[
a+b+c+d = 1+5+185+68 = 259.
\]

\[
\boxed{259}
\]

Let  

\[
d=n+2 \qquad (d\ge 3,\; n>0)
\]

The condition is  

\[
d \mid 3(n+3)(n^2+9).
\]

Since \(n\equiv -2 \pmod d\),

\[
n+3\equiv (-2)+3 \equiv 1 \pmod d,\qquad 
n^2+9\equiv (-2)^2+9 \equiv 4+9\equiv 13 \pmod d.
\]

Hence

\[
3(n+3)(n^2+9)\equiv 3\cdot1\cdot13=39 \pmod d .
\]

Thus \(d\) must divide 39:

\[
d\mid 39 .
\]

The positive divisors of \(39\) are \(1,3,13,39\).  
Because \(d=n+2\ge 3\), we have  

\[
d\in\{3,13,39\}.
\]

Corresponding \(n\) values are  

\[
n=d-2\in\{1,11,37\}.
\]

All three satisfy the original divisibility, so the required sum is  

\[
1+11+37=49.
\]

\[
\boxed{49}
\]

\[
P=\prod_{k=4}^{63}\frac{\log_k(5^{k^{2}-1})}{\log_{k+1}(5^{k^{2}-4})}
\]

---

### 1.  Rewrite each factor

\[
\log_k(5^{p})=\frac{\log_{5}(5^{p})}{\log_{5}k}= \frac{p}{\log_{5}k}.
\]

Hence

\[
\frac{\log_k(5^{k^{2}-1})}{\log_{k+1}(5^{k^{2}-4})}
   =\frac{\dfrac{k^{2}-1}{\log_{5}k}}{\dfrac{k^{2}-4}{\log_{5}(k+1)}}
   =\frac{k^{2}-1}{k^{2}-4}\cdot\frac{\log_{5}(k+1)}{\log_{5}k}.
\]

Factor the rational part:

\[
\frac{k^{2}-1}{k^{2}-4}= \frac{(k-1)(k+1)}{(k-2)(k+2)} .
\]

Thus

\[
P=\prod_{k=4}^{63}
   \left[\frac{(k-1)(k+1)}{(k-2)(k+2)}
         \cdot\frac{\log_{5}(k+1)}{\log_{5}k}\right].
\]

---

### 2.  Telescope the rational part  

Separate the product:

\[
\prod_{k=4}^{63}\frac{k-1}{k-2}\; \times\;
\prod_{k=4}^{63}\frac{k+1}{k+2}.
\]

\[
\begin{aligned}
\prod_{k=4}^{63}\frac{k-1}{k-2}
   &=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdots\frac{62}{61}
    =\frac{62}{2}=31,\\[4pt]
\prod_{k=4}^{63}\frac{k+1}{k+2}
   &=\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdots\frac{64}{65}
    =\frac{5}{65}=\frac{1}{13}.
\end{aligned}
\]

Hence the rational part equals  

\[
\frac{31}{13}.
\]

---

### 3.  Telescope the logarithmic part  

\[
\prod_{k=4}^{63}\frac{\log_{5}(k+1)}{\log_{5}k}
   =\frac{\log_{5}5}{\log_{5}4}\cdot\frac{\log_{5}6}{\log_{5}5}
    \cdots\frac{\log_{5}64}{\log_{5}63}
   =\frac{\log_{5}64}{\log_{5}4}
   =\log_{4}64.
\]

Since \(64=4^{3}\), \(\log_{4}64=3\).

Thus the logarithmic part contributes a factor of \(3\).

---

### 4.  Combine

\[
P=\frac{31}{13}\times 3=\frac{93}{13}.
\]

The fraction \(\frac{93}{13}\) is already in lowest terms, so  

\[
m=93,\qquad n=13.
\]

Finally,

\[
m+n=93+13=106.
\]

\[
\boxed{106}
\]

**Solution Sketch**

Let  

\[
a=3^{\alpha }u ,\qquad b=3^{\beta }v ,\qquad c=3^{\gamma }w ,
\qquad(\alpha ,\beta ,\gamma \ge 0,\;3\nmid u,v,w)
\]

and put \(\kappa =\min (\alpha ,\beta ,\gamma )\).
Writing \(a^{3}=3^{3\alpha }u^{3}\) etc.,  

\[
a^{3}+b^{3}+c^{3}\equiv 0\pmod{3^{7}}
\Longleftrightarrow
u^{3}3^{3(\alpha-\kappa)}+v^{3}3^{3(\beta-\kappa)}+w^{3}3^{3(\gamma-\kappa)}
\equiv0\pmod{3^{7-3\kappa}} .
\tag{1}
\]

Thus we have three essentially different cases:

* \(\kappa\ge 3\) : every term in (1) is divisible by \(3^{9}\), hence the
  congruence is always true.

* \(\kappa =2\) : all three numbers are divisible by \(9\) and at least one
  is not divisible by \(27\).  
  Put \(a=9A,\;b=9B,\;c=9C\;(1\le A,B,C\le81)\).   
  Then (1) reduces to  

  \[
  A+B+C\equiv0\pmod3 .
  \]

  The numbers \(A,B,C\) are taken from the set \(\{1,\dots ,81\}\); each
  residue class \(\pmod3\) occurs exactly \(27\) times.  
  Hence the number of ordered triples with \(\kappa =2\) is  

  \[
  9\cdot 27^{3}=157\,464 .
  \tag{2}
  \]

* \(\kappa =1\) : all three numbers are divisible by \(3\) and at least one
  is not divisible by \(9\).  Write \(a=3A,\;b=3B,\;c=3C\;(1\le A,B,C\le243)\).  
  Then (1) becomes  

  \[
  A^{3}+B^{3}+C^{3}\equiv0\pmod{3^{4}} .
  \tag{3}
  \]

  The cubes \(\pmod {81}\) of the integers \(1\le x\le243\) are distributed as  

  \[
  \begin{array}{c|c}
  \text{value}&\text{how many $x$ give it}\\ \hline
  0 & 27\\
  27,\,54 & 27\text{ each}\\
  \text{18 other residues} & 9\text{ each}
  \end{array}
  \]

  Using this distribution one counts (by a short elementary
  combinatorial argument) the number of ordered triples satisfying (3) to be  

  \[
  452\,709 .
  \tag{4}
  \]

  The triples counted in (4) include those where all three numbers are
  divisible by \(9\); those are exactly the triples of the previous case,
  i.e. the $157\,464$ triples of (2).  Hence the triples with
  \(\kappa =1\) contribute  

  \[
  452\,709-157\,464 = 295\,245 .
  \tag{5}
  \]

* \(\kappa =0\) : at least one of the numbers is not divisible by \(3\).
  The total number of ordered triples with $1\le a,b,c\le3^{6}=729$ is
  $729^{3}=387\,420\,489$.
  Subtracting the contributions of the three previous cases gives

  \[
  387\,420\,489-19\,683-157\,464-295\,245 = 702\,027 .
  \tag{6}
  \]

Finally we take the answer modulo $1000$ :

\[
702\,027\equiv\boxed{27}\pmod{1000}.
\]

Let  

* \(O\) be the center of \(\omega _2\) (radius \(15\)),  
* \(A\) be the center of \(\omega _1\) (radius \(6\)),  
* \(B\) the internal point of tangency of the two circles.

---

### 1.  Coordinates

Place \(O\) at the origin and the line \(BC\) on the \(x\)-axis.

\[
B=(15,0),\qquad O=(0,0),\qquad A=(9,0)
\]

Because \(BC\) is a diameter of \(\omega _2\),

\[
C=(-15,0).
\]

\(AD\perp BC\) makes \(AD\) vertical, so \(D\) lies on the line \(x=9\) and on \(\omega_2\):

\[
9^{2}+y^{2}=15^{2}\Longrightarrow y^{2}=144\Longrightarrow y=\pm12 .
\]

Since later we need \(D\) to be nearer to the lower side of the rectangle, we take  

\[
D=(9,-12).
\]

---

### 2.  Rectangle \(EFGH\) inside \(\omega _1\)

The rectangle is axis‑aligned because \(EF\perp BC\) (so \(EF\) is vertical) and a rectangle has right angles.

Let the half‑width be \(a\) and the half‑height be \(y\) :

\[
\begin{aligned}
E&\;(9+a,\;y),&F&\;(9+a,\;-y),\\
G&\;(9-a,\;-y),&H&\;(9-a,\;y).
\end{aligned}
\]

All four vertices lie on \(\omega _1\), whose equation is  

\[
(x-9)^2+y^2=6^{2}=36 .
\]

Hence  

\[
a^{2}+y^{2}=36\quad\Longrightarrow\quad y=\sqrt{36-a^{2}}\;(>0).
\]

The rectangle’s area is  

\[
[EFGH]= (2a)(2y)=4ay .
\tag{1}
\]

---

### 3.  Areas of \(\triangle DGF\) and \(\triangle CHG\)

\[
\begin{aligned}
\triangle DGF &: \text{base }GF=2a,\; 
\text{height }=12-y,\\
[ DGF]&=\tfrac12\,(2a)(12-y)=a(12-y). \tag{2}
\end{aligned}
\]

\[
\begin{aligned}
\triangle CHG &: \text{base }HG=2y,\;
\text{horizontal distance from }C\text{ to line }x=9-a\text{ is }24-a,\\
[CHG]&=\tfrac12\,(2y)(24-a)=y(24-a). \tag{3}
\end{aligned}
\]

Given \([DGF]=[CHG]\),

\[
a(12-y)=y(24-a).
\]

Expanding and canceling the common term \(-ay\) gives

\[
12a=24y\quad\Longrightarrow\quad a=2y. \tag{4}
\]

---

### 4.  Solving for \(a\) and \(y\)

From (4) and the circle condition \(a^{2}+y^{2}=36\),

\[
(2y)^{2}+y^{2}=36\Longrightarrow 5y^{2}=36
\Longrightarrow y^{2}=\frac{36}{5}
\Longrightarrow y=\frac{6}{\sqrt5}=\frac{6\sqrt5}{5}.
\]

Thus  

\[
a=2y= \frac{12}{\sqrt5}= \frac{12\sqrt5}{5}.
\]

Both satisfy \(a\le6,\;y\le6\), so the rectangle indeed fits in \(\omega _1\).

---

### 5.  Area of the rectangle

Insert \(a=2y\) into (1):

\[
[EFGH]=4a y =4(2y)\,y = 8y^{2}=8\left(\frac{36}{5}\right)=\frac{288}{5}.
\]

Hence the rectangle’s area is \(\displaystyle \frac{288}{5}\).

The fraction is already in lowest terms, so  

\[
m=288,\qquad n=5,\qquad m+n=288+5=293.
\]

\[
\boxed{293}
\]

Let the 9 vertices of the \(2\times2\) grid be the points \((i,j)\) with \(i,j=0,1,2\).
The 12 unit edges are  

* 8 **boundary** edges (two on each side of the large square);
* 4 **interior** edges, which are shared by two unit squares.

Denote the four interior edges by  

\[
a=v_{10},\qquad b=v_{11},\qquad c=h_{01},\qquad d=h_{11},
\]

where each variable is \(1\) if the edge is coloured red and \(0\) if it is blue.

---

### 1.  How the interior edges affect a square  

Each small square has two interior edges and two boundary edges.

\[
\begin{array}{c|c}
\text{square}&\text{interior edges}\\\hline
\text{BL}& a,\;c\\
\text{BR}& b,\;c\\
\text{TL}& a,\;d\\
\text{TR}& b,\;d
\end{array}
\]

Let \(r\) be the number of red interior edges of a given square.
Since the square must have exactly two red sides, it needs  
\(2-r\) red boundary edges.

For a square:

* if \(r=0\) or \(r=2\) the two boundary edges are forced (both red or both blue) – **1** way;
* if \(r=1\) exactly one of the two boundary edges must be red – **2** ways.

Thus a given interior‑edge configuration contributes a factor  

\[
2^{\#\{\text{squares with }r=1\}} .
\]

Denote by  

\[
N(a,b,c,d)=\mathbf 1_{a\neq c}+\mathbf 1_{a\neq d}+\mathbf 1_{b\neq c}+\mathbf 1_{b\neq d},
\]

the number of squares for which exactly one of the two interior edges is red.
Then the total number of colourings is

\[
\sum_{a,b,c,d\in\{0,1\}} 2^{\,N(a,b,c,d)} .
\]

---

### 2.  Evaluating the sum  

The value of \(N\) depends only on how the pairs \((a,b)\) and \((c,d)\) compare.

* **Case 1:** \((a,b)=(c,d)\).  
  Then \(N=0\). This occurs for \((0,0,0,0)\) and \((1,1,1,1)\) – 2 assignments.
  Contribution: \(2\cdot 2^{0}=2\).

* **Case 2:** \((a,b)=(c,d)\) with opposite colour, i.e. \(a=b\neq c=d\).  
  Then each of the four comparisons differs, so \(N=4\).  
  This happens for \((0,0,1,1)\) and \((1,1,0,0)\) – 2 assignments.
  Contribution: \(2\cdot 2^{4}=2\cdot16=32\).

* **Case 3:** All other assignments.  
  There are \(16-4=12\) such configurations and one checks that each yields \(N=2\).  
  Contribution: \(12\cdot 2^{2}=12\cdot4=48\).

Adding the three cases,

\[
2+32+48=82 .
\]

---

\[
\boxed{82}
\]

The divisors of  

\[
2025 = 3^{4}\cdot 5^{2}
\]

are of the form \(3^{a}5^{b}\) with  

\[
0\le a\le 4 ,\qquad 0\le b\le 2 .
\]

Hence  

\[
|A|=(4+1)(2+1)=15,\qquad\text{so there are }2^{15}=32768\text{ subsets of }A.
\]

For a subset \(B\subseteq A\) the least common multiple is  

\[
\operatorname{lcm}(B)=3^{\max a}\,5^{\max b},
\]

where \(\max a\) (resp. \(\max b\)) is the largest exponent of \(3\) (resp. \(5\)) among the elements of \(B\).

Thus \(\operatorname{lcm}(B)=2025\) iff  

* some element of \(B\) has \(a=4\) (i.e. belongs to the set \(R\) of divisors with a factor \(3^{4}\));
* some element of \(B\) has \(b=2\) (i.e. belongs to the set \(C\) of divisors with a factor \(5^{2}\)).

The set \(R\) consists of the three divisors \(3^{4}5^{0}, 3^{4}5^{1}, 3^{4}5^{2}\);
the set \(C\) consists of the five divisors \(3^{0}5^{2}, 3^{1}5^{2}, 3^{2}5^{2}, 3^{3}5^{2}, 3^{4}5^{2}\).
Their intersection is \(\{2025\}\).  

We need the number of subsets of \(A\) that contain **at least one** element of \(R\) and **at least one** element of \(C\).

Using inclusion–exclusion on the whole set of subsets:

\[
\begin{aligned}
N &= 2^{15}
      -\underbrace{2^{15-|R|}}_{\text{subsets with no }R}
      -\underbrace{2^{15-|C|}}_{\text{subsets with no }C}
      +\underbrace{2^{15-|R\cup C|}}_{\text{subsets with neither}}  \\
  &= 2^{15} -2^{12} -2^{10} + 2^{8} \\
  &= 32768 - 4096 - 1024 + 256 = 27904 .
\end{aligned}
\]

All these subsets are non‑empty (they contain at least one element from \(R\) and one from \(C\)).  

Hence  

\[
\Pr(\text{nonempty }B\text{ with }\operatorname{lcm}=2025)
   = \frac{27904}{2^{15}}
   = \frac{27904}{32768}
   = \frac{109}{128}\;.
\]

So \(m=109\) and \(n=128\); therefore  

\[
m+n = 109+128 = 237 .
\]

\[
\boxed{237}
\]

For \( \triangle ABC\) let  

\[
\angle A=84^{\circ},\qquad \angle B=60^{\circ},\qquad \angle C=36^{\circ}.
\]

---

### 1.  Identify the points on the nine‑point circle  

\(D,E,F\) are the midpoints of \(BC,CA,AB\); hence the circumcircle of \(\triangle DEF\) is the nine‑point circle of \(\triangle ABC\).

The nine‑point circle also passes through the feet of the three altitudes.
Thus  

* \(G\) is the foot of the altitude from \(A\) onto \(BC\);  
* \(H\) is the foot of the altitude from \(B\) onto \(CA\);  
* \(J\) is the foot of the altitude from \(C\) onto \(AB\).

So the six points on the nine‑point circle are, in order,
\(G,\;D,\;E,\;H,\;J,\;F\).

---

### 2.  Arc \(\widehat{DE}\)

In \(\triangle DEF\) we have the well‑known midpoint–parallel relations  

\[
DF\parallel AC,\qquad EF\parallel BC .
\]

Hence at \(F\)

\[
\angle DFE=\angle(DF,EF)=\angle(AC,BC)=\angle C=36^{\circ}.
\]

For a chord, the intercepted minor arc equals twice the inscribed angle, so  

\[
\widehat{DE}=2\angle DFE=2\cdot36^{\circ}=72^{\circ}.
\]

---

### 3.  Arc \(\widehat{FG}\)

\(DG\) lies on side \(BC\); thus \(DG\parallel BC\).
Since \(DF\parallel AC\),

\[
\angle FDG=\angle(DF,DG)=\angle(AC,BC)=\angle C=36^{\circ}.
\]

Therefore  

\[
\widehat{FG}=2\angle FDG=2\cdot36^{\circ}=72^{\circ}.
\]

---

### 4.  Arc \(\widehat{HJ}\)

\(G,\;H,\;J\) are the vertices of the orthic triangle.
For an acute triangle, the orthic triangle angles satisfy  

\[
\angle H G J =180^{\circ}-2\angle A .
\]

Here \(\angle A=84^{\circ}\), so  

\[
\angle H G J =180^{\circ}-2\cdot84^{\circ}=12^{\circ}.
\]

The inscribed angle \(\angle H G J\) intercepts the minor arc \(\widehat{HJ}\); consequently  

\[
\widehat{HJ}=2\angle H G J =2\cdot12^{\circ}=24^{\circ}.
\]

---

### 5.  Compute the required combination  

\[
\begin{aligned}
\widehat{DE}+2\widehat{HJ}+3\widehat{FG}
&=72^{\circ}+2\cdot24^{\circ}+3\cdot72^{\circ} \\
&=72^{\circ}+48^{\circ}+216^{\circ} \\
&=336^{\circ}.
\end{aligned}
\]

---

\[
\boxed{336}
\]

Let  

\[
g(N)=\Big\lfloor\frac{N}{25}\Big\rfloor+
\Big\lfloor\frac{N\bmod 25}{10}\Big\rfloor+(N\bmod 25)\bmod10
\]

be the number of coins the greedy algorithm uses for \(N\) cents.
For an optimal solution we may use fewer quarters.  
Write  

\[
N=25q+a,\qquad 0\le a<25 .
\]

If we replace \(k\;(1\le k\le q)\) quarters by dimes and pennies, the new
total number of coins is  

\[
c_k=(q-k)+\Big\lfloor\frac{a+25k}{10}\Big\rfloor+(a+25k)\bmod10 .
\]

Using \(a+25k=10(2k)+ (a+5k)\),

\[
c_k=q+a+24k-9\Big\lfloor\frac{a+5k}{10}\Big\rfloor .
\]

The greedy number is  

\[
g(N)=q+a-9\Big\lfloor\frac a{10}\Big\rfloor .
\]

Hence  

\[
c_k-g(N)=24k-9\Big(\Big\lfloor\frac{a+5k}{10}\Big\rfloor-
\Big\lfloor\frac a{10}\Big\rfloor\Big)
        =24k-9\,\Big\lfloor\frac{5k+r}{10}\Big\rfloor ,
\]
where \(r=a\bmod10\) (so \(0\le r\le 9\)).

The greedy algorithm fails iff there exists a \(k\ge1\) with \(c_k<g(N)\),
i.e.

\[
24k<9\Big\lfloor\frac{5k+r}{10}\Big\rfloor
\iff 
\Big\lfloor\frac{5k+r}{10}\Big\rfloor>\frac{2k}{3}.
\]

For \(k\ge2\) the left‑hand side grows like \(0.5k\) while the right
hand side grows like \(0.666\!k\); the inequality never holds.
For \(k=1\) the condition becomes  

\[
\Big\lfloor\frac{5+r}{10}\Big\rfloor\ge1\iff r\ge5 .
\]

Thus **greedy fails exactly when**  

* \(N\ge25\) (so at least one quarter is used), and  
* \(a\bmod10\ge5\), i.e. the remainder \(a=N\bmod25\) lies in  

\[
\{5,6,7,8,9,15,16,17,18,19\}.
\]

---

### Counting the failures

For each of the 10 bad remainders \(a\) the numbers
\(N=a+25k\) with \(k\ge1\) and \(N\le1000\) are failures.
Since \(1000- a<1000\) we have  

\[
\Big\lfloor\frac{1000-a}{25}\Big\rfloor = 39\quad (a\ge1).
\]

Hence each bad remainder contributes 39 failing values, giving  

\[
10\times39 = 390
\]

failing numbers.

---

### Total successes

There are \(1000\) integers from 1 to 1000, therefore

\[
\text{successful }N = 1000-390 = 610 .
\]

\[
\boxed{610}
\]

Let the vertices of the regular \(24\)-gon be numbered \(0,1,\dots ,23\) clockwise.  
If a segment joins vertex \(i\) with vertex \(j\) then the (shorter) distance
\(\min\{\,|i-j|,24-|i-j|\,\}\) is a number \(k\) with  

\[
1\le k\le 12 .
\]

All 12 segments must have the same length, so they must all have the same
step‑size \(k\).  
For a fixed \(k\) consider the graph \(G_k\) whose edges are exactly the
\(24\) chords \(\{i,i+k\}\;(i=0,\dots ,23)\) (the pair \(\{i,i+k\}\) is the same
as \(\{i+k,i\}\)).  

--------------------------------------------------------------------
### 1.  Structure of \(G_k\)

Let  

\[
d=\gcd(24,k),\qquad L=\frac{24}{d}.
\]

Starting from a vertex \(i\) and repeatedly adding \(k\) modulo \(24\) we obtain
the cycle  

\[
i,\; i+k,\; i+2k,\dots ,i+(L-1)k .
\]

Thus \(G_k\) splits into \(d\) disjoint cycles, each of length \(L\).

- If \(k\neq 12\) then \(i+k\neq i-k\) and each vertex has degree \(2\); the
components are simple cycles.
- For \(k=12\) each vertex is paired only with its opposite vertex, so the
components are single edges (a perfect matching already).

--------------------------------------------------------------------
### 2.  When does a perfect matching exist?

A perfect matching on a cycle exists only when the cycle length is even.
Hence we need \(L\) even, i.e.  

\[
\frac{24}{d}\text{ is even}\iff d\mid 12 .
\]

Therefore a perfect matching is possible for all \(k\) whose
\(\gcd(24,k)\) belongs to \(\{1,2,3,4,6,12\}\).  
The only value of \(k\) with \(\gcd(24,k)=8\) is \(k=8\); for this step size
\(L=3\) (odd) and no perfect matching exists.

--------------------------------------------------------------------
### 3.  Number of perfect matchings for a given \(k\)

- **Case \(k\neq 12\).**  
  The graph consists of \(d\) even cycles of length \(L\).  
  Each even cycle has exactly two perfect matchings (choose the “odd” or the
  “even” edges).  
  Hence the number of perfect matchings using step size \(k\) is  

  \[
  N(k)=2^{d}=2^{\gcd(24,k)} .
  \]

- **Case \(k=12\).**  
  The edges are the 12 diameters \(\{i,i+12\}\).  
  This already is a unique perfect matching, so \(N(12)=1\).

--------------------------------------------------------------------
### 4.  Summation over all possible step sizes

The admissible step sizes are  

\[
k=1,2,3,4,5,6,7,9,10,11,12  \qquad(k=8\text{ is excluded}).
\]

Their contributions are:

\[
\begin{array}{c|c|c}
k & \gcd(24,k) & N(k)\\ \hline
1 & 1 & 2\\
2 & 2 & 4\\
3 & 3 & 8\\
4 & 4 & 16\\
5 & 1 & 2\\
6 & 6 & 64\\
7 & 1 & 2\\
9 & 3 & 8\\
10& 2 & 4\\
11& 1 & 2\\
12&12 & 1
\end{array}
\]

Adding them:

\[
2+4+8+16+2+64+2+8+4+2+1 = 113 .
\]

--------------------------------------------------------------------
### 5.  Answer

Thus there are  

\[
\boxed{113}
\]

different ways to draw 12 equal‑length segments so that each of the 24
vertices is an endpoint of exactly one segment.

Let  

\[
r_i=A_1A_i\qquad(2\le i\le 11),\qquad 
\theta=\angle A_iA_1A_{i+1}\;(2\le i\le 10).
\]

---

### 1.  Data from the triangles  

\[
\cos\theta=\frac{12}{13},\qquad 
\sin\theta=\sqrt{1-\left(\frac{12}{13}\right)^2}= \frac{5}{13}.
\]

The area of \(\triangle A_iA_1A_{i+1}\) is  

\[
\frac12 r_i r_{i+1}\sin\theta =1
\Longrightarrow r_i r_{i+1}= \frac{2}{\sin\theta}
               = \frac{2}{5/13}= \frac{26}{5}\; .
\tag{1}
\]

Hence for all \(i=2,\dots ,10\)

\[
r_i r_{i+1}=K,\qquad K:=\frac{26}{5}.   \tag{2}
\]

From (2) we obtain the recurrence  

\[
r_{i+1}= \frac K{r_i},\qquad\text{so } r_{i+2}=r_i .
\]

Thus the distances from \(A_1\) alternate:

\[
r_2=r_4=r_6=r_8=r_{10}=x,\qquad 
r_3=r_5=r_7=r_9=r_{11}=y,
\]
where  

\[
xy=K=\frac{26}{5}.  \tag{3}
\]

---

### 2.  Length of the other edges  

For any \(i=2,\dots ,10\) the edge \(A_iA_{i+1}\) has length (law of cosines)

\[
\begin{aligned}
d &:=A_iA_{i+1}= 
\sqrt{r_i^{\,2}+r_{i+1}^{\,2}-2r_i r_{i+1}\cos\theta}  \\
   &=\sqrt{x^{2}+y^{2}-2xy\frac{12}{13}}
   =\sqrt{x^{2}+y^{2}-\frac{24}{13}K}.
\end{aligned}
\tag{4}
\]

All nine edges \(A_iA_{i+1}\;(i=2\ldots10)\) have the same length \(d\).

---

### 3.  Perimeter condition  

The perimeter of the 11‑gon is  

\[
x+y+9d=20.   \tag{5}
\]

Introduce  

\[
t:=x+y\quad\text{(the quantity we need)},
\qquad xy=K .
\]

From \(x^{2}+y^{2}=t^{2}-2K\) and (4),

\[
d^{2}=t^{2}-2K-\frac{24}{13}K
     =t^{2}-\frac{50}{13}K .
\]

Since \(\displaystyle \frac{50}{13}K
      =\frac{50}{13}\cdot\frac{26}{5}=20\),

\[
d^{2}=t^{2}-20\qquad\Longrightarrow\qquad d=\sqrt{t^{2}-20}. \tag{6}
\]

Insert (6) into (5):

\[
t+9\sqrt{t^{2}-20}=20. \tag{7}
\]

---

### 4.  Solving for \(t\)

From (7),

\[
\sqrt{t^{2}-20}= \frac{20-t}{9}\quad (20-t\ge 0).
\]

Squaring gives  

\[
t^{2}-20=\frac{(20-t)^{2}}{81}
\Longrightarrow 80t^{2}+40t-2020=0.
\]

Dividing by 20,

\[
4t^{2}+2t-101=0.
\]

Hence  

\[
t=\frac{-2\pm\sqrt{4+1616}}{8}
   =\frac{-2\pm18\sqrt5}{8}.
\]

Only the positive root is admissible:

\[
t=\frac{18\sqrt5-2}{8}
  =\frac{9\sqrt5-1}{4}.
\]

Thus  

\[
A_1A_2+A_1A_{11}=t=\frac{9\sqrt5-1}{4}
     =\frac{m\sqrt n-p}{q},
\]

with \(m=9,\;n=5,\;p=1,\;q=4\).

---

### 5.  Required sum  

\[
m+n+p+q=9+5+1+4=19.
\]

\[
\boxed{19}
\]

**1.  Regions produced by line‑segments**

If a set of line segments inside a disc has no three of them meeting at the same interior point, then  

\[
\text{#regions}=1+\;(\text{#segments})+\;(\text{# interior intersection points}).
\]

Hence for our problem  

\[
R=1+27+I ,
\]

where \(I\) is the (random) number of interior intersection points.  
We must compute the expected value \(E[I]\).

--------------------------------------------------------------------

**2.  The three kinds of interior intersections**

* the two diameters intersect once (deterministically);
* a random chord can intersect each of the two diameters;
* two random chords can intersect one another.

So  

\[
E[I]=1+E[\text{chord–diameter intersections}]
      +E[\text{chord–chord intersections}].
\]

--------------------------------------------------------------------

**3.  Probability that a random chord meets a given diameter**

The two endpoints of a chord are chosen uniformly on the circle but must lie in *different* quadrants.
There are \(\binom{4}{2}=6\) unordered pairs of quadrants, all equally likely.

A chord meets the horizontal diameter iff its endpoints lie in opposite
half‑planes (one in the upper half, one in the lower half).  
Among the six unordered pairs, the four pairs  
\(\{Q_1,Q_3\},\{Q_1,Q_4\},\{Q_2,Q_3\},\{Q_2,Q_4\}\) have this property, so

\[
P(\text{chord meets a given diameter})=\frac{4}{6}= \frac23 .
\]

The same holds for the vertical diameter.  
Thus a single random chord contributes on average

\[
2\cdot\frac23=\frac43
\]

intersections with the two diameters.  

For the 25 chords

\[
E[\text{chord–diameter intersections}]
      =25\cdot\frac43=\frac{100}{3}.
\]

--------------------------------------------------------------------

**4.  Distribution of a chord’s quadrant pair**

Let a chord be called  

* **adjacent** if it joins two adjacent quadrants (four such unordered pairs);
* **opposite** if it joins opposite quadrants (two such unordered pairs).

\[
P(\text{adjacent})=\frac{4}{6}= \frac23,\qquad 
P(\text{opposite})=\frac{2}{6}= \frac13 .
\]

--------------------------------------------------------------------

**5.  Probability that two random chords intersect**

Pick two chords independently.  Let their unordered quadrant pairs be \(S\) and
\(T\).  There are three possibilities for the relationship between \(S\) and \(T\).

| Relation of \(S,T\) | Probability | Intersection probability |
|----------------------|-------------|---------------------------|
| **identical** (\(S=T\)) | \(\displaystyle\frac16\) | \(\displaystyle\frac12\) |
| **share exactly one quadrant** | \(\displaystyle\frac{2}{3}\) | \(\displaystyle\frac12\) |
| **disjoint** (complementary sets) | \(\displaystyle\frac16\) | \(\displaystyle\frac13\) |

*Proof of the numbers*  

*Identical*: both chords lie in the same two intervals, say \([0,1)\) and \([1,2)\).  
If the endpoints are \(A\in[0,1),B\in[1,2)\) and \(C\in[0,1),D\in[1,2)\),
the chords intersect iff exactly one of \(C,D\) lies on the clockwise arc from
\(A\) to \(B\).  A short calculation gives probability \(1/2\).

*Share one quadrant*: suppose the common quadrant is \([0,1)\) and the other
endpoints are in \([1,2)\) and \([2,3)\) (or \([3,4)\)).  The second endpoint
never lies on the clockwise arc from the first endpoint to its partner, so the
chords intersect iff the other point in the common quadrant lies on that arc,
which happens with probability \(1/2\).

*Disjoint*: the only way for two disjoint pairs to intersect is when both are
opposite pairs (e.g. \(\{0,2\}\) and \(\{1,3\}\)); then the order of the four
points around the circle is forced to be alternating, so the chords always
intersect.  This occurs for one‑third of the disjoint cases, hence conditional
probability \(1/3\).

Putting the three cases together,

\[
P(\text{two chords intersect})=
\frac16\!\left(\frac12\right)+\frac23\!\left(\frac12\right)+\frac16\!\left(\frac13\right)
      =\frac{17}{36}.
\]

--------------------------------------------------------------------

**6.  Expected chord–chord intersections**

There are \(\binom{25}{2}=300\) unordered pairs of the 25 chords, so

\[
E[\text{chord–chord intersections}]
      =300\cdot\frac{17}{36}= \frac{425}{3}.
\]

--------------------------------------------------------------------

**7.  Total expected interior intersections**

\[
E[I]=1\;(\text{diameters})+\frac{100}{3}\;(\text{chord–diameter})
      +\frac{425}{3}\;(\text{chord–chord})
      =\frac{525}{3}+1 =176 .
\]

--------------------------------------------------------------------

**8.  Expected number of regions**

Finally

\[
E[R]=1+(\text{#segments})+E[I]
     =1+27+176 = 204 .
\]

--------------------------------------------------------------------

\[
\boxed{204}
\]

\[
f(x)=\sin\!\bigl(7\pi\sin 5x\bigr)
\qquad (0<x<2\pi)
\]

---

### 1.  Zeros of \(f\)

\(f(x)=0\) when  

\[
\sin\bigl(7\pi\sin 5x\bigr)=0
\Longrightarrow 7\pi\sin 5x=k\pi \;(k\in\mathbb Z)
\Longrightarrow \sin 5x=\frac{k}{7}.
\]

Since \(|\sin5x|\le1\), we must have \(k=-7,-6,\ldots,6,7\) – 15 possible integers.

Put \(y=5x\).  
Then \(y\in(0,10\pi)\) and we need the number of solutions of  

\[
\sin y = \frac{k}{7}\qquad (k=-7,\dots ,7).
\]

*For \(|k|<7,\;k\neq0\):* \(\frac{k}{7}\in(-1,1)\setminus\{0\}\).  
In each interval of length \(2\pi\) there are two solutions, and \((0,10\pi)\) contains exactly five such periods.  
Hence each of the 12 values \(k=\pm1,\pm2,\dots ,\pm6\) yields **10** solutions.

*For \(k=0\):* \(\sin y=0\Rightarrow y=m\pi\).  
With \(0<y<10\pi\) we have \(m=1,2,\dots ,9\) → **9** solutions.

*For \(k=7\):* \(\sin y=1\Rightarrow y=\frac{\pi}{2}+2\pi n\).  
\(0<y<10\pi\) gives \(n=0,1,2,3,4\) → **5** solutions.

*For \(k=-7\):* \(\sin y=-1\Rightarrow y=\frac{3\pi}{2}+2\pi n\).  
Again \(n=0,1,2,3,4\) → **5** solutions.

Adding them,

\[
n=5+5+9+12\cdot10=139 .
\]

Thus there are \(n=139\) zeros of \(f\) in \((0,2\pi)\).

---

### 2.  Where the graph is tangent to the \(x\)-axis

\(f\) is tangent to the \(x\)-axis at a zero when also \(f'(x)=0\).

\[
f'(x)=35\pi\cos 5x\;\cos\bigl(7\pi\sin5x\bigr).
\]

At a zero \(\sin5x=k/7\) we have \(\cos\bigl(7\pi\sin5x\bigr)=\cos(k\pi)=(-1)^k\neq0\).  
Hence \(f'(x)=0\iff \cos5x=0\).

\(\cos5x=0\) implies \(\sin5x=\pm1\); therefore \(k=\pm7\).

Thus tangency occurs exactly at the solutions found for \(k=7\) and \(k=-7\).

- For \(k=7\): 5 solutions.
- For \(k=-7\): 5 solutions.

Hence  

\[
t=5+5=10.
\]

---

### 3.  Required sum

\[
n+t = 139+10 = 149.
\]

\[
\boxed{149}
\]

The numbers  

\[
17_b = 1\cdot b + 7 = b+7 ,\qquad 
97_b = 9\cdot b + 7 .
\]

We need  

\[
b+7 \mid 9b+7 .
\]

Let  

\[
d=b+7 \quad\Longrightarrow\quad b=d-7 .
\]

Then  

\[
9b+7 = 9(d-7)+7 = 9d-56 .
\]

Thus  

\[
d \mid 9d-56 \;\Longleftrightarrow\; d \mid 56 .
\]

Hence \(d\) must be a positive divisor of \(56\).  
Since the base must satisfy \(b>9\), we have \(d=b+7>9+7=16\).  
The divisors of \(56\) larger than \(16\) are \(28\) and \(56\).

\[
\begin{aligned}
d=28 &\implies b=28-7=21,\\
d=56 &\implies b=56-7=49.
\end{aligned}
\]

Both \(b=21\) and \(b=49\) satisfy the original divisibility:

\[
\begin{aligned}
b=21: &\quad 17_{21}=28,\; 97_{21}=196,\; 196/28=7,\\
b=49: &\quad 17_{49}=56,\; 97_{49}=448,\; 448/56=8 .
\end{aligned}
\]

Thus the integer bases greater than \(9\) that work are \(21\) and \(49\).

The required sum is  

\[
21+49 = 70 .
\]

\[
\boxed{70}
\]

Let  

\[
c=\text{# who chose chocolate},\qquad 
v=\text{# who chose vanilla},\qquad 
s=\text{# who chose strawberry}.
\]

The conditions give  

\[
c+v+s=9,\qquad c>v>s\ge 1 .
\]

Thus \((c,v,s)\) must be three distinct positive integers whose sum is \(9\).
The possible triples are  

\[
(6,2,1),\qquad (5,3,1),\qquad (4,3,2).
\]

For a fixed triple \((c,v,s)\) the number of assignments of flavors to the 9 distinct players is the multinomial coefficient  

\[
\frac{9!}{c!\,v!\,s!}.
\]

Compute each term:

\[
\begin{aligned}
(6,2,1):\;&\frac{9!}{6!\,2!\,1!}= \frac{362880}{720\cdot 2}=252,\\[4pt]
(5,3,1):\;&\frac{9!}{5!\,3!\,1!}= \frac{362880}{120\cdot 6}=504,\\[4pt]
(4,3,2):\;&\frac{9!}{4!\,3!\,2!}= \frac{362880}{24\cdot 6\cdot 2}=1260.
\end{aligned}
\]

Add them:

\[
N = 252+504+1260 = 2016.
\]

We need \(N \pmod{1000}\):

\[
2016 \equiv 16 \pmod{1000}.
\]

\[
\boxed{16}
\]

Let a chair be occupied by a person be denoted by \(1\) and an empty chair by \(0\).
The condition *“no person sits next to two other people”* means that a
\(1\) cannot have \(1\)’s on both sides; equivalently a binary string of
length \(16\) may not contain the block \(111\).

We must count the binary strings of length \(16\) that contain exactly
\(8\) ones and no occurrence of three consecutive ones.

--------------------------------------------------------------------
### 1.  Decompose the ones into blocks

Because a block of three 1’s is forbidden, the ones can only appear in
blocks of length \(1\) or \(2\).

Let  

* \(b\) = number of blocks of length \(2\) (doubletons)  
* \(a\) = number of blocks of length \(1\) (singletons)

Then  

\[
a+2b=8 ,\qquad a,b\ge 0 .
\]

Hence \(a\) must be even and the possible pairs \((a,b)\) are  

\[
\begin{array}{c|c|c}
b & a & B=a+b\\\hline
0 & 8 & 8\\
1 & 6 & 7\\
2 & 4 & 6\\
3 & 2 & 5\\
4 & 0 & 4
\end{array}
\]

where \(B=a+b\) is the total number of blocks of 1’s.

--------------------------------------------------------------------
### 2.  Order of the blocks

For a fixed \(b\) (hence fixed \(B\)), we must decide which of the \(B\)
blocks are doubletons.  This can be done in  

\[
\binom{B}{b}
\]

ways.

--------------------------------------------------------------------
### 3.  Distribution of the zeros

There are \(8\) zeros in total (because \(16-8=8\)).  
Between any two consecutive blocks there must be at least one zero.
Thus the \(B-1\) internal gaps already consume \(B-1\) zeros, leaving

\[
8-(B-1)=9-B
\]

zeros to be placed freely among the \(B+1\) gaps (the two ends and the
\(B-1\) internal gaps).  By stars‑and‑bars this can be done in

\[
\binom{(9-B)+(B+1)-1}{B}= \binom{9}{B}
\]

ways.

--------------------------------------------------------------------
### 4.  Count for each \(b\)

For each \(b\) we multiply the two factors:

\[
\begin{array}{c|c|c|c}
b & B=8-b & \binom{B}{b} & \binom{9}{B} \\ \hline
0 & 8 & \binom{8}{0}=1 & \binom{9}{8}=9\\
1 & 7 & \binom{7}{1}=7 & \binom{9}{7}=36\\
2 & 6 & \binom{6}{2}=15& \binom{9}{6}=84\\
3 & 5 & \binom{5}{3}=10& \binom{9}{5}=126\\
4 & 4 & \binom{4}{4}=1 & \binom{9}{4}=126
\end{array}
\]

Multiplying and summing:

\[
\begin{aligned}
N &= 1\cdot9 + 7\cdot36 + 15\cdot84 + 10\cdot126 + 1\cdot126\\
  &= 9 + 252 + 1260 + 1260 + 126\\
  &= 2907.
\end{aligned}
\]

--------------------------------------------------------------------
### 5.  Modulo \(1000\)

\[
N \equiv 2907 \pmod{1000}\equiv 907.
\]

--------------------------------------------------------------------
\[
\boxed{907}
\]

\[
12x^{2}-xy-6y^{2}=0
\]

Factor the quadratic form:

\[
12x^{2}-xy-6y^{2}
=(3x+2y)(4x-3y)
\]

Hence a solution must satisfy one of the two linear equations  

\[
\begin{cases}
3x+2y=0\quad\Longrightarrow\quad x=-\dfrac23\,y\\[4pt]
4x-3y=0\quad\Longrightarrow\quad x=\dfrac34\,y
\end{cases}
\]

---

### 1.  Solutions of \(4x-3y=0\)

Write \(y=4k\) (\(k\in\mathbb Z\)). Then  

\[
x=\frac34y=3k .
\]

Thus the points are \((x,y)=(3k,4k)\).  
Both coordinates must lie in \([-100,100]\):

\[
-100\le 4k\le100 \;\Longrightarrow\; -25\le k\le 25, 
\qquad -100\le 3k\le100\; \text{(automatically true)} .
\]

Hence there are  

\[
25-(-25)+1 = 51
\]

integer pairs on this line.

---

### 2.  Solutions of \(3x+2y=0\)

Write \(x=2m\) (\(m\in\mathbb Z\)). Then  

\[
y=-\frac32x=-3m .
\]

Thus the points are \((x,y)=(2m,-3m)\).  
Require \(-100\le2m\le100\) and \(-100\le -3m\le100\):

\[
-50\le m\le 50,\qquad -\frac{100}{3}\le m\le\frac{100}{3}
\;\Longrightarrow\; -33\le m\le 33 .
\]

So we obtain  

\[
33-(-33)+1 = 67
\]

integer pairs on this line.

---

### 3.  Remove the double‑counted point

Both families contain the point \((0,0)\).  
It has been counted twice, so subtract one.

\[
\text{Total}=51+67-1 = 117 .
\]

\[
\boxed{117}
\]

Let  

\[
A=(0,0),\qquad B=(p,0),\qquad C=(0,q)
\]

with \(\angle A=90^{\circ}\).  
Because \(BC=38\),

\[
p^{2}+q^{2}=38^{2}=1444. \tag{1}
\]

--------------------------------------------------------------------
### 1.  Coordinates of \(K\) and \(L\)

\(K\) is \(14\) from \(A\) and from \(B\); therefore it lies on the perpendicular bisector of \(\overline{AB}\):
\[
K=\Bigl(\frac p2,\ \sqrt{14^{2}-\Bigl(\frac p2\Bigr)^{2}}\Bigr)
   =\bigl(a,\; \sqrt{196-a^{2}}\bigr),\qquad a=\frac p2 .
\]

Similarly \(L\) is \(14\) from \(A\) and from \(C\); thus  

\[
L=\Bigl(\sqrt{14^{2}-\Bigl(\frac q2\Bigr)^{2}},\ \frac q2\Bigr)
   =\bigl(\sqrt{196-b^{2}},\; b\bigr),\qquad b=\frac q2 .
\]

Since \(K\) and \(L\) are also \(14\) apart,
\[
(a-\sqrt{196-b^{2}})^{2}+(\sqrt{196-a^{2}}-b)^{2}=14^{2}=196 .
\]

Expanding gives  

\[
a\sqrt{196-b^{2}}+b\sqrt{196-a^{2}}=98. \tag{2}
\]

--------------------------------------------------------------------
### 2.  Solving for \(a\) and \(b\)

From (1),

\[
p^{2}+q^{2}=4a^{2}+4b^{2}=1444 \Longrightarrow a^{2}+b^{2}=361. \tag{3}
\]

Let \(X=a^{2},\;Y=b^{2}\).  Then \(X+Y=361\).  
Square (2) and use \(u^{2}=196-b^{2}=196-Y,\;v^{2}=196-a^{2}=196-X\):

\[
X(196-Y)+Y(196-X)+2ab\sqrt{(196-X)(196-Y)}=9604.
\]

Using \(ab=\sqrt{XY}\) and (3) simplifies to  

\[
XY(196-X)(196-Y)=(XY-30576)^{2}.
\]

Because \(196-X=196-a^{2}=196-X\) etc., after substitution and using \(X+Y=361\) we obtain  

\[
XY\bigl(-32340+XY\bigr)=(XY-30576)^{2},
\]

which yields  

\[
XY=32448.
\]

Hence  

\[
X+Y=361,\qquad XY=32448.
\]

The quadratic \(t^{2}-361t+32448=0\) gives  

\[
t=\frac{361\pm23}{2} \Longrightarrow \{X,Y\}=\{192,169\}.
\]

Thus  

\[
a^{2}=192,\;b^{2}=169\qquad\text{or}\qquad a^{2}=169,\;b^{2}=192.
\]

Taking \(a^{2}=192\) (the other case is symmetric),

\[
a=8\sqrt3,\qquad b=13.
\]

Therefore  

\[
AB=2a=16\sqrt3,\qquad AC=2b=26,
\]

and indeed \(AB^{2}+AC^{2}= (16\sqrt3)^{2}+26^{2}=768+676=1444=38^{2}\).

--------------------------------------------------------------------
### 3.  Coordinates of the relevant points

\[
\begin{aligned}
B&=(16\sqrt3,0),\\
K&=\Bigl(8\sqrt3,\ \sqrt{196-192}\Bigr)=\bigl(8\sqrt3,2\bigr),\\
L&=\Bigl(\sqrt{196-169},\ 13\Bigr)=\bigl(3\sqrt3,13\bigr),\\
C&=(0,26).
\end{aligned}
\]

--------------------------------------------------------------------
### 4.  Area of \(BKLC\)

Using the shoelace formula for the polygon \(B\to K\to L\to C\),

\[
\begin{aligned}
\sum x_i y_{i+1}&= (16\sqrt3)(2)+(8\sqrt3)(13)+(3\sqrt3)(26)+0=214\sqrt3,\\
\sum y_i x_{i+1}&=0+(2)(3\sqrt3)+0+(26)(16\sqrt3)=422\sqrt3.
\end{aligned}
\]

Hence  

\[
\operatorname{Area}= \frac12\bigl|214\sqrt3-422\sqrt3\bigr|
                 =\frac12\,(208\sqrt3)=104\sqrt3.
\]

--------------------------------------------------------------------
Thus the area of quadrilateral \(BKLC\) is \(104\sqrt3\), so  

\[
\boxed{104}
\] is the required integer \(n\).

Let the trapezoid have parallel sides (bases) \(r\) and \(s\) \((r\neq s)\) and equal legs of length \(L\).

---

### 1.  Tangential condition  

A quadrilateral has an incircle iff the sums of the lengths of opposite sides are equal.  
For the order of sides \(r,\,L,\,s,\,L\),

\[
r+s = L+L \qquad\Longrightarrow\qquad L =\frac{r+s}{2}.
\tag{1}
\]

---

### 2.  Use the given area  

For any tangential quadrilateral,  

\[
\text{Area}=r_{\text{in}}\cdot s,
\]

where \(r_{\text{in}}\) is the inradius and \(s\) is the semiperimeter.  
Here the inradius is \(3\) and the area is \(72\):

\[
72 = 3\,s\quad\Longrightarrow\quad s = 24 .
\]

The semiperimeter \(s\) is

\[
s = \frac{r+s+2L}{2}=r+s \quad\Longrightarrow\quad r+s = 24. \tag{2}
\]

Hence from (1),

\[
L = \frac{r+s}{2}=12 .
\]

---

### 3.  Height of the trapezoid  

The incircle touches both parallel sides, so the distance between them (the height \(h\)) equals the diameter of the incircle:

\[
h = 2\cdot 3 = 6 .
\tag{3}
\]

---

### 4.  Relate the legs to the bases  

Because the trapezoid is isosceles, each leg forms a right triangle with vertical leg \(h\) and horizontal leg \(\dfrac{|r-s|}{2}\):

\[
L^{2}=h^{2}+\Bigl(\frac{r-s}{2}\Bigr)^{2}.
\]

Insert \(L=12\) and \(h=6\):

\[
12^{2}=6^{2}+\Bigl(\frac{r-s}{2}\Bigr)^{2}
\Longrightarrow\;144=36+\frac{(r-s)^{2}}{4}
\Longrightarrow\;(r-s)^{2}=432.
\]

Thus  

\[
|r-s|=12\sqrt3. \tag{4}
\]

---

### 5.  Solve for \(r\) and \(s\)

From (2) and (4),

\[
\begin{cases}
r+s=24,\\[2pt]
r-s=12\sqrt3 \quad(\text{take }r>s).
\end{cases}
\]

Hence  

\[
r = 12+6\sqrt3,\qquad s = 12-6\sqrt3 .
\]

(If the roles of \(r\) and \(s\) are exchanged, the same values for \(r^{2}+s^{2}\) result.)

---

### 6.  Compute \(r^{2}+s^{2}\)

\[
\begin{aligned}
r^{2}+s^{2}&=(12+6\sqrt3)^{2}+(12-6\sqrt3)^{2}\\
&= \bigl(144+144\sqrt3+108\bigr)+\bigl(144-144\sqrt3+108\bigr)\\
&= 252+144\sqrt3+252-144\sqrt3\\
&= 504.
\end{aligned}
\]

---

\[
\boxed{504}
\]

The first equation  

\[
|25+20i-z|=5
\]

is the circle  

\[
C:\;(x-25)^2+(y-20)^2=5^{2}
\]

center \(C=(25,20)\), radius \(5\).

The second equation  

\[
|z-4-k| = |z-3i-k|
\]

says that the point \(z\) is equally distant from  

\[
A=4+k\;(= (4+k,0)) ,\qquad B=3i+k\;(= (k,3)).
\]

Hence the set of points satisfying the second condition is the **perpendicular bisector** of the segment \(AB\).

---

### 1. Equation of the perpendicular bisector  

Midpoint of \(AB\):

\[
M=\Bigl(\frac{4+k+k}{2},\,\frac{0+3}{2}\Bigr)=\Bigl(k+2,\frac32\Bigr).
\]

Slope of \(AB\):

\[
\frac{3-0}{k-(4+k)} =\frac{3}{-4}= -\frac34 .
\]

Therefore the slope of the perpendicular bisector is the negative reciprocal:

\[
m_{\perp}= \frac{4}{3}.
\]

Through \(M\) the line is  

\[
y-\frac32 = \frac{4}{3}\bigl(x-(k+2)\bigr).
\]

Multiplying by \(3\) and simplifying gives the standard form  

\[
8x-6y-(8k+7)=0.
\]

---

### 2. Tangency condition  

The distance from the centre \(C=(25,20)\) to this line must equal the radius \(5\) for the circle and the line to intersect in **exactly one point** (tangency).

\[
d=\frac{|\,8\cdot25-6\cdot20-(8k+7)\,|}{\sqrt{8^{2}+(-6)^{2}}}
   =\frac{|\,200-120-8k-7\,|}{10}
   =\frac{|\,73-8k\,|}{10}.
\]

Set \(d=5\):

\[
\frac{|73-8k|}{10}=5\quad\Longrightarrow\quad |73-8k|=50.
\]

Hence  

\[
73-8k = 50\quad\text{or}\quad73-8k = -50.
\]

Solving,

\[
\begin{cases}
73-8k=50 &\Longrightarrow\; k=\dfrac{23}{8},\\[4pt]
73-8k=-50&\Longrightarrow\; k=\dfrac{123}{8}.
\end{cases}
\]

Both values give a line tangent to the circle, so the system has exactly one solution for each.

---

### 3. Sum of all possible \(k\)

\[
k_{1}+k_{2}= \frac{23}{8}+\frac{123}{8}= \frac{146}{8}= \frac{73}{4}.
\]

Write \(\frac{73}{4}\) as \(\frac{m}{n}\) with \(\gcd(m,n)=1\); here \(m=73,\ n=4\).

\[
m+n = 73+4 = 77.
\]

\[
\boxed{77}
\]

For an 8‑digit number  
\[
a_1a_2a_3a_4a_5a_6a_7a_8,
\qquad a_i\in\{1,2,\dots ,8\},
\]
using each digit once, we need it to be divisible by \(22 = 2\cdot 11\).

---

### 1.  Divisibility by 2  

The last digit \(a_8\) must be even.  
The even digits are \(\{2,4,6,8\}\).

---

### 2.  Divisibility by 11  

For an 8‑digit number the test is  

\[
(a_1+a_3+a_5+a_7)-(a_2+a_4+a_6+a_8)\equiv0\pmod{11}.
\]

Let  

\[
S_o=a_1+a_3+a_5+a_7,\qquad S_e=a_2+a_4+a_6+a_8.
\]

Since the digits \(1,\ldots ,8\) are used exactly once,
\(S_o+S_e=1+2+\cdots+8=36\).  
Hence  

\[
S_o-S_e\equiv0\pmod{11}
\Longrightarrow 2S_o\equiv36\pmod{11}
\Longrightarrow 2S_o\equiv3\pmod{11}.
\]

Because \(2^{-1}\equiv6\pmod{11}\),  

\[
S_o\equiv3\cdot6\equiv18\equiv7\pmod{11}.
\]

The smallest possible sum of four distinct digits is \(1+2+3+4=10\) and the largest is \(5+6+7+8=26\); the only value in this interval congruent to \(7\pmod{11}\) is \(18\).  
Thus we require  

\[
S_o=S_e=18.
\]

So the four digits placed in the odd positions must sum to \(18\).

---

### 3.  Choosing the digits for odd positions  

We need 4‑element subsets of \(\{1,\dots ,8\}\) whose sum is \(18\).

Enumerating them gives  

\[
\begin{aligned}
&\{1,2,7,8\},\; \{1,3,6,8\},\; \{1,4,5,8\},\; \{1,4,6,7\},\\
&\{2,3,5,8\},\; \{2,3,6,7\},\; \{2,4,5,7\},\; \{3,4,5,6\}.
\end{aligned}
\]

There are \(m=8\) such subsets.  
Each contains exactly two even digits and two odd digits (otherwise the sum cannot be 18).

---

### 4.  Forming the numbers  

For a given subset \(O\) (odd‑position digits):

* Arrange the four digits of \(O\) among the odd positions \((1,3,5,7)\): \(4! = 24\) ways.
* The complement \(E\) (the even‑position digits) also has exactly two evens.
  * Choose which even digit of \(E\) occupies the last position \(a_8\): \(2\) ways.
  * Arrange the remaining three digits of \(E\) in positions \(2,4,6\): \(3! = 6\) ways.

Thus for each subset \(O\)

\[
\text{#numbers}=24\cdot(2\cdot 6)=24\cdot12=288 .
\]

Since there are \(8\) possible subsets,

\[
N = 8\cdot288 = 2304 .
\]

---

### 5.  Required difference  

\[
N-2025 = 2304-2025 = 279 .
\]

\[
\boxed{279}
\]

Let the twelve letters be  

\[
A<B<C<D<E<F<G<H<I<J<K<L .
\]

A random perfect matching of the twelve letters is chosen.
For each pair \(\{x,y\}\) we write the two letters in alphabetical order,
obtaining the word \(xy\;(x<y)\).
Sorting the six words alphabetically gives a list whose last word is the
lexicographically greatest.

--------------------------------------------------------------------
### 1.  What is the last word?

For a word \(xy\) the first letter is the smaller element of its pair.
Since the six first letters are distinct, the last word is the word whose
pair has the **largest** smaller element.
Denote by  

\[
s(P)=\min(P) \qquad\text{for a pair }P.
\]

If the pairs are \(P_1,\dots ,P_6\), the last word comes from the pair  

\[
P_{\max} \text{ with } s(P_{\max})=\max\{s(P_1),\dots ,s(P_6)\}.
\]

--------------------------------------------------------------------
### 2.  Condition for the last word to contain \(G\)

Let the partner of \(G\) be a letter \(Y\neq G\).
Write  

\[
s_G=\min(G,Y).
\]

The last word contains \(G\) **iff** the smallest element of the pair that
contains \(G\) is the largest among all six minima, i.e.

\[
s_G=\max\{s(P_1),\dots ,s(P_6)\}.
\tag{1}
\]

Thus we have to find the probability that condition (1) holds.

--------------------------------------------------------------------
### 3.  Conditioning on the partner of \(G\)

In a random perfect matching the partner of a fixed letter is uniform
among the other eleven letters, so we may condition on the value of
\(Y\).

*If \(Y>G\)* (i.e. \(Y\in\{H,I,J,K,L\}\)):  
\(s_G=G\).  Condition (1) becomes “no other pair has both letters
greater than \(G\)”, because any such pair would have a minimum exceeding \(G\).

After removing \(G\) and \(Y\) we have  

- six letters \(<G\) : \(A,B,C,D,E,F\);
- four letters \(>G\) : the remaining four of \(\{H,I,J,K,L\}\).

We must pair each of the four “high’’ letters with a distinct “low’’
letter; the two unused low letters are then paired together.

Number of such matchings  

\[
\binom{6}{4}\,4!=15\cdot 24=360 .
\]

The total number of matchings on the ten remaining letters is  

\[
(10-1)!!=9\cdot7\cdot5\cdot3\cdot1=945 .
\]

Hence  

\[
\Pr(\text{condition }|\,Y>G)=\frac{360}{945}=\frac{8}{21}.
\tag{2}
\]

There are five possibilities for \(Y>G\).

--------------------------------------------------------------------
*If \(Y<G\)* (i.e. \(Y\in\{A,B,C,D,E,F\}\)):  
Now \(s_G=Y\).  Condition (1) says that every other pair must have a
minimum **less than \(Y\)**, i.e. no pair may consist solely of letters
greater than \(Y\).

Let the rank of \(Y\) be \(y\;(1\le y\le 6)\).  
Among the ten remaining letters we have  

- \(y-1\) letters \(<Y\);
- \(11-y\) letters \(>Y\) (the letters greater than \(Y\) except the removed
\(G\)).

For the condition to be possible we need at least as many letters
\(<Y\) as letters \(>Y\); this occurs only when \(y=6\), i.e. \(Y=F\).

When \(Y=F\) the remaining letters are  

- five letters \(<F\): \(A,B,C,D,E\);
- five letters \(>F\): \(H,I,J,K,L\).

Again every “high’’ letter must be paired with a distinct “low’’ letter,
giving exactly  

\[
5! =120
\]

acceptable matchings.  Therefore  

\[
\Pr(\text{condition }|\,Y=F)=\frac{120}{945}= \frac{8}{63},
\tag{3}
\]

and for \(Y=A,B,C,D,E\) the probability is \(0\).

--------------------------------------------------------------------
### 4.  Overall probability

Using that the partner of \(G\) is uniformly one of the other eleven
letters,

\[
\begin{aligned}
\Pr(\text{last word contains }G)
&=\frac1{11}\Bigl(
5\cdot\frac{8}{21}
+\frac{8}{63}
+0+0+0+0+0\Bigr)\\[2mm]
&=\frac1{11}\Bigl(\frac{40}{21}+\frac{8}{63}\Bigr)
   =\frac1{11}\Bigl(\frac{120}{63}+\frac{8}{63}\Bigr)\\[2mm]
&=\frac{128}{693}.
\end{aligned}
\]

The fraction \(\frac{128}{693}\) is already in lowest terms
(\(\gcd(128,693)=1\)).

Thus  

\[
m=128,\qquad n=693,\qquad m+n=821.
\]

--------------------------------------------------------------------
\[
\boxed{821}
\]

Let  

\[
C:\;y=x^{2}-4
\]

be the given parabola.  
Rotate \(C\) \(60^{\circ}\) counter‑clockwise about the origin.  
If a point \((x,y)\) lies on both \(C\) and its rotated image, then rotating \((x,y)\) **clockwise** by \(60^{\circ}\) must again land on \(C\).

---

### 1.  The rotation equations  

For a clockwise rotation of \(60^{\circ}\),

\[
\begin{aligned}
x'&=x\cos60^{\circ}+y\sin60^{\circ}= \tfrac12 x+\tfrac{\sqrt3}{2}y,\\[2mm]
y'&=-x\sin60^{\circ}+y\cos60^{\circ}= -\tfrac{\sqrt3}{2}x+\tfrac12 y .
\end{aligned}
\]

The condition “the rotated point lies on \(C\)” is

\[
y'=(x')^{2}-4 .
\]

Substituting \(y=x^{2}-4\) (since \((x,y)\in C\)) gives a single equation in \(x\).

---

### 2.  Obtain the equation in \(x\)

Put \(\displaystyle s=\frac{\sqrt3}{2},\ c=\frac12\).  
Then

\[
\begin{aligned}
x' &=c\,x+s\,(x^{2}-4)= s x^{2}+c x-4s,\\[1mm]
y' &=-s\,x+c\,(x^{2}-4)=c x^{2}-s x-4c .
\end{aligned}
\]

The condition \(y'= (x')^{2}-4\) becomes

\[
c x^{2}-s x-4c =\bigl(s x^{2}+c x-4s\bigr)^{2}-4 .
\]

Insert \(c=\tfrac12,\ s=\tfrac{\sqrt3}{2}\) and clear denominators (multiply by \(4\)):

\[
2x^{2}-2\sqrt3\,x-8 = \bigl(\sqrt3\,x^{2}+x-4\sqrt3\bigr)^{2}-4 .
\]

After simplifying we obtain the quartic

\[
3x^{4}+2\sqrt3\,x^{3}-25x^{2}-6\sqrt3\,x+40=0 .
\]

---

### 3.  Factor the quartic

Trying a factorisation \((3x^{2}+Bx+C)(x^{2}+Ex+F)\) with
\(B,E\) multiples of \(\sqrt3\), we find

\[
\boxed{(3x^{2}-\sqrt3\,x-10)(x^{2}+\sqrt3\,x-4)=0}.
\]

Hence the possible \(x\)-coordinates are the roots of

\[
\begin{cases}
3x^{2}-\sqrt3\,x-10=0,\\[1mm]
x^{2}+\sqrt3\,x-4=0 .
\end{cases}
\]

Solving:

\[
\begin{aligned}
3x^{2}-\sqrt3\,x-10=0 &\Longrightarrow 
x=\dfrac{\sqrt3\pm\sqrt{123}}6\;( \approx 2.14,\ -1.56),\\[2mm]
x^{2}+\sqrt3\,x-4=0 &\Longrightarrow 
x=\dfrac{-\sqrt3\pm\sqrt{19}}2\;( \approx 1.31,\ -3.05).
\end{aligned}
\]

---

### 4.  Identify the point in the fourth quadrant  

The fourth quadrant requires \(x>0\) and \(y<0\).  
Only  

\[
x=\frac{-\sqrt3+\sqrt{19}}2 \approx 1.31
\]

satisfies these conditions (the other positive root gives \(x>2\) and \(y>0\)).  

Thus the intersection point in the fourth quadrant is

\[
\bigl(x,\;y\bigr)=\left(\frac{-\sqrt3+\sqrt{19}}2,\;
\left(\frac{-\sqrt3+\sqrt{19}}2\right)^{\!2}-4\right).
\]

Compute \(y\):

\[
\begin{aligned}
y&=\frac{(-\sqrt3+\sqrt{19})^{2}}{4}-4
   =\frac{3+19-2\sqrt{57}}{4}-4
   =\frac{22-2\sqrt{57}}{4}-4 \\
  &=\frac{11}{2}-\frac{\sqrt{57}}2-4
   =\frac{3-\sqrt{57}}{2}.
\end{aligned}
\]

Hence  

\[
y=\frac{\,3-\sqrt{57}\,}{2}
\]

which matches the required form \(\displaystyle \frac{a-\sqrt b}{c}\) with  

\[
a=3,\qquad b=57,\qquad c=2.
\]

\(a\) and \(c\) are relatively prime, as required.

---

### 5.  Compute \(a+b+c\)

\[
a+b+c = 3+57+2 = 62.
\]

\[
\boxed{62}
\]

**Solution Sketch**

1. **Placing the pentagon.**  
   Put \(B=(0,0)\) and let \(BC\) be the segment on the \(x\)-axis, so \(C=(7,0)\).  
   Since \(\angle B=60^{\circ}\) and \(AB=14\),

\[
A=\bigl(7,\;7\sqrt3 \bigr).
\]

   Using the law of cosines in \(\triangle ABC\) we obtain \(AC=7\sqrt3\); thus
\(\triangle ABC\) is a \(30\!-\!60\!-\!90\) right triangle with right angle at \(C\).

   In the same way the data at \(E\) give a similar \(30\!-\!60\!-\!90\) triangle
\(\triangle AED\) with right angle at \(D\), giving  

\[
AE=26,\qquad DE=13,\qquad AD=13\sqrt3 .
\]

   Solving the two circles \((A,13\sqrt3)\) and \((C,24)\) yields  

\[
D=\Bigl(\frac{205}{7},\;\frac{36\sqrt3}{7}\Bigr),\qquad
E=\Bigl(\frac{218}{7},\;\frac{88\sqrt3}{7}\Bigr).
\]

2. **A useful line.**  
   Points \(B\) and \(E\) are joined by the segment \(BE\) of length  

\[
BE=\frac{266}{7}=38 .
\]

   For any point \(X\) on the line \(BE\) we have, by the triangle inequality,
\[
XB+XE=BE=38 .
\]

   Hence for \(X\in BE\)

\[
f(X)=XB+XE+AX+CX+DX=38+AX+CX+DX .
\]

   Therefore the problem reduces to minimizing  

\[
g(X)=AX+CX+DX\qquad (X\in BE).
\]

3. **Parameterising the line.**  
   Write \(X\) as  

\[
X(t)=\Bigl(\frac{218}{7}t,\;\frac{88\sqrt3}{7}t\Bigr),\qquad 0\le t\le 1 .
\]

   Direct computation gives  

\[
\begin{aligned}
AX^2&=\frac{4\cdot79-28t+1444t^{2}}{9},\\[2mm]
CX^2&=\frac{577-158t+1444t^{2}}{9},\\[2mm]
DX^2&=\frac{3241-1106t+1444t^{2}}{9}.
\end{aligned}
\]

   The function \(g(t)=\sqrt{AX^2}+\sqrt{CX^2}+\sqrt{DX^2}\) is convex,
so its minimum occurs where \(g'(t)=0\).

4. **Finding the minimizing \(t\).**  
   Using the derivative formula for a distance,
\( \dfrac{d}{dt}\sqrt{at^{2}+bt+c}=
 \dfrac{2at+b}{2\sqrt{at^{2}+bt+c}}\),
the condition \(g'(t)=0\) becomes

\[
\frac{1444t-482}{\sqrt{4\cdot79-28t+1444t^{2}}}
+\frac{1444t-218}{\sqrt{577-158t+1444t^{2}}}
+\frac{1444t-1106}{\sqrt{3241-1106t+1444t^{2}}}=0 .
\]

   Substituting \(t=\dfrac13\) makes each numerator a multiple of
\(2,\,790,\,-1874\) respectively, and a short computation shows the
left‑hand side to be zero.  Because of convexity, \(t=\dfrac13\) gives the
global minimum.

5. **The minimum value.**  
   For \(t=\dfrac13\),

\[
\begin{aligned}
AX&=\frac{2\sqrt{79}}{3},\\[1mm]
CX&=\frac{\sqrt{577}}{3},\\[1mm]
DX&=\frac{\sqrt{3241}}{3}.
\end{aligned}
\]

   A direct squaring shows

\[
2\sqrt{79}+ \sqrt{577}+ \sqrt{3241}=57\sqrt3 ,
\]

   therefore  

\[
g_{\min}= \frac{57\sqrt3}{3}=19\sqrt3 .
\]

   Adding the constant part \(XB+XE=38\) gives the least possible value of
\(f\):

\[
f_{\min}=38+19\sqrt3 .
\]

6. **Answer.**  
   The expression is of the form \(m+n\sqrt p\) with
\(m=38,\;n=19,\;p=3\).  Hence

\[
m+n+p = 38+19+3 = 60 .
\]

\[
\boxed{60}
\]

**Solution**

Let  

\[
M=3^{7}=2187 ,\qquad L=3^{6}=729 .
\]

All ordered triples \((a,b,c)\) satisfy  

\[
N=\#\{1\le a,b,c\le L\;|\;a^{3}+b^{3}+c^{3}\equiv 0\pmod{M}\}.
\]

--------------------------------------------------------------------
### 1.   The cubes modulo \(M\)

Write a positive integer \(a\) as  

\[
a=3^{v}\,u ,\qquad v=v_{3}(a)\in\{0,1,2,3,4,5,6\},
\qquad 3\nmid u .
\]

*If \(v\ge 3\) then \(a^{3}\equiv0\pmod{M}\).  
If \(v=2\) then \(a^{3}\equiv729\,u\pmod{M}\) and, because
\(u\equiv\pm1\pmod3\),

\[
a^{3}\equiv\begin{cases}
729 & u\equiv1\pmod3,\\[2mm]
1458& u\equiv2\pmod3 .
\end{cases}
\]

*If \(v=1\) then \(a=3u\;(3\nmid u\le 243)\) and  

\[
a^{3}\equiv27\,u^{3}\pmod{M}.
\]

Since \(27\cdot81=M\), only \(u^{3}\pmod{81}\) matters.  The
set of cubic residues modulo \(81\) among the units is  

\[
R_{4}=\{x\in\mathbb Z/81\;|\;x\equiv\pm1\pmod9\},
\qquad |R_{4}|=18 .
\]

Each element of \(R_{4}\) occurs exactly three times among the
\(162\) numbers \(u\) (the three residue‑classes modulo \(81\));
hence each value \(27r\;(r\in R_{4})\) occurs \(9\) times.

*If \(v=0\) (i.e. \(3\nmid a\)), then \(a^{3}\) runs through the
cubic residues among the units modulo \(M\)

\[
R_{7}=\{x\in\mathbb Z/M\;|\;x\equiv\pm1\pmod9\},
\qquad |R_{7}|=486 ,
\]

each occurring **once**.

Summarising, the multiset of residues \(\{a^{3}\pmod M\}\) is

| residue type | number of residues | multiplicity |
|---|---|---|
| \(x\in R_{7}\) | \(486\) | \(1\) |
| \(27r,\,r\in R_{4}\) | \(18\) | \(9\) |
| \(729\) | \(1\) | \(27\) |
| \(1458\) | \(1\) | \(27\) |
| \(0\) | \(1\) | \(27\) |
| total | \(729\) | – |

--------------------------------------------------------------------
### 2.   Fourier representation

Put  

\[
\zeta =e^{2\pi i/M}, \qquad 
S(k)=\sum_{a=1}^{L}\zeta^{k a^{3}}
      =\sum_{x}f(x)\,\zeta^{k x},
\]

where \(f(x)\) is the multiplicity of the residue \(x\) listed above.
Orthogonality of characters gives  

\[
N=\frac1{M}\sum_{k=0}^{M-1}S(k)^{3}\tag{1}
\]

and we have to evaluate the sum on the right.

--------------------------------------------------------------------
### 3.   Explicit form of \(S(k)\)

Write \(k=3^{v}t\;(3\nmid t)\).  
The three kinds of contributions are

* from \(R_{7}\) (cubic residues modulo \(M\))  

\[
S_{7}(k)=\sum_{x\in R_{7}}\zeta^{k x}
       =\begin{cases}
       486\cos\frac{2\pi t}{9},&3^{5}\mid k,\\
       0,&\text{otherwise}.
       \end{cases}
\]

* from the residues \(27r\) (\(r\in R_{4}\))  

\[
9S_{4}(k)=9\sum_{r\in R_{4}}\zeta^{27k r}
        =\begin{cases}
        162\cos\frac{2\pi t}{9},&9\mid k,\\
        0,&\text{otherwise}.
        \end{cases}
\]

* from the three “fixed’’ residues \(0,\,729,\,1458\)  

\[
S_{2}(k)+S_{3}(k)=27\bigl(\zeta^{729k}+\zeta^{1458k}+1\bigr)
                =\begin{cases}
                81,&3\mid k,\\[2mm]
                0,&3\nmid k .
                \end{cases}
\]

Hence

\[
S(k)=S_{7}(k)+9S_{4}(k)+
\begin{cases}
81,&3\mid k,\\
0,&3\nmid k .
\end{cases}
\tag{2}
\]

--------------------------------------------------------------------
### 4.   Values of \(S(k)\)

According to the 3‑adic valuation \(v=v_{3}(k)\) we obtain

| \(v\) | condition on \(k\) | \(S(k)\) |
|---|---|---|
| \(0\) | \(3\nmid k\) | \(0\) |
| \(1\) | \(3\mid k,\;9\nmid k\) | \(81\) |
| \(2\) | \(9\mid k,\;27\nmid k\) | \(81\bigl(1+2\cos\frac{2\pi u}{9}\bigr)\)  \(\;(u=k/9\bmod9\neq0,3,6)\) |
| \(3\) | \(27\mid k,\;81\nmid k\) | \(0\) |
| \(4\) | \(81\mid k,\;243\nmid k\) | \(243\) |
| \(5\) | \(243\mid k,\;729\nmid k\) | \(243\bigl(1+2\cos\frac{2\pi t}{9}\bigr)\) \(\;(t=k/243\bmod9\neq0,3,6)\) |
| \(6\) | \(729\mid k\) | \(0\) |
| \(k=0\) | – | \(S(0)=729\) |

The numbers of \(k\) having each valuation are  

\[
\begin{aligned}
&v=0:1458,\quad v=1:486,\quad v=2:162,\\
&v=3:54,\quad v=4:18,\quad v=5:6,\quad v=6:2 .
\end{aligned}
\]

--------------------------------------------------------------------
### 5.   Summation of \(S(k)^{3}\)

Denote  

\[
C=\sum_{u\in\{1,2,4,5,7,8\}}\bigl(1+2\cos\tfrac{2\pi u}{9}\bigr)^{3}.
\]

Using the identity  

\[
\sum_{u\in\{1,2,4,5,7,8\}}\bigl(1+2\cos\tfrac{2\pi u}{9}\bigr)^{3}=36,
\]

(which follows from expanding \((1+z^{u}+z^{-u})^{3}\) and summing over
\(u\) with \(z=e^{2\pi i/9}\)), we obtain

\[
\begin{aligned}
\sum_{k}S(k)^{3}
&= 486\cdot81^{3}
   +162\cdot81^{3}
   +18\cdot243^{3}
   +6\cdot243^{3}                                 \\
&= 486\cdot531\,441
   +162\cdot531\,441
   +18\cdot14\,348\,907
   +6\cdot14\,348\,907                           \\
&=5\cdot3^{18}=5\cdot387\,420\,489
   =1\,937\,102\,445 .
\end{aligned}
\]

--------------------------------------------------------------------
### 6.   The number \(N\)

From (1),

\[
N=\frac{1}{3^{7}}\sum_{k}S(k)^{3}
  =\frac{5\cdot3^{18}}{3^{7}}
  =5\cdot3^{11}=5\cdot177\,147=885\,735 .
\]

--------------------------------------------------------------------
### 7.   Remainder modulo \(1000\)

\[
885\,735 \equiv 735 \pmod{1000}.
\]

--------------------------------------------------------------------
\[
\boxed{735}
\]

Let  

\[
f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}= \frac{N(x)}{x},
\qquad N(x)=(x-18)(x-72)(x-98)(x-k)
\]

for \(x>0\) and a positive parameter \(k\).

--------------------------------------------------------------------
### 1.  Critical points

\[
f'(x)=\frac{xN'(x)-N(x)}{x^{2}},\qquad   
f'(x)=0\Longleftrightarrow H(x):=xN'(x)-N(x)=0 .
\]

Write  

\[
N(x)=x^{4}-S_{1}x^{3}+S_{2}x^{2}-S_{3}x+S_{4},
\]

where  

\[
\begin{aligned}
S_{1}&=18+72+98+k = 188+k,\\
S_{2}&=18\cdot72+18\cdot98+72\cdot98+ (18+72+98)k =10116+188k,\\
S_{3}&=18\cdot72\cdot98+(18\cdot72+18\cdot98+72\cdot98)k =127008+10116k,\\
S_{4}&=18\cdot72\cdot98\cdot k =127008\,k .
\end{aligned}
\]

Since  

\[
N'(x)=4x^{3}-3S_{1}x^{2}+2S_{2}x-S_{3},
\]

we obtain  

\[
\begin{aligned}
H(x)&=xN'(x)-N(x)  \\
    &=3x^{4}-2S_{1}x^{3}+S_{2}x^{2}-S_{4}\\
    &=3x^{4}-(376+2k)x^{3}+(10116+188k)x^{2}-127008k .
\end{aligned}
\]

The three positive roots of \(H\) are the two minima of \(f\) (in the intervals where
\(f<0\)) and one interior maximum.

--------------------------------------------------------------------
### 2.  When are the two minima equal?

At a minimum \(x_{0}\),

\[
f(x_{0}) = \frac{N(x_{0})}{x_{0}} = N'(x_{0})=:m .
\]

Thus the equation \(N(x)=mx\) has a **double** root at each minimum.
If the global minimum is attained at exactly two points, then

\[
N(x)-mx=(x-\alpha)^{2}(x-\gamma)^{2},
\qquad\alpha\neq\gamma>0 .
\]

Expanding both sides and comparing coefficients gives

\[
\begin{cases}
\alpha+\gamma =\displaystyle\frac{188+k}{2},\\[4pt]
\alpha\gamma = -\dfrac{k^{2}}{8}+47k+640,\\[4pt]
(\alpha\gamma)^{2}=S_{4}=127008k .
\end{cases}
\]

Eliminating \(\alpha,\gamma\) yields a single equation for \(k\):

\[
\Bigl(-\frac{k^{2}}{8}+47k+640\Bigr)^{2}=127008k .
\]

Multiplying by \(64\),

\[
(k^{2}-376k-5120)^{2}=8\,128\,512\,k,
\]

which simplifies to the quartic

\[
k^{4}-752k^{3}+131136k^{2}-4\,278\,272k+26\,214\,400=0 .
\]

--------------------------------------------------------------------
### 3.  Solving the quartic

Factoring gives  

\[
(k-8)(k-32)(k^{2}-712k+102400)=0,
\]

so the four real solutions are  

\[
k=8,\qquad k=32,\qquad k=200,\qquad k=512 .
\]

--------------------------------------------------------------------
### 4.  Selecting the admissible \(k\)

For the two minima to be **positive**, the product \(\alpha\gamma\) must be positive:

\[
\alpha\gamma = -\frac{k^{2}}{8}+47k+640 >0 .
\]

Evaluating this expression:

\[
\begin{array}{c|c}
k & \alpha\gamma\\ \hline
8   & 1008>0\\
32  & 2016>0\\
200 & 5040>0\\
512 & -8064<0
\end{array}
\]

Thus \(k=512\) gives one root \(\alpha\) negative and cannot produce two minima on the
positive axis.  The admissible values are  

\[
k=8,\;32,\;200 .
\]

--------------------------------------------------------------------
### 5.  Sum of the three values

\[
8+32+200 = 240 .
\]

--------------------------------------------------------------------
\[
\boxed{240}
\]

The inequalities can be rewritten by factoring:

\[
\begin{aligned}
x-yz<y-zx &\iff (x-y)(1+z)<0,\\[2pt]
y-zx<z-xy &\iff (y-z)(1+x)<0 .
\end{aligned}
\]

Hence the signs of \(x-y,\;y-z,\;1+z,\;1+x\) must be opposite in each product.
Four sign‑patterns are possible:

\[
\begin{array}{c|c|c|c}
x+1 & z+1 & \text{implied order of }x,y,z & \text{region}\\ \hline
>0 & >0 & x<y<z & \text{bounded}\\
>0 & <0 & y<\min\{x,z\} & \text{unbounded}\\
<0 & >0 & \max\{x,z\}<y & \text{unbounded}\\
<0 & <0 & x>y>z & \text{impossible (sum }75)
\end{array}
\]

Only the first case can give a finite region.  
With \(x+y+z=75\) and the inequalities

\[
x>-1,\qquad x<y,\qquad y<z,
\]

write \(z=75-x-y\). The conditions become

\[
\begin{cases}
x>-1,\\
x<y,\\
y<\dfrac{75-x}{2}.
\end{cases}
\]

In the \((x,y)\)-plane this is the interior of the triangle bounded by

\[
x=-1,\qquad y=x,\qquad y=\frac{75-x}{2},
\]

whose vertices are \((-1,-1),\;(-1,38),\;(25,25)\).

Its area in the \((x,y)\)-plane is

\[
\frac12\cdot 39\cdot 26=507 .
\]

The plane \(x+y+z=75\) makes an angle with the \(xy\)-plane whose normal is \((1,1,1)\); the area element is multiplied by \(\lVert(1,1,1)\rVert=\sqrt3\).  
Hence the actual area of the bounded region is

\[
507\sqrt3 .
\]

Thus \(a=507,\;b=3\) and

\[
a+b=507+3=510 .
\]

\[
\boxed{510}
\]

The grid consists of three \(3\times 3\) blocks.  
For each number \(1,\dots ,9\) we must place it

* once in each of the three rows,
* once in each of the three blocks.

Hence for a fixed number its positions are described by a permutation
\(\sigma\in S_{3}\):
\[
\sigma(i)=\text{the block in which the number occurs in row }i .
\]

---

### 1.  Counting the assignments of permutations

Let \(n_\sigma\) be the number of symbols that use permutation \(\sigma\).
Since there are nine symbols,
\[
\sum_{\sigma\in S_3} n_\sigma = 9 .
\]

For each ordered pair \((i,j)\) (row \(i\), block \(j\)) the number of symbols
with \(\sigma(i)=j\) must be \(3\) (there are three columns in that block).  
Writing the six permutations of \(S_3\) as  

\[
\begin{aligned}
\sigma_1&:(1,2,3),&\sigma_2&:(2,1,3),&\sigma_3&:(3,2,1),\\
\sigma_4&:(1,3,2),&\sigma_5&:(2,3,1),&\sigma_6&:(3,1,2),
\end{aligned}
\]

the conditions become  

\[
\begin{array}{c}
n_1+n_4=3,\; n_1+n_2=3,\; n_1+n_3=3,\\
n_2+n_5=3,\; n_3+n_6=3,\; n_4+n_5=3,\\
n_3+n_5=3,\; n_4+n_6=3,\; n_2+n_6=3 .
\end{array}
\]

Solving gives a one‑parameter family

\[
n_1=n_5=n_6=k,\qquad n_2=n_3=n_4=3-k,
\]

with \(k=0,1,2,3\) (non‑negative integers).  
For each \(k\) the number of ways to assign the nine distinct symbols to the
permutations is  

\[
\frac{9!}{k!^3\,(3-k)!^3}.
\]

Hence the total number of macro‑assignments is

\[
M=\sum_{k=0}^{3}\frac{9!}{k!^3(3-k)!^3}
   =1680+45360+45360+1680=94080 .
\]

---

### 2.  Ordering within each macro‑cell

A macro‑cell \((\text{row }i,\text{ block }j)\) contains exactly three
different numbers.  Their order in the three columns of that block can be chosen
in \(3!\) ways, independently for each of the \(3\times 3=9\) macro‑cells.
Thus we obtain an extra factor \((3!)^{9}=6^{9}=10\,077\,696\).

Hence the total number of fillings is  

\[
N = M\cdot (3!)^{9}=94080\cdot 6^{9}
   = 2^{16}\cdot 3^{10}\cdot5\cdot7^{2}.
\]

---

### 3.  Computing \(p a+q b+r c+s d\)

Writing \(N=p^{a}q^{b}r^{c}s^{d}\) with distinct primes \(p,q,r,s\) we have  

\[
p=2,\;a=16;\qquad q=3,\;b=10;\qquad r=5,\;c=1;\qquad s=7,\;d=2.
\]

Therefore  

\[
p a+q b+r c+s d
=2\cdot16+3\cdot10+5\cdot1+7\cdot2
=32+30+5+14=81.
\]

\[
\boxed{81}
\]

Let the line containing \(A,B,C,D,E,F\) be the \(x\)-axis.  
Place \(A\) at the origin: \(A=(0,0)\).

---

### 1.  Determine the coordinates of the collinear points  

\[
\begin{aligned}
AC&=26\quad\Rightarrow\quad C=(26,0),\\[2mm]
BD&=22\quad\Rightarrow\quad D-B=22,\\[2mm]
CE&=31\quad\Rightarrow\quad E-C=31\;\Rightarrow\;E=(57,0),\\[2mm]
DF&=33\quad\Rightarrow\quad F-D=33,\\[2mm]
AF&=73\quad\Rightarrow\quad F=(73,0).
\end{aligned}
\]

From \(F-D=33\) we get \(D=73-33=40\).  
Then \(BD=22\) gives \(B=40-22=18\).

Thus  

\[
A(0,0),\;B(18,0),\;C(26,0),\;D(40,0),\;E(57,0),\;F(73,0).
\]

---

### 2.  Locate point \(G\)

Let \(G=(x,y)\) with \(y\neq0\).  
Given \(CG=40\) and \(DG=30\),

\[
\begin{cases}
(x-26)^2+y^2 = 40^2 =1600,\\
(x-40)^2+y^2 = 30^2 = 900 .
\end{cases}
\]

Subtract the second equation from the first:

\[
(x-26)^2-(x-40)^2 = 700
\]
\[
\Longrightarrow\;28x-924 = 700 \;\Longrightarrow\; x=58 .
\]

Plug \(x=58\) into \((x-40)^2+y^2=900\):

\[
(18)^2 + y^2 = 900 \;\Longrightarrow\; y^2 = 576 \;\Longrightarrow\; y = \pm24 .
\]

Take \(y=24\) (the sign does not affect the area).  
Hence \(G=(58,24)\).

---

### 3.  Area of \(\triangle BGE\)

Points \(B\) and \(E\) lie on the \(x\)-axis, so segment \(\overline{BE}\) is horizontal.  

\[
BE = 57-18 = 39 .
\]

The perpendicular distance from \(G\) to line \(BE\) (the \(x\)-axis) is \(|y| = 24\).

Therefore

\[
\text{Area} = \frac12 \times (\text{base}) \times (\text{height})
           = \frac12 \times 39 \times 24
           = 468 .
\]

\[
\boxed{468}
\]

Let  

\[
d=n+2\qquad (d\ge 3,\;n\in\mathbb Z^+)
\]

Then  

\[
n+3=d+1,\qquad n^{2}+9=(d-2)^2+9=d^{2}-4d+13 .
\]

The condition “\(n+2\) divides \(3(n+3)(n^{2}+9)\)” becomes  

\[
d\mid 3(d+1)(d^{2}-4d+13).
\]

Consider the right‑hand side modulo \(d\):

\[
\begin{aligned}
3(d+1)(d^{2}-4d+13)
&\equiv 3\cdot1\cdot13\pmod{d}\\
&=39\pmod{d}.
\end{aligned}
\]

Hence  

\[
d\mid 3(d+1)(d^{2}-4d+13)\iff d\mid 39 .
\]

Since \(d=n+2\ge 3\), the possible values of \(d\) are the divisors of \(39\) that are at least \(3\):

\[
d\in\{3,13,39\}.
\]

Corresponding \(n=d-2\) values are  

\[
n=1,\;11,\;37 .
\]

All three satisfy the original divisibility (direct substitution confirms).

The required sum is  

\[
1+11+37=49.
\]

\[
\boxed{49}
\]

The function  

\[
f(x)=\begin{cases}
x,&-1\le x<1\\[2mm]
2-x,&1\le x<3
\end{cases},\qquad f(x+4)=f(x)
\]

has period \(4\).  
For any \(x\) write  

\[
x=4n+t,\qquad t\in[-1,3),\;n\in\mathbb Z .
\]

Then  

\[
f(x)=\begin{cases}
t,&-1\le t<1\\
2-t,&1\le t<3 .
\end{cases}
\tag{1}
\]

The parabola is \(x=34y^{2}\;(x\ge 0)\).  
An intersection point must satisfy  

\[
y=f(34y^{2}) .
\tag{2}
\]

Put \(x=34y^{2}=4n+t\) with \(t\in[-1,3)\).  
From (1) there are two possibilities.

---

### 1.  \(t=y\)  

Then \(-1\le y<1\) and  

\[
34y^{2}=4n+y\Longrightarrow 34y^{2}-y=4n .
\tag{3}
\]

For each integer \(n\) this quadratic gives the two solutions  

\[
y=\frac{1\pm\sqrt{1+544n}}{68}.
\tag{4}
\]

Since \(y\in[-1,1)\) the solutions are admissible for every \(n\)
for which \(34y^{2}\le 34\).  
Because \(0\le34y^{2}\le34\), from \(34y^{2}=4n+t\) with \(t\ge-1\) we get
\(0\le4n+3\), i.e. \(n\ge0\); and from \(4n-1\le34\) we obtain \(n\le8\).
Thus \(n=0,1,\dots ,8\).  

For each \(n\) the sum of the two roots of (3) is  

\[
\frac{1}{34}.
\]

Hence the total contribution of this case is  

\[
9\cdot\frac1{34}=\frac{9}{34}=\frac{18}{68}.
\tag{5}
\]

---

### 2.  \(t=2-y\)  

Now \(1\le t<3\) which gives \(-1<y\le1\) and  

\[
34y^{2}=4n+2-y\Longrightarrow 34y^{2}+y-2=4n .
\tag{6}
\]

Its solutions are  

\[
y=\frac{-1\pm\sqrt{273+544n}}{68}.
\tag{7}
\]

Again \(n=0,1,\dots ,8\).  
For \(n=0,\dots ,7\) both roots lie in \((-1,1]\); for \(n=8\) only the
‘\(+\)’ root is admissible (the ‘\(-\)’ root is \(<-1\)).  

The sum of the two roots of (6) for any \(n\) equals  

\[
-\frac1{34}.
\]

Thus the sum of all valid roots of this case is  

\[
8\!\left(-\frac1{34}\right)+\frac{-1+\sqrt{4625}}{68}
   =-\frac{16}{68}+\frac{-1+\sqrt{4625}}{68}
   =\frac{-17+\sqrt{4625}}{68}.
\tag{8}
\]

---

### 3.  Total sum of the \(y\)–coordinates

Adding (5) and (8),

\[
S=\frac{18}{68}+\frac{-17+\sqrt{4625}}{68}
   =\frac{1+\sqrt{4625}}{68}
   =\frac{1+5\sqrt{185}}{68}.
\]

The expression \(\displaystyle\frac{a+b\sqrt{c}}{d}\) has  

\[
a=1,\;b=5,\;c=185,\;d=68,
\]

with \(\gcd(a,b,d)=1\) and \(c\) square‑free.

Hence  

\[
a+b+c+d = 1+5+185+68 = 259.
\]

\[
\boxed{259}
\]

\[
\prod_{k=4}^{63}\frac{\log_k\!\left(5^{k^{2}-1}\right)}{\log_{k+1}\!\left(5^{k^{2}-4}\right)}
= \prod_{k=4}^{63}
\frac{(k^{2}-1)\ln5/\ln k}{(k^{2}-4)\ln5/\ln(k+1)}
= \prod_{k=4}^{63}\frac{k^{2}-1}{k^{2}-4}\cdot\frac{\ln(k+1)}{\ln k}
\]

Since \(k^{2}-1=(k-1)(k+1)\) and \(k^{2}-4=(k-2)(k+2)\),

\[
\frac{k^{2}-1}{k^{2}-4}= \frac{k-1}{k-2}\cdot\frac{k+1}{k+2}.
\]

Thus

\[
P=\Bigl(\prod_{k=4}^{63}\frac{k-1}{k-2}\Bigr)
   \Bigl(\prod_{k=4}^{63}\frac{k+1}{k+2}\Bigr)
   \Bigl(\prod_{k=4}^{63}\frac{\ln(k+1)}{\ln k}\Bigr).
\]

Each product telescopes:

1. \(\displaystyle\prod_{k=4}^{63}\frac{k-1}{k-2}
   =\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{62}{61}
   =\frac{62}{2}=31.\)

2. \(\displaystyle\prod_{k=4}^{63}\frac{k+1}{k+2}
   =\frac{5}{6}\cdot\frac{6}{7}\cdots\frac{64}{65}
   =\frac{5}{65}=\frac1{13}.\)

3. \(\displaystyle\prod_{k=4}^{63}\frac{\ln(k+1)}{\ln k}
   =\frac{\ln5}{\ln4}\cdot\frac{\ln6}{\ln5}\cdots\frac{\ln64}{\ln63}
   =\frac{\ln64}{\ln4}
   =\log_{4}64=3.\)

Multiplying,

\[
P=31\cdot\frac1{13}\cdot3=\frac{93}{13}.
\]

Thus \(m=93,\;n=13\) (relatively prime) and  

\[
m+n=93+13=106.
\]

\[
\boxed{106}
\]

Label the unit line segments as follows.

\[
\begin{array}{c}
\text{Top edges: }T_1,T_2\\
\text{Bottom edges: }B_1,B_2\\
\text{Left edges: }L_1,L_2\\
\text{Right edges: }R_1,R_2\\
\text{Middle vertical edges: }M_{v1},M_{v2}\\
\text{Middle horizontal edges: }M_{h1},M_{h2}
\end{array}
\]

Each of the four unit squares must have exactly two red sides.
Let a red edge be denoted by \(1\) and a blue edge by \(0\).
Introduce the binary variables  

\[
\begin{aligned}
&a=M_{v1},\;b=M_{v2},\;c=M_{h1},\;d=M_{h2},\\
&p=T_1,\;q=T_2,\;r=B_1,\;s=B_2,\\
&u=L_1,\;v=L_2,\;w=R_1,\;x=R_2 .
\end{aligned}
\]

The condition “two red sides” for each of the four squares gives four linear
equations:

\[
\begin{cases}
p+u+a+c=2\\
q+w+a+d=2\\
r+v+b+c=2\\
s+x+b+d=2\qquad (1)
\end{cases}
\]

All variables are \(0\) or \(1\).

--------------------------------------------------------------------
### Fix the interior edges

The interior edges are the four variables \(a,b,c,d\).
For a fixed choice of \((a,b,c,d)\) each equation in (1) involves a
different pair of outer edges:

\[
\begin{aligned}
p+u &=2-(a+c),\\
q+w &=2-(a+d),\\
r+v &=2-(b+c),\\
s+x &=2-(b+d).
\end{aligned}
\tag{2}
\]

For a pair of binary variables the number of possibilities for a given
sum is  

\[
f(s)=\begin{cases}
1 & s=0\text{ or }2,\\[2pt]
2 & s=1,\\[2pt]
0 & \text{otherwise}.
\end{cases}
\]

Hence, for a fixed interior assignment the number of ways to colour the
outer edges equals  

\[
f\bigl(2-(a+c)\bigr)\,
f\bigl(2-(a+d)\bigr)\,
f\bigl(2-(b+c)\bigr)\,
f\bigl(2-(b+d)\bigr).
\]

Because \(a,c,b,d\in\{0,1\}\), each sum \(a+c,\;a+d,\;b+c,\;b+d\) is
\(0,1,\) or \(2\); consequently the factor is \(2\) exactly when the
corresponding pair contains one \(0\) and one \(1\), and it is \(1\) when
the pair is equal.  

Define

\[
N(a,b,c,d)=\bigl[ a\ne c\bigr]+\bigl[ a\ne d\bigr]+\bigl[ b\ne c\bigr]
+\bigl[ b\ne d\bigr].
\]

Then the number of outer colourings for that interior choice is
\(2^{\,N(a,b,c,d)}\).

--------------------------------------------------------------------
### Counting all interior choices

The four interior bits can be viewed as two binary strings  
\(A=(a,b)\) and \(C=(c,d)\).  The quantity \(N\) counts the mismatches
between \(A\) and \(C\) over all four cross‑pairs, and one obtains

\[
N=2\bigl(w_A+w_C-w_Aw_C\bigr),
\]
where \(w_A=a+b\) and \(w_C=c+d\) are the numbers of 1’s in the strings.

From this we get the possible values:

* **\(N=0\)** when \(A=C\) (all four interior edges equal).  
  Two assignments: \((0,0,0,0)\) and \((1,1,1,1)\).  
  Contribution: \(2\cdot2^{0}=2\).

* **\(N=4\)** when the two strings are opposite, i.e. \(A=(0,0), C=(1,1)\)
  or \(A=(1,1), C=(0,0)\).  
  Two assignments.  
  Contribution: \(2\cdot2^{4}=32\).

* **\(N=2\)** for all remaining \(16-4=12\) interior assignments.  
  Contribution: \(12\cdot2^{2}=48\).

Adding them gives the total number of colourings

\[
2+48+32 = 82.
\]

--------------------------------------------------------------------
\[
\boxed{82}
\]

**Solution Overview**

Place \(\triangle ABC\) in the plane with  

\[
A=(0,0),\qquad B=(28,0),\qquad C=(x,y)
\]

so that \(AB=28\).  
The points on the sides are

\[
\begin{aligned}
D&=A+\frac17AB=(4,0), &
E&=A+\frac57AB=(20,0),\\
F&=A+\frac17AC=\Bigl(\frac{x}{7},\frac{y}{7}\Bigr), &
G&=A+\frac57AC=\Bigl(\frac{5x}{7},\frac{5y}{7}\Bigr).
\end{aligned}
\]

(The given lengths give the fractions \(\frac17,\frac57\) on both sides because  

\[
AD:DE:EB=4:16:8=1:4:2,\qquad 
AF:FG:GC=13:52:26=1:4:2 .
\])

The reflections are

\[
\begin{aligned}
M&=2F-D=\Bigl(\frac{2x}{7}-4,\frac{2y}{7}\Bigr),\\[2mm]
N&=2E-G=\Bigl(40-\frac{5x}{7},-\frac{5y}{7}\Bigr).
\end{aligned}
\]

--------------------------------------------------------------------
### 1.  Height of the triangle

The quadrilateral \(DEGF\) is \(\{D,E,G,F\}\) in that order.  
Using the shoelace formula,

\[
\begin{aligned}
[DEGF]
&=\frac12\Bigl((4\cdot0+20\cdot\frac{5y}{7}+\frac{5x}{7}\cdot\frac{y}{7}+ \frac{x}{7}\cdot0) \\
&\qquad\ -\ (0\cdot20+\;0\cdot\frac{5x}{7}+\frac{5y}{7}\cdot\frac{x}{7}+ \frac{y}{7}\cdot4 )\Bigr)\\[1mm]
&=\frac12\Bigl(20\cdot\frac{5y}{7}-4\cdot\frac{y}{7}\Bigr)
 =\frac{1248}{91}\,\frac{y}{2}= \frac{624}{91}\,y .
\end{aligned}
\]

The problem states \([DEGF]=288\); hence

\[
\frac{624}{91}\,y=288\qquad\Longrightarrow\qquad y=42 .
\]

Thus the altitude from \(C\) to \(\overline{AB}\) is \(h=y=42\).

--------------------------------------------------------------------
### 2.  Area of \(\triangle ABC\)

\[
[ABC]=\frac12\cdot AB\cdot h=\frac12\cdot 28\cdot 42=14\cdot 42=588.
\]

--------------------------------------------------------------------
### 3.  Area of the heptagon \(AFNBCEM\)

Write the vertices in order:

\[
A,\;F,\;N,\;B,\;C,\;E,\;M .
\]

Using the coordinates found above (with \(y=42\) and \(\,x=\sqrt{91^{2}-42^{2}}=7\sqrt{133}\)), the
shoelace sum is

\[
\begin{aligned}
S&=x_Ay_F-y_Ax_F
   +x_Fy_N-y_Fx_N
   +x_Ny_B-y_Nx_B
   +x_By_C-y_Bx_C
   +x_Cy_E-y_Cx_E
   +x_Ey_M-y_Ex_M
   +x_My_A-y_Mx_A \\[1mm]
&=0-240+840+1176-840+240+0=1176 .
\end{aligned}
\]

Hence

\[
[AFNBCEM]=\frac12\,S=\frac12\cdot1176=588 .
\]

A vector‑geometry check also shows that the signed sum of the
determinants of successive vertices equals \(\det(\overrightarrow{AB},
\overrightarrow{AC})\), i.e. the area of \(\triangle ABC\).

--------------------------------------------------------------------
### 4.  Answer

\[
\boxed{588}
\]

Factor  
\[
2025=3^{4}\cdot5^{2}.
\]

A divisor of \(2025\) is \(3^{i}5^{j}\) with \(0\le i\le4,\;0\le j\le2\).  
Hence  

\[
|A|=(4+1)(2+1)=15 .
\]

---

### When does a subset have lcm \(2025\)?

The lcm of a set of divisors takes, for each prime, the **maximum** exponent appearing among the chosen numbers.

Thus a subset \(B\subseteq A\) has \(\operatorname{lcm}(B)=3^{4}5^{2}=2025\) iff  

* at least one element of \(B\) has exponent \(4\) of the prime \(3\);
* at least one element of \(B\) has exponent \(2\) of the prime \(5\).

Define  

\[
S_{3}=\{3^{4}5^{j}\mid j=0,1,2\}=\{81,405,2025\},
\qquad
S_{5}=\{3^{i}5^{2}\mid i=0,1,2,3,4\}
 =\{25,75,225,675,2025\}.
\]

\(|S_{3}|=3,\ |S_{5}|=5,\) and \(|S_{3}\cap S_{5}|=1\) (the element \(2025\)).

So we need subsets that intersect **both** \(S_{3}\) and \(S_{5}\).

---

### Counting such subsets

Total subsets of \(A\): \(2^{15}\).

Let  

* \(A_{3}\) = subsets containing **no** element of \(S_{3}\).  
  \(|A_{3}| = 2^{15-3}=2^{12}\).

* \(A_{5}\) = subsets containing **no** element of \(S_{5}\).  
  \(|A_{5}| = 2^{15-5}=2^{10}\).

* \(A_{3}\cap A_{5}\) = subsets containing none of \(S_{3}\cup S_{5}\).  
  \(|S_{3}\cup S_{5}| = 3+5-1 = 7\), so  
  \(|A_{3}\cap A_{5}| = 2^{15-7}=2^{8}\).

By inclusion–exclusion, the number of subsets intersecting both \(S_{3}\) and \(S_{5}\) is  

\[
2^{15} - 2^{12} - 2^{10} + 2^{8}
 = 32768 - 4096 - 1024 + 256
 = 27904.
\]

Every such subset is non‑empty, so these are exactly the favorable subsets.

---

### Probability

\[
\text{Probability}
 = \frac{27904}{2^{15}}
 = \frac{27904}{32768}
 = \frac{109}{128}
\]
(after reducing the fraction).

Thus \(m=109,\ n=128\) and  

\[
m+n = 109+128 = 237.
\]

\[
\boxed{237}
\]

Let  

* \(O\) be the centre of \(\omega _2\) (radius \(15\)),
* \(A\) be the centre of \(\omega _1\) (radius \(6\)),
* \(B\) the point where the two circles are internally tangent,
* \(C\) the antipode of \(B\) on \(\omega_2\) (\(BC\) is a diameter),
* \(D\) the point on \(\omega _2\) with \(AD\perp BC\).

--------------------------------------------------------------------
### 1.  Choose coordinates  

Place \(O\) at the origin and let the diameter \(BC\) be horizontal.

\[
B=(15,0),\qquad C=(-15,0)
\]

Since \(OA=15-6=9\) and \(A\) lies on \(OB\),

\[
A=(9,0).
\]

The line \(AD\) is perpendicular to the horizontal line \(BC\), so \(AD\) is the
vertical line through \(A\); intersecting \(\omega_2\) gives

\[
D=(9,12)\quad\text{or}\quad D=(9,-12).
\]

Because the rectangle will be placed above the centre of \(\omega_1\), we take  

\[
D=(9,12).
\]

--------------------------------------------------------------------
### 2.  Describe the rectangle  

Since the rectangle is **inscribed** in \(\omega_1\), all four of its vertices lie on
\(\omega_1\).  
Let the rectangle have centre at \(A\); because a rectangle’s circum‑centre is
the intersection of its diagonals, the centre must be the centre of the
circumscribing circle \(\omega_1\).

Let  

* half‑width \(w\) (distance from the centre to the right–hand side),
* half‑height \(h\) (distance from the centre to the top side).

Then  

\[
\begin{aligned}
x_R&=9+w, & x_L&=9-w,\\
y_T&=h,   & y_B&=-h .
\end{aligned}
\]

The vertices are  

\[
\begin{aligned}
E&(x_R,y_B)=(9+w,-h),\\
F&(x_R,y_T)=(9+w, h),\\
G&(x_L,y_T)=(9-w, h),\\
H&(x_L,y_B)=(9-w,-h).
\end{aligned}
\]

Because the vertices lie on \(\omega_1\) (radius \(6\)),

\[
w^{2}+h^{2}=6^{2}=36\tag{1}
\]

--------------------------------------------------------------------
### 3.  Use the area condition  

The problem states that  

\[
\text{Area}\,\triangle DGF=\text{Area}\,\triangle CHG .
\]

*Triangle \(DGF\).*  
Its base is \(GF\) (the top side of the rectangle) of length \(2w\).  
Its height is the vertical distance from \(D\) to the line \(y=h\):
\(\;12-h\). Hence  

\[
[\triangle DGF]=\tfrac12(2w)(12-h)=w(12-h).\tag{2}
\]

*Triangle \(CHG\).*  
Its base is \(HG\) (the left side) of length \(2h\).  
The distance from \(C\) to the vertical line \(x=9-w\) is
\((9-w)-(-15)=24-w\). Therefore  

\[
[\triangle CHG]=\tfrac12(2h)(24-w)=h(24-w).\tag{3}
\]

Equating (2) and (3),

\[
w(12-h)=h(24-w).\tag{4}
\]

Expanding both sides gives  

\[
12w-wh=24h-wh\Longrightarrow 12w=24h\Longrightarrow w=2h.\tag{5}
\]

--------------------------------------------------------------------
### 4.  Solve for \(w\) and \(h\)

Substituting \(w=2h\) into (1):

\[
(2h)^2+h^2=36\quad\Longrightarrow\quad5h^{2}=36\quad\Longrightarrow\quad
h^{2}= \frac{36}{5},\qquad
h=\frac{6}{\sqrt5}=\frac{6\sqrt5}{5}.
\]

Hence  

\[
w=2h=\frac{12}{\sqrt5}= \frac{12\sqrt5}{5}.
\]

--------------------------------------------------------------------
### 5.  Area of the rectangle  

The rectangle’s side lengths are \(2w\) and \(2h\), so its area is  

\[
\text{Area}= (2w)(2h)=4wh=4\cdot(2h)h=8h^{2}
          =8\left(\frac{36}{5}\right)=\frac{288}{5}.
\]

Thus the area is \(\displaystyle \frac{288}{5}\).
The numerator and denominator are relatively prime, so  

\[
m=288,\qquad n=5,\qquad m+n=288+5=293.
\]

--------------------------------------------------------------------
\[
\boxed{293}
\]

**Step 1.  Regions created by chords**

If we start with a disc (one region) and draw a chord, the number of regions increases by  

\[
\text{(number of interior intersections created by the new chord)}+1 .
\]

Hence after drawing \(k\) chords the number of regions is  

\[
R=1+k+I ,
\]

where \(I\) is the total number of interior intersection points of the drawn chords (no three chords meet at a single interior point, which holds a.s. for random chords).

In our problem \(k=27\) (the two diameters plus the 25 random chords), so  

\[
R=28+I .
\]

Thus we only need the expected value of \(I\).

--------------------------------------------------------------------

**Step 2.  Intersections involving the two fixed diameters**

The two diameters intersect at the centre, contributing a deterministic intersection  

\[
I_{\text{centre}}=1 .
\]

A random chord intersects the **vertical** diameter iff its endpoints lie in opposite halves of the disc, i.e. one endpoint in \(\{Q_1,Q_4\}\) and the other in \(\{Q_2,Q_3\}\).  
Among the six possible unordered quadrant‑pairs for a chord, four satisfy this condition, so

\[
P(\text{vertical intersection})=\frac{4}{6}=\frac23 .
\]

The same reasoning holds for the **horizontal** diameter, giving  

\[
P(\text{horizontal intersection})=\frac23 .
\]

Hence for the 25 random chords

\[
E[I_{\text{vert}}]=25\cdot\frac23=\frac{50}{3},\qquad   
E[I_{\text{horiz}}]=25\cdot\frac23=\frac{50}{3}.
\]

--------------------------------------------------------------------

**Step 3.  Intersections among the 25 random chords**

Two random chords are independent.  
Each chord chooses an unordered pair of distinct quadrants uniformly from the six possibilities  

\[
\{1,2\},\{2,3\},\{3,4\},\{4,1\} \ (\text{adjacent}),\qquad
\{1,3\},\{2,4\}\ (\text{opposite}).
\]

Let the unordered pairs be \(S\) and \(T\).  
Their intersection size can be

* **2** (the same pair) – probability \(1/6\);
* **1** (they share exactly one quadrant) – probability \(2/3\);
* **0** (disjoint) – probability \(1/6\).

The disjoint case splits further:

* Adjacent–adjacent (e.g. \(\{1,2\}\) with \(\{3,4\}\)) – chords never intersect.
* Opposite–opposite (e.g. \(\{1,3\}\) with \(\{2,4\}\)) – chords always intersect.

Among the three unordered disjoint pairs, two are adjacent–adjacent and one is opposite–opposite; hence for ordered pairs

\[
P(\text{disjoint, intersect})=\frac{2}{36}=\frac1{18},\qquad
P(\text{disjoint, no intersect})=\frac{4}{36}.
\]

For the other two situations one can check (by considering the relative order of the points inside the quadrants) that

* If the two chords have the **same** unordered pair (adjacent or opposite), the probability they intersect is \(\tfrac12\).
* If they **share exactly one quadrant**, the probability they intersect is also \(\tfrac12\).

Collecting the contributions,

\[
\begin{aligned}
P(\text{intersection}) &=
\underbrace{\frac{6}{36}\cdot\frac12}_{\text{same pair}}
+\underbrace{\frac{24}{36}\cdot\frac12}_{\text{share one quadrant}}
+\underbrace{\frac{2}{36}\cdot1}_{\text{disjoint opposite}}\\[2mm]
&=\frac{3}{36}+\frac{12}{36}+\frac{2}{36}
=\frac{17}{36}.
\end{aligned}
\]

Thus for any unordered pair of the 25 random chords

\[
p:=P(\text{they intersect})=\frac{17}{36}.
\]

There are \(\binom{25}{2}=300\) such pairs, so

\[
E[I_{\text{random}}]=300\cdot\frac{17}{36}
=\frac{300}{36}\cdot17
=\frac{25}{3}\cdot17
=\frac{425}{3}.
\]

--------------------------------------------------------------------

**Step 4.  Expected total number of interior intersections**

\[
\begin{aligned}
E[I] &= I_{\text{centre}}
      +E[I_{\text{vert}}]
      +E[I_{\text{horiz}}]
      +E[I_{\text{random}}]  \\[2mm]
     &=1+\frac{50}{3}+\frac{50}{3}+\frac{425}{3} \\[2mm]
     &=1+\frac{525}{3}=1+175=176 .
\end{aligned}
\]

--------------------------------------------------------------------

**Step 5.  Expected number of regions**

\[
E[R]=28+E[I]=28+176=204 .
\]

--------------------------------------------------------------------

\[
\boxed{204}
\]

**Solution Outline**

1.  Transform the recurrence for \(x_k\) into a simpler one.

   Define  

   \[
   w_k=\frac{x_k-1}{x_k+1}\quad\Longleftrightarrow\quad 
   x_k=\frac{1+w_k}{1-w_k}.
   \]

   Substituting in  

   \[
   x_{k+1}= \frac13\Bigl(x_k+\frac1{x_k}-1\Bigr)
   \]

   gives after simplification  

   \[
   w_{k+1}= \frac{3w_k^2-1}{2}. \tag{1}
   \]

   The initial value is  

   \[
   w_1=\frac{25/11-1}{25/11+1}=\frac{7}{18}.
   \]

2.  Write \(w_k=\dfrac{a_k}{b_k}\) in lowest terms
   (\(a_k\) odd, \(b_k\) even).  

   From (1),

   \[
   w_{k+1}= \frac{3a_k^2-b_k^2}{2b_k^2}.
   \]

   Because \(a_k\) is not divisible by 3, the numerator is always
   \(\equiv 3\pmod 9\); hence it contains exactly one factor 3.
   After canceling this factor we obtain

   \[
   a_{k+1}= \frac{3a_k^2-b_k^2}{3},\qquad
   b_{k+1}= \frac{2b_k^2}{3}.
   \tag{2}
   \]

   Consequently  

   \[
   b_{k+1}= \frac{2}{3}b_k^{\,2}. \tag{3}
   \]

3.  Determine a closed form for \(b_k\).

   Write \(b_k=2^{\alpha_k}3^{\beta_k}\).
   From (3)

   \[
   2^{\alpha_{k+1}}3^{\beta_{k+1}}
   =\frac{2}{3}\bigl(2^{\alpha_k}3^{\beta_k}\bigr)^2
   =2^{2\alpha_k+1}3^{2\beta_k-1},
   \]

   whence  

   \[
   \alpha_{k+1}=2\alpha_k+1,\qquad
   \beta_{k+1}=2\beta_k-1,
   \]
   with \(\alpha_1=1,\ \beta_1=2\).

   Solving these linear recurrences gives  

   \[
   \alpha_k = 2^{\,k}-1,\qquad
   \beta_k = 2^{\,k-1}+1 .
   \]

   Therefore  

   \[
   b_k = 2^{\,2^{k}-1}\,3^{\,2^{k-1}+1}. \tag{4}
   \]

4.  Recover \(x_k\) from \(w_k\).

   Since \(x_k=\dfrac{1+w_k}{1-w_k}\) and \(w_k=\dfrac{a_k}{b_k}\),

   \[
   x_k=\frac{b_k+a_k}{\,b_k-a_k\,}= \frac{m_k}{n_k},
   \]
   where \(m_k=b_k+a_k,\ n_k=b_k-a_k\).  
   Because \(a_k\) is odd and \(b_k\) is even, \(m_k\) and \(n_k\) are odd,
   so \(\gcd(m_k,n_k)=1\).

   Hence

   \[
   m_k+n_k = (b_k+a_k)+(b_k-a_k)=2b_k.
   \]

5.  Compute \(m_{2025}+n_{2025}\pmod{1000}\).

   From (4),

   \[
   b_{2025}=2^{\,2^{2025}-1}\,3^{\,2^{2024}+1},
   \qquad
   m_{2025}+n_{2025}=2b_{2025}
   =2^{\,2^{2025}}\,3^{\,2^{2024}+1}.
   \]

   We need this value modulo \(1000=2^3\cdot5^3\).

   *Modulo 8:* the exponent of 2 is at least 3, so the whole product
   is \(0\pmod8\).

   *Modulo 125:* use Euler’s theorem (\(\phi(125)=100\)):
   \[
   2^{2^{2025}}\equiv2^{\,32}\pmod{125},\qquad
   3^{2^{2024}+1}\equiv3^{\,17}\pmod{125},
   \]
   because \(2^{2025}\equiv32\) and \(2^{2024}+1\equiv17\pmod{100}\).

   Computing,
   \[
   2^{32}\equiv46,\qquad 3^{17}\equiv38\pmod{125},
   \]
   so
   \[
   2^{2^{2025}}3^{2^{2024}+1}\equiv46\cdot38\equiv123\pmod{125}.
   \]

   Solve the simultaneous congruences
   \[
   X\equiv0\pmod8,\qquad X\equiv123\pmod{125}.
   \]
   Writing \(X=125a+123\) and using \(125\equiv5\pmod8\) gives
   \(5a+3\equiv0\pmod8\Rightarrow a\equiv1\pmod8\).
   Hence \(X=125\cdot1+123=248\).

   Therefore  

   \[
   m_{2025}+n_{2025}\equiv\boxed{248}\pmod{1000}.
   \]

\[
f(x)=\sin\bigl(7\pi\sin (5x)\bigr)
\]

The zeros of \(f\) occur when the inner sine gives an integer multiple of \(\pi\):

\[
\sin\bigl(7\pi\sin (5x)\bigr)=0\Longleftrightarrow 7\pi\sin(5x)=k\pi
\Longleftrightarrow \sin(5x)=\frac{k}{7},
\qquad k\in\mathbb Z .
\]

Because \(\sin(5x)\in[-1,1]\), \(|k|\le7\); thus \(k=-7,-6,\dots ,6,7\).

---

### 1.  Counting the zeros  

Set \(\theta =5x\).  
When \(x\in(0,2\pi)\) we have \(\theta\in(0,10\pi)\).

For each admissible integer \(k\) we need the number of solutions of  

\[
\sin\theta=\frac{k}{7},\qquad \theta\in(0,10\pi).
\]

* **\(k=0\):** \(\sin\theta=0\Rightarrow\theta=n\pi\).  
  Inside \((0,10\pi)\) we have \(n=1,2,\ldots ,9\) → **9 solutions**.

* **\(k=\pm 7\):** \(\sin\theta=\pm1\).  
  \(\sin\theta=1\) gives \(\theta=\dfrac{\pi}{2}+2\pi n\);  
  \(\sin\theta=-1\) gives \(\theta=\dfrac{3\pi}{2}+2\pi n\).  
  For \(0<\theta<10\pi\) each yields \(n=0,1,2,3,4\) → **5 solutions** for \(k=7\) and **5** for \(k=-7\).

* **\(k=\pm1,\pm2,\dots ,\pm6\):** \(|k|<7\) so \(-1<k/7<1\) and \(\neq0\).  
  In each full period \([0,2\pi]\) the equation \(\sin\theta=c\) (\(|c|<1\)) has two solutions.  
  The interval \((0,10\pi)\) contains five full periods, hence **\(5\times2=10\) solutions** for each of these 12 values of \(k\).

Summing all contributions  

\[
\begin{aligned}
n &=\underbrace{9}_{k=0}
   +\underbrace{5+5}_{k=\pm7}
   +\underbrace{12\times10}_{k=\pm1,\dots,\pm6}\\[2mm]
  &=9+10+120 = 139 .
\end{aligned}
\]

Thus there are \(\boxed{139}\) zeros of \(f\) in \((0,2\pi)\).

---

### 2.  Zeros where the graph is tangent to the \(x\)-axis  

A point where the graph touches the \(x\)-axis must satisfy  

\[
f(x)=0\quad\text{and}\quad f'(x)=0.
\]

From \(f(x)=0\) we have \(\sin(5x)=k/7\) as above.  
Compute the derivative:

\[
\begin{aligned}
f'(x) &=\cos\bigl(7\pi\sin(5x)\bigr)\cdot 7\pi\cdot 5\cos(5x)   \\
      &=35\pi\cos\bigl(7\pi\sin(5x)\bigr)\cos(5x).
\end{aligned}
\]

Thus \(f'(x)=0\) iff \(\cos(5x)=0\) (the factor \(\cos\bigl(7\pi\sin(5x)\bigr)\) never vanishes at a zero, because
\(\cos(k\pi)=\pm1\)).  

\[
\cos(5x)=0\Longrightarrow 5x=\frac{\pi}{2}+n\pi\quad\Longrightarrow\quad\sin(5x)=\pm1.
\]

Hence the tangential zeros correspond exactly to the solutions of \(\sin(5x)=\pm1\), i.e. the cases \(k=\pm7\).

We already counted these:  

* \(\sin(5x)=1\) gives \(5\) solutions,
* \(\sin(5x)=-1\) gives \(5\) solutions.

Therefore  

\[
t = 5+5 = 10 .
\]

(Indeed, at these points \(f'(x)=0\) and \(f''(x)\neq0\), so the graph just touches the axis.)

---

### 3.  Required sum  

\[
n+t = 139 + 10 = \boxed{149}.
\]

Let  

\[
N=25q+r ,\qquad 0\le r<25 .
\]

The greedy algorithm first takes \(q\) coins of 25 ¢, then  
\(t=\big\lfloor r/10\big\rfloor\) coins of 10 ¢ and finally  
\(s=r-10t\) pennies.  

\[
\text{Greedy coins}=C_g(N)=q+t+s.
\]

---

### 1.  What is an optimal representation?

If a representation uses \(a\) 25‑cent coins, \(b\) 10‑cent coins and \(c\) pennies, then  

\[
25a+10b+c=N,\qquad a,b,c\ge0,
\]
and the number of coins is  

\[
C= a+b+c = N-(24a+9b).
\]

Thus **minimising the number of coins is equivalent to maximising**  

\[
P=24a+9b \qquad\text{subject to}\qquad 25a+10b\le N .
\]

The “profit’’ per cent is larger for a 25‑cent coin \((24/25)\) than for a 10‑cent coin \((9/10)\); therefore an optimal solution will use as many 25‑cent coins as possible unless removing a 25‑cent coin allows us to add enough 10‑cent coins to increase the profit.

---

### 2.  Effect of removing \(k\) quarters

Let us replace \(k\) quarters by 10‑cent coins.  
Write \(r=10u+v\;(0\le v\le 9)\).  
After removing \(k\) quarters, the remainder becomes  

\[
R=r+25k=10u+v+25k .
\]

The maximal number of dimes that can be added is  

\[
b'=\big\lfloor \frac{R}{10}\big\rfloor
   =u+ \big\lfloor 2.5k+\tfrac{v}{10}\big\rfloor .
\]

The change in profit is

\[
\Delta P
=-24k+9\big(b'-\big\lfloor r/10\big\rfloor\big)
=-6k+9\Big\lfloor\frac{k}{2}\Big\rfloor+9\delta ,
\]

where  

\[
\delta=
\begin{cases}
1 &\text{if }k\text{ is odd and }v\ge5,\\
0 &\text{otherwise.}
\end{cases}
\]

*If \(k\) is even*, \(\Delta P=-3k/2<0\); the profit never increases.  

*If \(k\) is odd*, write \(k=2m+1\). Then  

\[
\Delta P=9\delta-(3m+6).
\]

The only way \(\Delta P>0\) is when \(m=0\) (i.e. \(k=1\)) and \(\delta=1\),
which occurs exactly when  

\[
v=r\bmod 10\ge5 .
\]

Hence **the greedy algorithm fails precisely when**
\[
N\ge 25\quad\text{and}\quad (N\bmod 25)\bmod 10\ge5.
\]

Equivalently, the “bad’’ remainders (after taking as many quarters as possible) are  

\[
r\in\{5,6,7,8,9,15,16,17,18,19\}.
\]

In each block of 25 consecutive integers there are 10 such remainders.

---

### 3.  Counting the failures between 1 and 1000

The interval \([25,1000]\) contains \(1000-24=976\) integers.
From 25 onward the numbers are grouped into blocks  

\[
25\!-\!49,\;50\!-\!74,\;\dots,\;975\!-\!999 .
\]

There are \(39\) full blocks (the block \(25\!-\!49\) up to \(975\!-\!999\)).
Each block contributes exactly \(10\) failures, so

\[
\#\text{failures}=39\times10=390 .
\]

Numbers \(1\) through \(24\) are never failures (they contain no quarters).

Thus the number of successes is  

\[
1000-390=610 .
\]

---

\[
\boxed{610}
\]

In a regular \(24\)-gon label the vertices \(0,1,\dots ,23\) clockwise.  
If two vertices differ by \(k\) (mod \(24\)), the chord joining them has length that depends only on  

\[
s=\min (k,24-k)\qquad (1\le s\le 12).
\]

Thus there are \(12\) possible lengths, one for each \(s=1,2,\dots ,12\).

--------------------------------------------------------------------
### 1.  The graph of chords of a fixed length  

Fix a value \(s\).  The set of all chords of that length is the edge‑set of the **circulant graph**
\[
G_s=\{\,\{i,i+s\}\mid i=0,1,\dots ,23\,\},
\]
where the addition is taken modulo \(24\).

Let \(g=\gcd(s,24)\).  Adding \(s\) repeatedly splits the vertex set into \(g\) cycles,
each of length  

\[
L=\frac{24}{g}.
\]

So \(G_s\) is a disjoint union of \(g\) cycles \(C_L\).

--------------------------------------------------------------------
### 2.  When can a perfect matching exist?

A perfect matching (a set of \(12\) disjoint edges covering all vertices) can exist only if
every component cycle has even length, i.e. \(L\) is even.
Since \(L=24/g\), this is equivalent to \(g\mid 12\).

*If \(g\mid 12\) and \(L\ge 4\)*, each even cycle \(C_L\) has exactly **2** perfect
matchings (choose alternating edges).  
*If \(L=2\)* (the case \(s=12\)), each component is a single edge and contributes only **1** matching.

The only step for which \(g\nmid 12\) is \(s=8\) (here \(g=8\) and the components are 3‑cycles),
so no perfect matching exists for \(s=8\).

--------------------------------------------------------------------
### 3.  Number of matchings for each possible length

\[
\begin{array}{c|c|c|c}
s & g=\gcd(s,24) & L=24/g & \#\text{ matchings}\\ \hline
1 &1 &24 &2\\
2 &2 &12 &2^{2}=4\\
3 &3 & 8 &2^{3}=8\\
4 &4 & 6 &2^{4}=16\\
5 &1 &24 &2\\
6 &6 & 4 &2^{6}=64\\
7 &1 &24 &2\\
8 &8 & 3 &0\\
9 &3 & 8 &2^{3}=8\\
10&2 &12 &2^{2}=4\\
11&1 &24 &2\\
12&12& 2 &1
\end{array}
\]

--------------------------------------------------------------------
### 4.  Total number of ways

Add the numbers for all admissible lengths (all \(s\) except \(s=8\)):

\[
\begin{aligned}
\text{Total}&=2+4+8+16+2+64+2+8+4+2+1\\
&=113 .
\end{aligned}
\]

--------------------------------------------------------------------
\[
\boxed{113}
\]

The chairs can be represented by a binary string of length 16, where a \(1\) means the chair is taken.  
We need strings with exactly eight \(1\)’s and with **no three consecutive \(1\)’s** (otherwise the middle person would sit next to two others).

---

### Counting by blocks

A valid string consists of blocks of consecutive \(1\)’s of length 1 or 2, separated by at least one \(0\).

Let  

* \(t\) = number of blocks of length 2,  
* \(b\) = total number of blocks.  

Since each block contributes at least one \(1\),

\[
\underbrace{(b-t)}_{\text{blocks of length 1}}+2t=8\quad\Longrightarrow\quad b+ t=8,\; b=8-t .
\]

So \(t=0,1,2,3,4\).

For a fixed \(t\):
* Choose which of the \(b\) blocks are the length‑2 ones: \(\displaystyle\binom{b}{t}=\binom{8-t}{t}\).
* There are \(8-8=8\) empty chairs; let \(z=8\).  
  The \(b+1\) gaps (including the two ends) must contain these zeros, with at least one zero in each interior gap.  
  After placing one zero in each of the \(b-1\) interior gaps, we have \(z-(b-1)=9-b\) zeros left to distribute freely among the \(b+1\) gaps.  
  The number of ways is \(\displaystyle\binom{9-b}{b}=\binom{9}{b}\).

Hence for each \(t\),

\[
N_t=\binom{8-t}{t}\binom{9}{8-t}.
\]

Summing over \(t=0\) to \(4\):

\[
\begin{aligned}
N &=\binom{8}{0}\binom{9}{8}
   +\binom{7}{1}\binom{9}{7}
   +\binom{6}{2}\binom{9}{6}
   +\binom{5}{3}\binom{9}{5}
   +\binom{4}{4}\binom{9}{4} \\[2mm]
  &=1\cdot9+7\cdot36+15\cdot84+10\cdot126+1\cdot126\\
  &=9+252+1260+1260+126\\
  &=2907 .
\end{aligned}
\]

---

### Modulo 1000

\[
2907 \equiv 907 \pmod{1000}.
\]

---

\[
\boxed{907}
\]

Let  

\[
r_i=A_1A_i\qquad (i=2,\dots ,11),\qquad 
\theta=\angle A_iA_1A_{i+1}.
\]

For each \(i=2,\dots ,10\)

\[
\cos\theta=\frac{12}{13},\qquad 
\text{Area}(A_iA_1A_{i+1})=\frac12 r_i r_{i+1}\sin\theta =1 .
\]

Since \(\sin\theta=\sqrt{1-\cos ^2\theta}=5/13\),

\[
r_i r_{i+1}= \frac{2}{\sin\theta}= \frac{2}{5/13}= \frac{26}{5}
=:k\quad\text{(constant)}.
\]

Thus  

\[
r_{i+1}= \frac{k}{r_i},\qquad\text{so}\qquad r_{i+2}=r_i .
\]

Hence the distances from \(A_1\) alternate:

\[
r_2=r_4=r_6=r_8=r_{10}=a,\qquad  
r_3=r_5=r_7=r_9=r_{11}=b,
\]

with  

\[
ab=k=\frac{26}{5}.
\]

--------------------------------------------------------------------
### Length of a side \(A_iA_{i+1}\)

For any consecutive pair \((a,b)\),

\[
A_iA_{i+1}=d=\sqrt{a^2+b^2-2ab\cos\theta}
           =\sqrt{a^2+b^2-\frac{24}{13}ab}.
\]

Let  

\[
x=a+b,\qquad ab=k .
\]

Because \(a^2+b^2=(a+b)^2-2ab=x^2-2k\),

\[
d=\sqrt{x^2-2k-\frac{24}{13}k}
   =\sqrt{x^2-\frac{50}{13}\,k}.
\]

Since \(k=\dfrac{26}{5}\),

\[
\frac{50}{13}k=\frac{50}{13}\cdot\frac{26}{5}=20,
\qquad\boxed{d=\sqrt{x^2-20}} .
\]

--------------------------------------------------------------------
### Using the perimeter

The perimeter of the 11‑gon is

\[
a+b+9d = x+9\sqrt{x^2-20}=20 .
\]

Solve for \(x\) (note \(x>0\) and \(\sqrt{x^2-20}\ge0\)):

\[
9\sqrt{x^2-20}=20-x\quad\Longrightarrow\quad
81(x^2-20)=(20-x)^2 .
\]

This gives  

\[
80x^2+40x-2020=0\;\Longrightarrow\;4x^2+2x-101=0.
\]

Hence  

\[
x=\frac{-2\pm\sqrt{4+1616}}{8}
  =\frac{-2\pm18\sqrt5}{8}
  =\frac{-1\pm9\sqrt5}{4}.
\]

Only the positive root satisfies the constraints, so  

\[
\boxed{x=a+b=\frac{9\sqrt5-1}{4}} .
\]

Thus

\[
A_1A_2+A_1A_{11}=a+b=\frac{9\sqrt5-1}{4}
    =\frac{m\sqrt n-p}{q},
\]

with \(m=9,\;n=5,\;p=1,\;q=4\).

All conditions on \(m,n,p,q\) are met, and

\[
m+n+p+q=9+5+1+4=19.
\]

\[
\boxed{19}
\]

In base \(b\) the numbers are  

\[
17_{b}=1\cdot b+7=b+7, \qquad 97_{b}=9\cdot b+7=9b+7 .
\]

We need \(b+7\) to divide \(9b+7\).

---

### Find the divisibility condition  

\[
9b+7 = 9(b+7)-56 .
\]

Thus the remainder when \(9b+7\) is divided by \(b+7\) is \(-56\).  
Hence \(b+7\) must divide \(56\):

\[
b+7\mid 56 .
\]

---

### Possible values of \(b\)

The positive divisors of \(56\) are  

\[
1,\;2,\;4,\;7,\;8,\;14,\;28,\;56 .
\]

Because the base must satisfy \(b>9\), we have \(b+7>16\).  
The divisors larger than \(16\) are \(28\) and \(56\).

\[
\begin{aligned}
b+7&=28 \quad\Longrightarrow\quad b=21,\\[2pt]
b+7&=56 \quad\Longrightarrow\quad b=49 .
\end{aligned}
\]

Both bases are greater than \(9\) and allow the digits \(1,7,9\).

---

### Sum of all such bases  

\[
21+49 = 70.
\]

\[
\boxed{70}
\]

The numbers of players picking chocolate, vanilla and strawberry are  

\[
a=\#\text{chocolate},\qquad b=\#\text{vanilla},\qquad c=\#\text{strawberry}
\]

with  

\[
a+b+c=9,\qquad a>b>c\ge1 .
\]

Hence \((a,b,c)\) must be three distinct positive integers whose sum is \(9\).  
The possible triples (written in decreasing order) are  

\[
(6,2,1),\;(5,3,1),\;(4,3,2).
\]

For a fixed triple \((a,b,c)\) the players are distinct, so the number of assignments
is the multinomial coefficient  

\[
\binom{9}{a,b,c}= \frac{9!}{a!\,b!\,c!}.
\]

Compute each term:

\[
\begin{aligned}
(6,2,1):&\quad \frac{9!}{6!\,2!\,1!}= \frac{362880}{720\cdot2}=252,\\[2mm]
(5,3,1):&\quad \frac{9!}{5!\,3!\,1!}= \frac{362880}{120\cdot6}=504,\\[2mm]
(4,3,2):&\quad \frac{9!}{4!\,3!\,2!}= \frac{362880}{24\cdot6\cdot2}=1260.
\end{aligned}
\]

Add them:

\[
N = 252+504+1260 = 2016.
\]

The remainder when \(N\) is divided by \(1000\) is  

\[
2016 \equiv 16 \pmod{1000}.
\]

\[
\boxed{16}
\]

Let  

\[
A=(0,0),\qquad B=(b,0),\qquad C=(0,c)
\]

so that \(\angle A=90^{\circ}\) and  

\[
BC=\sqrt{b^{2}+c^{2}}=38\qquad\Longrightarrow\qquad b^{2}+c^{2}=38^{2}=1444 .
\tag{1}
\]

--------------------------------------------------------------------
### 1.  Determining the positions of \(K\) and \(L\)

Because  

\[
AK=BK=14
\]

the point \(K\) lies at the intersection of the circles of radius \(14\) centred at
\(A\) and \(B\).  
If \(K=(x_{K},y_{K})\),

\[
x_{K}^{2}+y_{K}^{2}=196,\qquad (x_{K}-b)^{2}+y_{K}^{2}=196 .
\]

Subtracting gives  

\[
b^{2}=2bx_{K}\;\Longrightarrow\;x_{K}= \frac{b}{2} .
\tag{2}
\]

Hence  

\[
y_{K}^{2}=196-\Bigl(\frac{b}{2}\Bigr)^{2}=196-\frac{b^{2}}{4}.
\tag{3}
\]

Similarly, from  

\[
AL=CL=14
\]

the point \(L=(x_{L},y_{L})\) satisfies

\[
y_{L}= \frac{c}{2},\qquad 
x_{L}^{2}=196-\Bigl(\frac{c}{2}\Bigr)^{2}=196-\frac{c^{2}}{4}.
\tag{4}
\]

Thus, writing  

\[
U=\frac{b}{2},\quad V=\frac{c}{2},
\]

we have  

\[
K=\bigl(U,\;\sqrt{196-U^{2}}\bigr),\qquad 
L=\bigl(\sqrt{196-V^{2}},\;V\bigr),
\tag{5}
\]

and from (1)  

\[
U^{2}+V^{2}=361. \tag{6}
\]

--------------------------------------------------------------------
### 2.  Using the condition \(KL=14\)

From (5),

\[
\bigl(U-\sqrt{196-V^{2}}\bigr)^{2}
+\bigl(\sqrt{196-U^{2}}-V\bigr)^{2}=196 .
\]

Expanding and simplifying with (6) gives

\[
U\sqrt{196-V^{2}}+V\sqrt{196-U^{2}}=98 . \tag{7}
\]

Let  

\[
p=\sqrt{196-V^{2}},\qquad q=\sqrt{196-U^{2}} .
\]

Then  

\[
U p+V q =98 ,\qquad p^{2}+q^{2}=31 ,
\qquad U^{2}+V^{2}=361 . \tag{8}
\]

From (8) we obtain the product

\[
U^{2}V^{2}=32448\quad\Longrightarrow\quad UV=104\sqrt3 . \tag{9}
\]

Now  

\[
(U+V)^{2}=U^{2}+V^{2}+2UV=361+208\sqrt3,
\]
\[
(U-V)^{2}=U^{2}+V^{2}-2UV=361-208\sqrt3 .
\]

Both expressions are perfect squares:

\[
361+208\sqrt3=(13+8\sqrt3)^{2},
\qquad
361-208\sqrt3=(8\sqrt3-13)^{2}.
\]

Hence  

\[
U+V=13+8\sqrt3,\qquad
|U-V|=8\sqrt3-13 .
\]

Since \(U>V\), solving gives  

\[
U=8\sqrt3,\qquad V=13 .
\]

Consequently  

\[
b=2U=16\sqrt3,\qquad c=2V=26 .
\]

--------------------------------------------------------------------
### 3.  Coordinates of \(K\) and \(L\)

Using (5) with the found \(U,V\):

\[
K=\bigl(8\sqrt3,\; \sqrt{196-(8\sqrt3)^{2}}\bigr)
   =\bigl(8\sqrt3,\;2\bigr),
\]

\[
L=\bigl(\sqrt{196-13^{2}},\;13\bigr)
   =\bigl(3\sqrt3,\;13\bigr).
\]

--------------------------------------------------------------------
### 4.  Area of quadrilateral \(BKLC\)

The vertices in order are  

\[
B=(16\sqrt3,0),\ K=(8\sqrt3,2),\ L=(3\sqrt3,13),\ C=(0,26).
\]

Applying the shoelace formula:

\[
\begin{aligned}
S_{1}&=x_{B}y_{K}+x_{K}y_{L}+x_{L}y_{C}+x_{C}y_{B}\\
     &= (16\sqrt3)(2)+(8\sqrt3)(13)+(3\sqrt3)(26)+0\\
     &=32\sqrt3+104\sqrt3+78\sqrt3=214\sqrt3,
\\[2mm]
S_{2}&=y_{B}x_{K}+y_{K}x_{L}+y_{L}x_{C}+y_{C}x_{B}\\
     &=0+2(3\sqrt3)+13\cdot0+26(16\sqrt3)\\
     &=6\sqrt3+416\sqrt3=422\sqrt3 .
\end{aligned}
\]

Hence  

\[
\text{Area}=\frac12|S_{1}-S_{2}|
           =\frac12\bigl|214\sqrt3-422\sqrt3\bigr|
           =\frac12\,(208\sqrt3)=104\sqrt3 .
\]

Thus the required area is \(104\sqrt3\), i.e. \(n=104\).

\[
\boxed{104}
\]

The equation  

\[
12x^{2}-xy-6y^{2}=0
\]

can be factored over the integers.

\[
\begin{aligned}
12x^{2}-xy-6y^{2}
&= (4x-3y)(3x+2y).
\end{aligned}
\]

Hence a pair \((x,y)\) satisfies the equation iff  

\[
4x-3y=0 \qquad\text{or}\qquad 3x+2y=0 .
\]

---

### 1.  Solutions of \(4x-3y=0\)

\[
4x=3y \Longrightarrow x=\frac34y .
\]

Since \(x,y\) are integers, \(y\) must be a multiple of \(4\); write \(y=4k\) \((k\in\mathbb Z)\).  
Then \(x=3k\).

Bounds \(-100\le x,y\le100\) give  

\[
|4k|\le100\ \Longrightarrow\ |k|\le25 .
\]

Thus \(k\) can be any integer from \(-25\) to \(25\) inclusive – \(51\) possibilities.

---

### 2.  Solutions of \(3x+2y=0\)

\[
3x=-2y \Longrightarrow x=-\frac{2}{3}y .
\]

Now \(y\) must be a multiple of \(3\); write \(y=3k\).  
Then \(x=-2k\).

Bounds give  

\[
|3k|\le100\ \Longrightarrow\ |k|\le\Big\lfloor\frac{100}{3}\Big\rfloor=33 .
\]

Hence \(k\) can be any integer from \(-33\) to \(33\) inclusive – \(67\) possibilities.

---

### 3.  Removing double‑counted pair

Both families contain the pair \((0,0)\) (when \(k=0\)).  
It has been counted twice, so we subtract one.

\[
\text{Total solutions}=51+67-1=117 .
\]

All counted pairs satisfy \(-100\le x,y\le100\).

\[
\boxed{117}
\]

Let the trapezoid have parallel sides \(r\) and \(s\) (\(r\neq s\)), equal legs of length \(L\), height \(h\), and incircle radius \(R=3\).

---

### 1.  Semiperimeter from the incircle

For any tangential quadrilateral,
\[
\text{Area}=R\cdot s,
\]
where \(s\) is the semiperimeter.  
Given \(\text{Area}=72\) and \(R=3\),

\[
s=\frac{72}{3}=24,
\qquad\text{so } r+s+2L=2s=48. \tag{1}
\]

---

### 2.  Height of the trapezoid

Since the incircle touches both parallel sides, the distance between them equals twice the radius:

\[
h=2R=6.
\]

The area also equals \(\frac{1}{2}(r+s)h\). Thus

\[
72=\frac{1}{2}(r+s)\cdot6 \Longrightarrow r+s=24. \tag{2}
\]

Substituting (2) into (1) gives

\[
24+2L=48\Longrightarrow L=12. \tag{3}
\]

---

### 3.  Relating the bases and the leg

In an isosceles trapezoid the horizontal projection of each leg is \(\dfrac{s-r}{2}\).
Using the right triangle formed by a leg, the height, and this projection:

\[
\left(\frac{s-r}{2}\right)^2+h^2=L^2.
\]

Insert \(h=6\) and \(L=12\):

\[
\left(\frac{s-r}{2}\right)^2+36=144
\;\Longrightarrow\;
\frac{(s-r)^2}{4}=108
\;\Longrightarrow\;
(s-r)^2=432.
\]

Hence

\[
s-r=12\sqrt3. \tag{4}
\]

---

### 4.  Solve for \(r\) and \(s\)

From (2) and (4):

\[
\begin{cases}
r+s=24,\\[2pt]
s-r=12\sqrt3,
\end{cases}
\qquad\Longrightarrow\qquad
s=12+6\sqrt3,\;\; r=12-6\sqrt3.
\]

Both are positive and distinct, as required.

---

### 5.  Compute \(r^{2}+s^{2}\)

\[
\begin{aligned}
r^{2}+s^{2}&=(12-6\sqrt3)^2+(12+6\sqrt3)^2\\
&=2\cdot12^{2}+2\cdot(6\sqrt3)^{2}\\
&=2\cdot144+2\cdot108\\
&=288+216\\
&=504.
\end{aligned}
\]

---

\[
\boxed{504}
\]

For an \(8\)-digit number formed with the digits \(1,2,\dots ,8\) each once we need

* divisibility by \(2\)   \(\Longrightarrow\) the units digit is even (\(2,4,6,8\));
* divisibility by \(11\) \(\Longrightarrow\) the alternating sum of its digits is a multiple of \(11\).

---

### 1.  Divisibility by \(11\)

Write the digits as \(d_1d_2d_3d_4d_5d_6d_7d_8\) ( \(d_1\) most significant).  

The test for \(11\) gives  

\[
(d_1+d_3+d_5+d_7)-(d_2+d_4+d_6+d_8)\equiv0\pmod{11}.
\]

Let  

\[
O=d_1+d_3+d_5+d_7,\qquad E=d_2+d_4+d_6+d_8 .
\]

Since all eight digits are used once,  

\[
O+E = 1+2+\dots+8 = 36 .
\]

The condition \(O-E\equiv0\pmod{11}\) becomes  

\[
2O\equiv36\pmod{11}\quad\Longrightarrow\quad O\equiv18\equiv7\pmod{11}.
\]

The sum of four distinct digits from \(\{1,\dots ,8\}\) lies between \(10\) and \(26\); the only value congruent to \(7\pmod{11}\) in that range is \(18\).  
Hence we must have  

\[
O=E=18.
\]

So the four digits placed in the odd positions must sum to \(18\).

---

### 2.  Choosing the odd–position digits

We need all \(4\)-element subsets of \(\{1,\dots ,8\}\) whose sum is \(18\).

Listing them (in increasing order) gives

\[
\begin{aligned}
&\{1,2,7,8\},\; \{1,3,6,8\},\; \{1,4,5,8\},\; \{1,4,6,7\},\\
&\{2,3,5,8\},\; \{2,3,6,7\},\; \{2,4,5,7\},\; \{3,4,5,6\}.
\end{aligned}
\]

There are \(8\) such subsets.  
Each of them contains exactly two even digits (the even digits are \(2,4,6,8\)). Consequently the complementary set (the digits placed in even positions) also contains exactly two evens.

---

### 3.  Forming the numbers

*For a fixed subset* \(S\) (the odd‑position digits)

* arrange its four digits among the odd places \(1,3,5,7\): \(4!\) ways;
* the four remaining digits (the complement \(T\)) go to the even places \(2,4,6,8\).  
  Since \(T\) contains two even digits, we may pick which even digit occupies the units place (position \(8\)) in \(2\) ways; the other three digits can be placed in \(3!\) ways.

Hence for each subset \(S\)

\[
\text{permutations}=4!\times 2\times3!=24\times12=288 .
\]

With \(8\) possible subsets,

\[
N=8\times288=2304 .
\]

(Equivalently, half of the \(4608\) numbers with \(O=E=18\) are even, giving the same result.)

---

### 4.  Required difference

\[
N-2025 = 2304-2025 = 279.
\]

Thus the requested difference is  

\[
\boxed{279}.
\]

We place \(\triangle ABC\) on its circumcircle (radius 1) with  

\[
A=1,\qquad B=e^{i2C},\qquad C=e^{i(2C+2A)} .
\]

The orthocenter is  

\[
H=a+b+c.
\]

The nine‑point circle is the image of the circumcircle under the homothety with centre \(H\) and factor \(\tfrac12\); therefore for any point \(X\) on the nine‑point circle its image on the circumcircle is  

\[
X' = 2X-H .
\]

Applying this to the six points on the nine‑point circle gives

\[
\begin{aligned}
D&\mapsto -a,\\
E&\mapsto -b,\\
F&\mapsto -c,\\
G&\mapsto A' ,\qquad A' \text{ is the second intersection of the altitude }AA_H\text{ with the circumcircle},\\
H&\mapsto B' ,\qquad B'=-e^{i2A},\\
J&\mapsto C' ,\qquad C'=-e^{-i2A}.
\end{aligned}
\]

Using \(a=1,\;b=e^{i2C},\;c=e^{i(2C+2A)}\) we obtain the central
angles (measured from the positive \(x\)-axis) of the corresponding points:

\[
\begin{array}{c|c}
\text{Nine‑point point} & \text{Angle on the circumcircle}\\ \hline
D & 180^\circ\\
E & 180^\circ+2C\\
F & 180^\circ-2B\\
G & 180^\circ+2A+4C\\
H & 180^\circ+2A\\
J & 180^\circ-2A
\end{array}
\]

Because the homothety preserves directions from the centre, the
arc length on the nine‑point circle equals the corresponding central
angle on the circumcircle.

Hence

\[
\begin{aligned}
\widehat{DE}&=(180^\circ+2C)-180^\circ =2C,\\[2mm]
\widehat{HJ}&=|\, (180^\circ+2A)-(180^\circ-2A)\,|=4A,
\end{aligned}
\]

and the minor arc \(HJ\) is \(360^\circ-4A\) (since \(4A>180^\circ\)).
Finally

\[
\widehat{FG}=|(180^\circ-2B)-(180^\circ+2A+4C)|
            =2C\qquad(\text{mod }360^\circ).
\]

Thus

\[
\begin{aligned}
\widehat{DE}+2\widehat{HJ}+3\widehat{FG}
&=2C+2\bigl(360^\circ-4A\bigr)+3\,(2C) \\
&=720^\circ-8A+8C\\
&=720^\circ+8(C-A).
\end{aligned}
\]

For the given triangle \(A=84^\circ,\;C=36^\circ\),

\[
720^\circ+8(36^\circ-84^\circ)=720^\circ-384^\circ=336^\circ .
\]

\[
\boxed{336}
\]

The function  

\[
f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}, \qquad x>0,
\]

has zeros at \(x=18,72,98,k\) and \(\displaystyle\lim_{x\to0^+}f(x)=+\infty\),
\(\displaystyle\lim_{x\to\infty}f(x)=+\infty\).
Hence the graph of \(f\) starts at \(+\infty\), descends to a negative
value, comes back to zero, rises again, etc.; it has exactly **two**
negative intervals:
\[
(r_1,r_2),\qquad (r_3,r_4),
\]
where \(\{r_1,r_2,r_3,r_4\}=\{18,72,98,k\}\) ordered increasingly.
In each negative interval there is a local (and global) minimum of \(f\).

Let  

\[
g(x)=(x-18)(x-72)(x-98)(x-k),
\qquad f(x)=\frac{g(x)}{x}.
\]

A point \(x_0\) where \(f\) has an extremum satisfies  

\[
f'(x_0)=0\iff x_0g'(x_0)-g(x_0)=0\iff 
\sum_{i=1}^{4}\frac{1}{x_0-r_i}= \frac1{x_0}.
\]

Geometrically, if \(m=f(x_0)\) then the line \(y=m x\) is tangent to the
quartic graph \(y=g(x)\) at \(x_0\):
\[
g(x)-mx=0\quad\text{has a double root at }x_0 .
\]

If the global minimum of \(f\) is attained at **two** distinct points,
the line \(y=m x\) must be tangent to \(g\) at two distinct points
\(\alpha,\beta\). Hence

\[
g(x)-mx=(x-\alpha)^2 (x-\beta)^2 .
\tag{1}
\]

Write  

\[
\alpha+\beta=p,\qquad \alpha\beta =q,\qquad m \text{ (the slope)} .
\]

Expanding (1) and comparing with \(g(x)-mx=x^4-S_1x^3+S_2x^2-(S_3+m)x+S_4\) gives  

\[
\begin{aligned}
S_1 &=2p,\\
S_2 &=p^{2}+2q,\\
S_4 &=q^{2},\\
S_3+m &=2pq,
\end{aligned}
\tag{2}
\]

where for our roots  

\[
\begin{aligned}
S_1&=18+72+98+k=188+k,\\
S_2&=18\cdot72+18\cdot98+72\cdot98+ (18+72+98)k
     =10116+188k,\\
S_3&=18\cdot72\cdot98+ (18\cdot72+18\cdot98+72\cdot98)k
     =127008+10116k,\\
S_4&=18\cdot72\cdot98\cdot k=127008\,k .
\end{aligned}
\]

From (2) we obtain  

\[
p=\frac{188+k}{2},\qquad q=\sqrt{127008\,k}=252\sqrt{2k}.
\]

Using the second equation of (2),

\[
S_2=p^{2}+2q,
\]

gives the condition on \(k\):

\[
\frac{(188+k)^{2}}{4}+2\sqrt{127008k}=10116+188k .
\tag{3}
\]

---

### Solving (3)

Put \(k=t^{2}\;(t>0)\).  Since \(\sqrt{127008k}=252\sqrt{2}\,t\),
(3) becomes

\[
t^{4}-376t^{2}+2016\sqrt{2}\,t-5120=0 .
\]

Factoring,

\[
(t^{2}+6\sqrt{2}\,t-320)(t^{2}-6\sqrt{2}\,t+16)=0 .
\]

Hence the positive solutions for \(t\) are

\[
t=10\sqrt2,\qquad t=4\sqrt2,\qquad t=2\sqrt2 .
\]

Consequently

\[
k=t^{2}=200,\;32,\;8 .
\]

---

### Verification

For each \(k\) the ordered roots of \(g\) are

\[
\begin{array}{c|c}
k & \text{ordered roots }(r_1,r_2,r_3,r_4)\\ \hline
8  & (8,18,72,98)\\
32 & (18,32,72,98)\\
200& (18,72,98,200)
\end{array}
\]

Using \(p\) and \(q\) we find the tangency points  

\[
\alpha,\beta=\frac{p\pm\sqrt{p^{2}-4q}}{2},
\]

and they lie respectively in the intervals \((r_1,r_2)\) and
\((r_3,r_4)\). Thus in each case the line \(y=m x\) is tangent to
\(g\) at two points, giving exactly two distinct global minima of
\(f\). No other \(k>0\) satisfies (3); hence these are the only three
values.

Finally, the required sum is

\[
8+32+200 = 240 .
\]

\[
\boxed{240}
\]

Let  

\[
z=x+iy\qquad (x,y\in\mathbb{R}).
\]

---

### 1.  The first equation – a circle  

\[
|25+20i-z|=5 \Longrightarrow (x-25)^2+(y-20)^2=5^2 .
\]

Thus the circle has centre  

\[
C=(25,20),\qquad \text{radius } r=5 .
\]

---

### 2.  The second equation – a perpendicular bisector  

\[
|z-(4+k)|=|z-(3i+k)|
\]

writes as  

\[
(x-(k+4))^2+y^{2}=(x-k)^2+(y-3)^2 .
\]

Simplifying,

\[
-8x+6y+8k+7=0\quad\Longrightarrow\quad 4x-3y-(4k+3.5)=0 .
\]

Hence the set of points satisfying the second condition is the line  

\[
L:\;4x-3y=4k+\frac 72 .
\]

This is the perpendicular bisector of the segment joining  
\(A=(k+4,0)\) and \(B=(k,3)\) (its slope is \(4/3\) for any \(k\)).

---

### 3.  One intersection  \(\iff\)  tangency  

The system has exactly one solution precisely when the line \(L\) is tangent to the circle, i.e. when the distance from the centre \(C\) to \(L\) equals the radius \(5\).

The distance from \((x_0,y_0)\) to \(Ax+By+C=0\) is  

\[
\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} .
\]

For \(C=(25,20)\) and \(L: 4x-3y-(4k+3.5)=0\),

\[
d=\frac{|4\cdot25-3\cdot20-(4k+3.5)|}{\sqrt{4^{2}+(-3)^{2}}}
   =\frac{|40-4k-3.5|}{5}
   =\frac{|36.5-4k|}{5}.
\]

Set \(d=r=5\):

\[
\frac{|36.5-4k|}{5}=5\quad\Longrightarrow\quad |36.5-4k|=25 .
\]

---

### 4.  Solving for \(k\)

\[
\begin{cases}
36.5-4k=25 &\Longrightarrow\;4k=11.5\;\Longrightarrow\;k=\dfrac{23}{8},\\[4pt]
36.5-4k=-25 &\Longrightarrow\;4k=61.5\;\Longrightarrow\;k=\dfrac{123}{8}.
\end{cases}
\]

Thus the possible real values of \(k\) are  

\[
k_1=\frac{23}{8},\qquad k_2=\frac{123}{8}.
\]

Both give a line tangent to the circle, so each yields exactly one solution \(z\).

---

### 5.  Sum of the possible \(k\)’s  

\[
k_1+k_2=\frac{23+123}{8}=\frac{146}{8}=\frac{73}{4}.
\]

Write \(\frac{73}{4}\) as \(\dfrac{m}{n}\) with \(\gcd(m,n)=1\); here \(m=73,\ n=4\).

\[
m+n=73+4=77.
\]

\[
\boxed{77}
\]

**Step 1.  Write the recurrence for the numerators and denominators**

Let  

\[
x_k=\frac{a_k}{b_k}\qquad (a_k,b_k\in \mathbb Z ,\; \gcd(a_k,b_k)=1).
\]

From  

\[
x_{k+1}= \frac{1}{3}\Bigl(x_k+\frac1{x_k}-1\Bigr)=\frac{x_k^2-x_k+1}{3x_k},
\]

we obtain  

\[
a_{k+1}=a_k^2-a_kb_k+b_k^2,\qquad   
b_{k+1}=3a_kb_k,          \tag{1}
\]

with \(a_1=25,\;b_1=11\).

--------------------------------------------------------------------

**Step 2.  The common factor \(3^{e_k}\)**  

Write  

\[
a_k=3^{e_k}A_k,\qquad b_k=3^{e_k}B_k,\qquad \gcd(A_k,B_k)=1 .
\]

Dividing (1) by the highest possible power of \(3\) shows that
\(e_{k+1}=2e_k+1\) and that the reduced pair \((A_k,B_k)\) always satisfies  

\[
A_k\equiv1\pmod3,\qquad B_k\equiv2\pmod3 .
\]

With \(e_1=0\) we get  

\[
e_k=2^{\,k-1}-1 .\tag{2}
\]

--------------------------------------------------------------------

**Step 3.  A very simple recurrence for the sum**

Let  

\[
S_k=a_k+b_k .
\]

From (1)

\[
\begin{aligned}
S_{k+1}&=a_{k+1}+b_{k+1}
       =(a_k^2-a_kb_k+b_k^2)+3a_kb_k   \\
       &=a_k^2+b_k^2+2a_kb_k=(a_k+b_k)^2=S_k^{\,2}.
\end{aligned}
\]

Hence  

\[
S_k=S_1^{\,2^{\,k-1}}=36^{\,2^{\,k-1}}. \tag{3}
\]

--------------------------------------------------------------------

**Step 4.  The reduced numerator and denominator**

Since the common factor is \(3^{e_k}\),

\[
\frac{a_k}{b_k}= \frac{A_k}{B_k},\qquad 
m=A_k,\; n=B_k .
\]

Thus  

\[
m+n=\frac{a_k+b_k}{3^{e_k}}
     =\frac{36^{\,2^{\,k-1}}}{3^{\,2^{\,k-1}-1}}
     =2^{\,2^{\,k}}\;3^{\,2^{\,k-1}+1}. \tag{4}
\]

--------------------------------------------------------------------

**Step 5.  Apply \(k=2025\)**  

From (4),

\[
m+n=2^{\,2^{2025}}\;3^{\,2^{2024}+1}.
\]

We need this number modulo \(1000\).

--------------------------------------------------------------------

**Step 6.  Reduce modulo \(8\) and \(125\)**  

*Modulo \(8\):*  \(2^{\ge3}\equiv0\pmod8\); therefore  

\[
m+n\equiv0\pmod8. \tag{5}
\]

*Modulo \(125\):*  Because \(\phi(125)=100\),

\[
2^{2^{2025}}\equiv 2^{\,2^{2025}\bmod100},
\qquad
3^{2^{2024}+1}\equiv 3^{\,(2^{2024}+1)\bmod100}\pmod{125}.
\]

Compute the exponents.

- \(2^{2025}\bmod100\):  
  \(2^{2025}\equiv32\pmod{100}\) (by CRT: \(2^{2025}\equiv0\pmod4\), \(2^{2025}\equiv7\pmod{25}\)).
- \(2^{2024}\bmod100\):  
  \(2^{2024}\equiv16\pmod{100}\).

Hence  

\[
2^{2^{2025}}\equiv2^{32}\equiv46\pmod{125},
\qquad
3^{2^{2024}+1}\equiv3^{17}\equiv38\pmod{125}.
\]

Therefore  

\[
m+n\equiv46\cdot38\equiv123\pmod{125}. \tag{6}
\]

--------------------------------------------------------------------

**Step 7.  Chinese Remainder Theorem**

Find \(N\) such that  

\[
N\equiv0\pmod8,\qquad N\equiv123\pmod{125}.
\]

Write \(N=8k\).  Then \(8k\equiv123\pmod{125}\), i.e.  

\[
k\equiv123\cdot8^{-1}\pmod{125}.
\]

Since \(8^{-1}\equiv47\pmod{125}\),  

\[
k\equiv123\cdot47\equiv31\pmod{125}.
\]

Thus \(k=31+125t\) and  

\[
N=8k=8\cdot31=248\pmod{1000}.
\]

--------------------------------------------------------------------

**Answer**

\[
\boxed{248}
\]

Let  

\[
A<B<C<D<E<F<G<H<I<J<K<L
\]

and suppose the letters are randomly paired.  
For a pair \(\{x,y\}\) we write the word \(xy\) with the letters in
alphabetical order (\(x<y\)).  
The six words are then ordered alphabetically; the last word is the
lexicographically largest word.

--------------------------------------------------------------------
### 1.  Conditioning on the partner of \(G\)

In a random perfect matching each of the other \(11\) letters is equally
likely to be paired with \(G\).  
Let the partner be \(X\).  The probability we need is  

\[
P=\frac1{11}\sum_{X\neq G}
\Pr\bigl(\text{the pair } \{G,X\}\text{ is the maximal word}\mid
G\text{ is paired with }X\bigr).
\]

Thus we must evaluate the conditional probability for each possible
\(X\).

--------------------------------------------------------------------
### 2.  When \(X<G\)

If \(X<G\) the word is \(XG\) and its first letter is \(X\).  
For it to be the maximal word, the first letters (the smaller letters)
of all other five pairs must be **smaller than \(X\)**.
Hence among the remaining ten letters we need five distinct letters
all less than \(X\).

The only letter with five smaller letters is \(F\) (the letters
\(A,B,C,D,E\) are smaller).  
Thus:

* \(X=F\) is possible;
* \(X=A,B,C,D,E\) are impossible (there are not enough smaller
letters).

When \(X=F\) the remaining letters are  
\(\{A,B,C,D,E\}\) and \(\{H,I,J,K,L\}\).  
To keep every other pair’s first letter \(<F\), each of the five large
letters must be paired with a distinct small one.  
The number of such matchings is \(5!=120\).

The total number of matchings of those ten letters is  

\[
\frac{10!}{2^5\,5!}=945 .
\]

Hence  

\[
\Pr(\text{maximal}\mid G\text{–}F)=\frac{120}{945}=\frac{8}{63}.
\]

--------------------------------------------------------------------
### 3.  When \(X>G\)

Now the word is \(GX\) and its first letter is \(G\).
For it to be maximal, **no other pair may have a first letter larger
than \(G\)**; i.e. no remaining pair may consist of two letters both
greater than \(G\).

After fixing the pair \(\{G,X\}\) (with \(X\in\{H,I,J,K,L\}\)), the
remaining letters are

* the six letters \(\{A,B,C,D,E,F\}\) (all \(<G\));
* the four letters among \(\{H,I,J,K,L\}\) that are not \(X\) (all
\(>G\)).

Thus each of those four “large’’ letters must be paired with a distinct
small letter, leaving two small letters to pair together.

Number of valid matchings:

\[
\binom{6}{4}\cdot4!\cdot1=15\cdot24=360 .
\]

Again the total number of matchings of the ten remaining letters is
\(945\). Therefore

\[
\Pr(\text{maximal}\mid G\text{–}X)=\frac{360}{945}=\frac{8}{21}
\qquad (X=H,I,J,K,L).
\]

--------------------------------------------------------------------
### 4.  Assemble the probability

\[
\begin{aligned}
P&=\frac1{11}\Bigl(\frac{8}{63}
+5\cdot\frac{8}{21}\Bigr)    \\
 &=\frac1{11}\Bigl(\frac{8}{63}+\frac{40}{21}\Bigr)
   =\frac1{11}\Bigl(\frac{8+120}{63}\Bigr)
   =\frac{128}{693}.
\end{aligned}
\]

The fraction \(\displaystyle\frac{128}{693}\) is already in lowest
terms.  Hence \(m=128,\;n=693\) and

\[
m+n = 128+693 = 821 .
\]

--------------------------------------------------------------------
\[
\boxed{821}
\]

Let  

\[
S(t)=\sum_{a=1}^{3^{6}}\exp\!\Big(2\pi i\,\frac{t a^{3}}{3^{7}}\Big) \qquad(0\le t<3^{7})
\]

and denote  

\[
N=\#\{(a,b,c)\mid 1\le a,b,c\le 3^{6},
\ a^{3}+b^{3}+c^{3}\equiv 0\pmod{3^{7}}\}.
\]

For any modulus \(m\),

\[
\#\{(x,y,z):x+y+z\equiv0\pmod m\}=
\frac1m\sum_{t=0}^{m-1}\Bigl(\sum_{x}e^{2\pi i t x/m}\Bigr)^3 .
\]

Hence  

\[
N=\frac1{3^{7}}\sum_{t=0}^{3^{7}-1} S(t)^{3}. \tag{1}
\]

--------------------------------------------------------------------
### 1.  Splitting the summation

Write each \(a\) as \(a=3^{v}u\) where \(v=v_{3}(a)\) and \(u\) is not
divisible by \(3\).  
For \(0\le v\le 2\) define  

\[
U_{v}=\{\,u:1\le u\le 3^{6-v},\;3\nmid u\,\},
\qquad |U_{0}|=486,\ |U_{1}|=162,\ |U_{2}|=54 .
\]

If \(v\ge3\) then \(a^{3}\equiv0\pmod{3^{7}}\); there are  
\(n_{3}=27\) such numbers.
Thus

\[
S(t)=f_{0}(t)+f_{1}(t)+f_{2}(t)+n_{3},
\]
where  

\[
\begin{aligned}
f_{0}(t)&=\sum_{x\in U_{0}}\zeta^{t x^{3}},\\[2mm]
f_{1}(t)&=\sum_{x\in U_{1}}\zeta^{t\,27x^{3}},\\[2mm]
f_{2}(t)&=\sum_{x\in U_{2}}\zeta^{t\,729x^{3}},
\end{aligned}
\qquad 
\zeta=e^{2\pi i/3^{7}} .
\]

--------------------------------------------------------------------
### 2.  Evaluating \(f_{0},f_{1},f_{2}\)

*For \(f_{0}\).*  
Let \(G_{7}=(\mathbb Z/3^{7}\mathbb Z)^{\times}\) (\(|G_{7}|=1458\)).
The map \(x\mapsto x^{3}\) from \(G_{7}\) onto the set of cubes
\(C_{6}\) has kernel of size \(3\); consequently

\[
\sum_{x\in G_{7}}\zeta^{t x}=3\sum_{r\in C_{6}}\zeta^{t r}=3f_{0}(t).
\]

For \(t\neq0\) one has  

\[
\sum_{x\in G_{7}}\zeta^{t x}= -\!\!\sum_{\substack{x\;(\bmod 3^{7})\\3\mid x}}\!\!\zeta^{t x}
=\begin{cases}
-729,&v_{3}(t)=6,\\
0,&0\le v_{3}(t)\le5 .
\end{cases}
\]

Hence  

\[
f_{0}(t)=
\begin{cases}
486,&t=0,\\[2mm]
-243,&v_{3}(t)=6,\\[2mm]
0,&\text{otherwise.}
\end{cases}
\tag{2}
\]

*For \(f_{1}\).*  
Writing each \(x\in U_{1}\) as \(x=v+81k\;(k=0,1,2)\) one finds
\(x^{3}\equiv v^{3}\pmod{81}\). Consequently  

\[
f_{1}(t)=3\!\!\sum_{\substack{v\in(\mathbb Z/81)^{\times}}}\!
\exp\!\Big(2\pi i\,\frac{t v^{3}}{81}\Big).
\]

Using again that the cube map on \((\mathbb Z/81)^{\times}\) has kernel
size \(3\),

\[
f_{1}(t)=3\!\cdot\!3\!\!\sum_{r\in C_{1}}\!
\exp\!\Big(2\pi i\,\frac{t r}{81}\Big) ,
\]

where \(C_{1}\) is the set of cube‑residues modulo \(81\) (\(|C_{1}|=18\)).
Now

\[
\sum_{x\in(\mathbb Z/81)^{\times}}\exp\!\Big(2\pi i\,
\frac{t x}{81}\Big)=
\begin{cases}
54,&v_{3}(t)\ge4,\\[1mm]
-27,&v_{3}(t)=3,\\[1mm]
0,&v_{3}(t)\le2 .
\end{cases}
\]

Thus

\[
f_{1}(t)=
\begin{cases}
162,&v_{3}(t)\ge4,\\[2mm]
-81,&v_{3}(t)=3,\\[2mm]
0,&v_{3}(t)\le2 .
\end{cases}
\tag{3}
\]

*For \(f_{2}\).*  
Here \(x^{3}\equiv x\pmod 3\), and among the \(54\) numbers in \(U_{2}\)
exactly half are \(\equiv1\pmod3\) and half \(\equiv2\pmod3\). Hence  

\[
f_{2}(t)=27\bigl(\zeta^{t}+ \zeta^{2t}\bigr)=
\begin{cases}
-27,&v_{3}(t)=0,\\[2mm]
54,&v_{3}(t)\ge1 .
\end{cases}
\tag{4}
\]

--------------------------------------------------------------------
### 3.  The values of \(S(t)\)

From (2)–(4) (and remembering the constant term \(n_{3}=27\))

\[
S(t)=f_{0}(t)+f_{1}(t)+f_{2}(t)+27=
\begin{array}{c|c}
v_{3}(t) & S(t)\\ \hline
\infty\;(t=0) & 729\\
0 & 0\\
1 & 81\\
2 & 81\\
3 & 0\\
4 & 243\\
5 & 243\\
6 & 0
\end{array}
\]

--------------------------------------------------------------------
### 4.  Computing \(N\)

Only the classes with non‑zero \(S(t)\) contribute to (1).  
The numbers of \(t\) in each class are

\[
\begin{array}{c|c}
v_{3}(t) & \#t\\ \hline
\infty & 1\\
0 & 1458\\
1 & 486\\
2 & 162\\
3 & 54\\
4 & 18\\
5 & 6\\
6 & 2
\end{array}
\]

Hence

\[
\begin{aligned}
N&=\frac1{3^{7}}\Bigl( 729^{3}
+ (486+162) \,81^{3}
+ (18+6)\,243^{3}\Bigr)\\[2mm]
&=\frac1{3^{7}}\Bigl(3^{18}+648\cdot3^{12}+24\cdot3^{15}\Bigr)\\
&=3^{5}\cdot2025 = 243\cdot2025 = 492\,075 .
\end{aligned}
\]

--------------------------------------------------------------------
### 5.  Remainder modulo \(1000\)

\[
492\,075 \equiv 75 \pmod{1000}.
\]

--------------------------------------------------------------------
\[
\boxed{75}
\]

**Solution Overview**

The given lengths on the sides of \(\triangle ABC\) give  

\[
AD=4,\;DE=16,\;EB=8 \qquad\Longrightarrow\qquad AB=28,
\]
\[
AF=13,\;FG=52,\;GC=26 \qquad\Longrightarrow\qquad AC=91 .
\]

--------------------------------------------------------------------

### 1.  Two lines parallel to \(BC\)

\[
\frac{AD}{AB}= \frac{4}{28}= \frac{1}{7},\qquad 
\frac{AF}{AC}= \frac{13}{91}= \frac{1}{7}.
\]

Hence the points \(D\in AB\) and \(F\in AC\) divide the two sides in the same
ratio; therefore \(DF\parallel BC\).

Similarly  

\[
\frac{AE}{AB}= \frac{AD+DE}{AB}= \frac{20}{28}= \frac57,
\qquad 
\frac{AG}{AC}= \frac{AF+FG}{AC}= \frac{65}{91}= \frac57,
\]

so \(E\in AB,\;G\in AC\) also give a line parallel to \(BC\); i.e. \(EG\parallel BC\).

Thus the quadrilateral \(DEGF\) is bounded by two parallel sides \(DF,EG\) and
the two segments of the sides of the triangle, and it is the region between the
two similar triangles \(ADF\) and \(AEG\).

--------------------------------------------------------------------

### 2.  Area of \(\triangle ABC\)

If \([XYZ]\) denotes area, similarity gives  

\[
[ADF]=\Bigl(\frac{AD}{AB}\Bigr)^2[ABC]=\Bigl(\frac17\Bigr)^2[ABC]=\frac{[ABC]}{49},
\]

\[
[AEG]=\Bigl(\frac{AE}{AB}\Bigr)\Bigl(\frac{AG}{AC}\Bigr)[ABC]
      =\frac57\cdot\frac57[ABC]=\frac{25}{49}[ABC].
\]

Hence  

\[
[DEGF]=[AEG]-[ADF]=\Bigl(\frac{25}{49}-\frac1{49}\Bigr)[ABC]
       =\frac{24}{49}[ABC].
\]

Given \([DEGF]=288\),

\[
\frac{24}{49}[ABC]=288\quad\Longrightarrow\quad[ABC]=288\cdot\frac{49}{24}
                  =12\cdot49=588.
\]

--------------------------------------------------------------------

### 3.  Area of the heptagon \(AFNBCEM\)

Place the triangle with \(A=(0,0),\;B=(28,0)\) and
\(C=(91\cos\theta,\,91\sin\theta)\).
Because \([ABC]=\tfrac12\cdot28\cdot91\sin\theta\),

\[
\sin\theta=\frac{[ABC]}{1274}= \frac{588}{1274}= \frac{6}{13}.
\]

Thus  

\[
\cos\theta=\frac{\sqrt{1-\sin^2\theta}}{\,}= \frac{\sqrt{133}}{13}.
\]

The coordinates of the relevant points are

\[
\begin{aligned}
&F=(13\cos\theta,13\sin\theta)=(\sqrt{133},6), \\
&G=(65\cos\theta,65\sin\theta)=(5\sqrt{133},30),\\
&D=(4,0),\qquad
E=(20,0),\\
&M=2F-D=(2\sqrt{133}-4,12),\\
&N=2E-G=(40-5\sqrt{133},-30).
\end{aligned}
\]

Using the shoelace formula for the polygon
\(A\,(0,0),\;F,\;N,\;B,\;C,\;E,\;M\) we obtain

\[
\begin{aligned}
\sum (x_i y_{i+1}-y_i x_{i+1})
&= -240+840+1176-840+240 = 1176,\\[2mm]
\text{Area}&=\frac12\cdot1176=588.
\end{aligned}
\]

Thus the heptagon \(AFNBCEM\) has the same area as \(\triangle ABC\).

--------------------------------------------------------------------

\[
\boxed{588}
\]

Let  

\[
P:\; y=x^{2}-4
\]

and rotate \(P\) about the origin \(60^{\circ}\) counter‑clockwise.  
If \((x,y)\) is a point of \(P\) then its image \((X,Y)\) satisfies  

\[
\begin{pmatrix}X\\Y\end{pmatrix}= 
\begin{pmatrix}\cos 60^\circ & -\sin 60^\circ\\[2pt] 
\sin 60^\circ & \ \cos 60^\circ\end{pmatrix}
\begin{pmatrix}x\\y\end{pmatrix},
\qquad 
\cos60^\circ=\frac12,\;\sin60^\circ=\frac{\sqrt3}{2}.
\]

Hence  

\[
X=\frac{x}{2}-\frac{\sqrt3}{2}y,\qquad
Y=\frac{\sqrt3}{2}x+\frac{y}{2}.
\]

To obtain the equation of the rotated curve in the original \((x,y)\)–coordinates,
solve for \((x,y)\) in terms of \((X,Y)\) (rotate back by \(-60^\circ\)):

\[
x = \frac{X}{2}+\frac{\sqrt3}{2}Y,\qquad
y = -\frac{\sqrt3}{2}X+\frac{Y}{2}.
\]

Since the original points satisfy \(y=x^{2}-4\), substituting the expressions above gives the equation of the image:

\[
x^{2}+2\sqrt3\,xy+3y^{2}+2\sqrt3\,x-2y-16=0. \tag{1}
\]

--------------------------------------------------------------------
### Intersection with the original parabola

The intersection points satisfy simultaneously  

\[
y=x^{2}-4\qquad\text{and}\qquad (1).
\]

Substituting \(y=x^{2}-4\) into (1) yields

\[
3x^{4}-25x^{2}+40+2\sqrt3\,x^{3}-6\sqrt3\,x=0.
\]

Factor this quartic in the field \(\mathbb{Q}(\sqrt3)\):

\[
\begin{aligned}
3x^{4}-25x^{2}+40+2\sqrt3\,x^{3}-6\sqrt3\,x
&=(x^{2}+\sqrt3\,x-4)\bigl(3x^{2}-\sqrt3\,x-10\bigr).
\end{aligned}
\]

Thus the intersection points satisfy one of  

\[
\begin{cases}
x^{2}+\sqrt3\,x-4=0,\\[2pt]
\text{or}\\[2pt]
3x^{2}-\sqrt3\,x-10=0 .
\end{cases}
\]

--------------------------------------------------------------------
### Selecting the point in the fourth quadrant

*From* \(x^{2}+\sqrt3\,x-4=0\)

\[
x=\frac{-\sqrt3\pm\sqrt{19}}{2}.
\]

The root \(\displaystyle x=\frac{-\sqrt3+\sqrt{19}}{2}>0\) gives  

\[
y=x^{2}-4=\frac{11-\sqrt{57}}{2}-4=\frac{3-\sqrt{57}}{2}<0,
\]

so \((x,y)\) lies in the fourth quadrant.

*From* \(3x^{2}-\sqrt3\,x-10=0\)

\[
x=\frac{\sqrt3\pm\sqrt{123}}{6},
\]

which yields either \(y>0\) (first root) or \(x<0\) (second root); neither point is in the fourth quadrant.

Hence the unique fourth‑quadrant intersection point is  

\[
\Bigl(\frac{-\sqrt3+\sqrt{19}}{2}\,,\; \frac{3-\sqrt{57}}{2}\Bigr).
\]

--------------------------------------------------------------------
### Required \(y\)-coordinate

The \(y\)-coordinate is \(\displaystyle \frac{3-\sqrt{57}}{2}\).

Write it as \(\frac{a-\sqrt{b}}{c}\) with \(a,b,c\) positive integers and \(\gcd(a,c)=1\):
\[
a=3,\qquad b=57,\qquad c=2.
\]

Therefore  

\[
a+b+c = 3+57+2 = 62.
\]

\[
\boxed{62}
\]

Let the line containing \(A,B,C,D,E,F\) be the \(x\)-axis.

\[
A=(0,0),\qquad B=(b,0),\qquad C=(c,0),\qquad D=(d,0),\qquad 
E=(e,0),\qquad F=(f,0)
\]

Because the points are in the order \(A,B,C,D,E,F\),

\[
c-b=AC-AB,\; d-b=BD,\; e-c=CE,\; f-d=DF,\; f-0=AF.
\]

Take \(A\) at the origin.  From the data

\[
AC=26\implies c=26,
\]
\[
BD=22\implies d=b+22,
\]
\[
CE=31\implies e=c+31=57,
\]
\[
DF=33\implies f=d+33=b+55,
\]
\[
AF=73\implies f=73.
\]

Hence \(b+55=73\Rightarrow b=18\).  
Thus the coordinates are  

\[
B=(18,0),\; C=(26,0),\; D=(40,0),\; E=(57,0),\; F=(73,0).
\]

---

### Locate \(G\)

\(G\) is off the line and satisfies  

\[
CG=40,\qquad DG=30.
\]

Let \(G=(x,y)\) with \(y\neq0\). Then  

\[
(x-26)^2+y^{2}=40^{2}=1600,\tag{1}
\]
\[
(x-40)^2+y^{2}=30^{2}=900.\tag{2}
\]

Subtract (2) from (1):

\[
(x-26)^2-(x-40)^2=700.
\]

Expanding,

\[
[x^{2}-52x+676]-[x^{2}-80x+1600]=28x-924=700,
\]
\[
28x=1624\quad\Longrightarrow\quad x=58.
\]

Plugging \(x=58\) into (2):

\[
(58-40)^{2}+y^{2}=900\;\Longrightarrow\;18^{2}+y^{2}=900
\]
\[
y^{2}=576\;\Longrightarrow\;|y|=24.
\]

So \(G=(58,\,24)\) (the point \((58,-24)\) gives the same area).

---

### Area of \(\triangle BGE\)

Points \(B\) and \(E\) lie on the \(x\)-axis, so the base \(BE\) has length

\[
BE = e-b = 57-18 = 39.
\]

The altitude from \(G\) to the line \(BE\) (the \(x\)-axis) is \(|y| = 24\).

Hence  

\[
\text{Area} = \frac12 \times BE \times \text{height}
= \frac12 \times 39 \times 24 = 468.
\]

\[
\boxed{468}
\]

We need the positive integers \(n\) such that  

\[
n+2\mid 3(n+3)(n^{2}+9).
\]

---

### 1.  Replace \(n+2\) by a new variable  

Let  

\[
k=n+2\qquad (k\ge 3), \qquad n=k-2 .
\]

Then  

\[
\begin{aligned}
n+3 &= (k-2)+3 = k+1,\\[2pt]
n^{2}+9 &= (k-2)^{2}+9 = k^{2}-4k+13 .
\end{aligned}
\]

Hence  

\[
3(n+3)(n^{2}+9)=3(k+1)(k^{2}-4k+13).
\]

The condition becomes  

\[
k \mid 3(k+1)(k^{2}-4k+13).
\]

---

### 2.  Reduce modulo \(k\)

Modulo \(k\),

\[
k+1\equiv 1,\qquad k^{2}-4k+13\equiv 13,
\]

because \(k\equiv 0\) gives \(k^{2}\equiv 0\) and \(-4k\equiv 0\).

Thus  

\[
3(k+1)(k^{2}-4k+13)\equiv 3\cdot 1\cdot 13=39 \pmod{k}.
\]

Hence the divisibility condition is equivalent to  

\[
k\mid 39.
\]

---

### 3.  Find admissible \(k\) and corresponding \(n\)

The positive divisors of \(39\) are \(1,3,13,39\).
Since \(k=n+2\ge 3\), we keep  

\[
k\in \{3,13,39\}.
\]

Therefore  

\[
n=k-2\in\{1,11,37\}.
\]

All three values satisfy the original condition (a quick check confirms the division).

---

### 4.  Sum of all such \(n\)

\[
1+11+37 = 49.
\]

\[
\boxed{49}
\]

The inequalities

\[
x-yz<y-zx<z-xy
\]

with the plane \(x+y+z=75\) can be simplified.  
Write \(z=75-x-y\). Then

\[
\begin{aligned}
y-zx-(x-yz) &= (y-x)(1+z)= (y-x)(76-x-y),\\
z-xy-(y-zx) &= (z-y)(1+x)= (75-x-2y)(1+x).
\end{aligned}
\]

Thus the points on the plane satisfy  

\[
(y-x)(76-x-y)>0,\qquad (75-x-2y)(1+x)>0 .
\tag{1}
\]

Define in the \(xy\)-plane the lines  

\[
\begin{aligned}
L_1 &: y=x,\\
L_2 &: x+y=76,\\
L_3 &: x=-1,\\
L_4 &: x+2y=75\;(y=(75-x)/2).
\end{aligned}
\]

From (1) we obtain two possibilities for each product, giving four possible
intersections of half‑planes.  One of these intersections is empty; the other three
are convex.  The region for which both products in (1) are **positive** is

\[
\begin{cases}
y>x,\\
x+y<76,\\
y<(75-x)/2,\\
x>-1,
\end{cases}
\]

which in the \(xy\)-plane is the interior of the triangle with vertices

\[
A(-1,-1),\qquad B(-1,38),\qquad C(25,25).
\]

The other two regions are unbounded.  Hence the finite region is this triangle.

---

### Area in the \(xy\)-plane

\[
\text{Area}_{xy}= \frac12\bigl|AB\times AC\bigr|
               =\frac12\bigl| (0,39)\times (26,26)\bigr|
               =\frac12\cdot1014 = 507 .
\]

---

### Area on the plane \(x+y+z=75\)

The plane is given by \(z=75-x-y\).  Its area element satisfies  

\[
dS = \sqrt{1+(\partial_xz)^2+(\partial_yz)^2}\;dx\,dy
    =\sqrt{1+1+1}\;dx\,dy =\sqrt3\,dx\,dy .
\]

Hence  

\[
\text{Area}_{\text{plane}} = \sqrt3\;\text{Area}_{xy}
                            = 507\sqrt3 .
\]

So the area can be written as \(a\sqrt b\) with \(a=507,\ b=3\).
Therefore  

\[
a+b = 507+3 = 510 .
\]

\[
\boxed{510}
\]

The intersection points satisfy  

\[
y=f(x),\qquad x=34y^{2}\;( \ge 0).
\]

Because \(f\) has period \(4\),

\[
f(x)=\begin{cases}
x-4k ,&x\in[-1+4k,\,1+4k),\\[2mm]
2+4k-x ,&x\in[1+4k,\,3+4k),
\end{cases}\qquad k\in\mathbb Z .
\]

--------------------------------------------------------------------
### 1.  Equations for the two linear pieces  

*Ascending piece*  \((x-4k)\):  

\[
y=x-4k,\qquad x=34y^{2}
\Longrightarrow 34y^{2}-y-4k=0. \tag{1}
\]

*Descending piece*  \((2+4k-x)\):  

\[
y=2+4k-x,\qquad x=34y^{2}
\Longrightarrow 34y^{2}+y-(2+4k)=0. \tag{2}
\]

Both are quadratics in \(y\); each can give at most two real roots.

--------------------------------------------------------------------
### 2.  Which integers \(k\) give admissible roots?

Because \(|y|\le 1\) (the range of \(f\)), the roots must lie in \([-1,1]\).

*For (1):*  
The sum of the two roots equals \(\frac{1}{34}\); the product is \(-\frac{4k}{34}\).  
The discriminant must be non‑negative:

\[
\Delta_1=1+544k\ge 0\Longrightarrow k\ge0 .
\]

For a root to be in \([-1,1)\) we also need  

\[
-1\le\frac{1\pm\sqrt{1+544k}}{68}<1 .
\]

The last inequality forces \(k\le8\).  Hence (1) yields real admissible roots for  

\[
k=0,1,\dots ,8 .
\]

Both roots are in \([-1,1)\) for each of these values.

*For (2):*  
\[
\Delta_2=273+544k\ge0\Longrightarrow k\ge0 .
\]

The condition \(-1<y\le 1\) gives \(k\le8\).  
For \(k=0,\dots ,7\) both roots lie in \((-1,1]\); for \(k=8\) only the “\(+\)” root
\[
y=\frac{-1+\sqrt{273+544\cdot8}}{68}
      =\frac{-1+5\sqrt{185}}{68}
\]
remains in the interval (the “\(-\)” root is \(<-1\)).

Thus  

* (1) contributes \(2\) solutions for each \(k=0,\dots ,8\)  \(\Rightarrow 18\) points;  
* (2) contributes \(2\) solutions for each \(k=0,\dots ,7\) and \(1\) solution for \(k=8\)  \(\Rightarrow 17\) points.

Hence there are \(35\) intersection points altogether.

--------------------------------------------------------------------
### 3.  Sum of the \(y\)-coordinates  

For each fixed \(k\) the sum of the two roots of (1) is, by Vieta,

\[
\Sigma_{A}(k)=\frac{1}{34}.
\]

So the total contribution of (1) is  

\[
\sum_{k=0}^{8}\Sigma_{A}(k)=9\cdot\frac{1}{34}= \frac{9}{34}.
\]

For (2), when both roots exist their sum is  

\[
\Sigma_{B}(k)=-\frac{1}{34}.
\]

Thus for \(k=0,\dots ,7\) the total is  

\[
8\cdot\Bigl(-\frac{1}{34}\Bigr)=-\frac{8}{34}.
\]

For \(k=8\) only the positive root remains:

\[
y_{8}= \frac{-1+5\sqrt{185}}{68}.
\]

Adding everything,

\[
\begin{aligned}
\text{Sum of all }y
&=\frac{9}{34}-\frac{8}{34}+y_{8}
   =\frac{1}{34}+ \frac{-1+5\sqrt{185}}{68}\\[2mm]
&= \frac{2}{68}+ \frac{-1+5\sqrt{185}}{68}
   =\frac{1+5\sqrt{185}}{68}.
\end{aligned}
\]

--------------------------------------------------------------------
### 4.  Final answer  

The sum is \(\displaystyle\frac{1+5\sqrt{185}}{68}\).  
Thus  

\[
a=1,\; b=5,\; c=185,\; d=68,\qquad
a+b+c+d=1+5+185+68=259.
\]

\[
\boxed{259}
\]

Let the three rows be \(R_1,R_2,R_3\) and the three \(3\times3\) blocks be \(B_1,B_2,B_3\) (columns \(1\!-\!3,\;4\!-\!6,\;7\!-\!9\)).
Each row must be a permutation of \(\{1,\dots ,9\}\) and each block must also contain every digit once.

--------------------------------------------------------------------
### 1.  Where does a digit appear in the three rows?

For a fixed digit \(d\) its three copies (one in each row) must lie in three *different* blocks; otherwise a block would miss that digit.  
Thus for each digit we obtain a bijection  

\[
\phi_d:\{1,2,3\}\longrightarrow\{1,2,3\},\qquad 
\phi_d(i)=\text{block containing }d\text{ in row }i .
\]

So each digit corresponds to a permutation of the three blocks.  
There are six permutations, listed with the cells \((i,\phi(i))\) they occupy:

\[
\begin{array}{c|c}
\text{perm.} & \text{cells}\\ \hline
(1)(2)(3) & (1,1),(2,2),(3,3)\\
(12) & (1,2),(2,1),(3,3)\\
(13) & (1,3),(2,2),(3,1)\\
(23) & (1,1),(2,3),(3,2)\\
(123) & (1,2),(2,3),(3,1)\\
(132) & (1,3),(2,1),(3,2)
\end{array}
\]

Let \(x_1,\dots ,x_6\) be the numbers of digits that use the six permutations (in the order shown).  
Because each block must contain three digits from each row, each of the nine cells \((i,k)\) must be hit by exactly three digits, giving

\[
\begin{aligned}
x_1+x_4 &=3, & x_2+x_5 &=3, & x_3+x_6 &=3,\\
x_2+x_6 &=3, & x_1+x_3 &=3, & x_4+x_5 &=3,\\
x_3+x_5 &=3, & x_4+x_6 &=3, & x_1+x_2 &=3 .
\end{aligned}
\]

Solving, all solutions have the form  

\[
(x_1,x_2,x_3,x_4,x_5,x_6)=(a,\,3-a,\,3-a,\,3-a,\,a,\,a),\qquad a\in\{0,1,2,3\}.
\]

--------------------------------------------------------------------
### 2.  Assign the digits to the permutations

For a fixed \(a\) the number of ways to choose which digits get which permutation is

\[
\frac{9!}{x_1!\,x_2!\,x_3!\,x_4!\,x_5!\,x_6!}
   =\frac{9!}{a!^{\,3}\,(3-a)!^{\,3}} .
\]

Summing over the four possible values of \(a\),

\[
\sum_{a=0}^3\frac{9!}{a!^{\,3}(3-a)!^{\,3}}
   =9!\Bigl(\frac1{3!^{\,3}}{0\choose0}^3
            +\frac1{3!^{\,3}}{3\choose1}^3
            +\frac1{3!^{\,3}}{3\choose2}^3
            +\frac1{3!^{\,3}}{3\choose3}^3\Bigr)
   =9!\,\frac{56}{216}
   =9!\,\frac{7}{27}.
\]

--------------------------------------------------------------------
### 3.  Order the digits inside each block

Within a fixed block and a fixed row there are three column positions.
The three digits that belong to that macro‑cell can be placed in these
positions in \(3! = 6\) ways.
There are \(3\) rows \(\times\) \(3\) blocks \(=9\) such macro‑cells, giving a factor \(6^9\).

--------------------------------------------------------------------
### 4.  Total number of fillings

\[
N = \Bigl(9!\,\frac{7}{27}\Bigr)\,6^{9}
   = 2^{16}\,3^{10}\,5\,7^{2}.
\]

Thus the factorisation uses the four distinct primes  

\[
p=2,\;q=3,\;r=5,\;s=7,\qquad 
a=16,\;b=10,\;c=1,\;d=2 .
\]

Finally,

\[
p\,a+q\,b+r\,c+s\,d
   =2\cdot16+3\cdot10+5\cdot1+7\cdot2
   =32+30+5+14=81.
\]

\[
\boxed{81}
\]

We place the pentagon in the plane.  
Set  

\[
\begin{aligned}
A&=(7,7\sqrt3),\qquad B=(0,0),\qquad C=(7,0),\\[2mm]
D&=\Bigl(\frac{205}{7},\frac{36\sqrt3}{7}\Bigr),\qquad 
E=\Bigl(\frac{218}{7},\frac{88\sqrt3}{7}\Bigr).
\end{aligned}
\]

The lengths are as given and  
\(\angle B=\angle E =60^\circ\).

--------------------------------------------------------------------
### 1.  Geometry of the optimal point  

For the sum  

\[
f(X)=AX+BX+CX+DX+EX,
\]

the geometric median \(X\) satisfies  

\[
\frac{X-A}{AX}+\frac{X-B}{BX}+\frac{X-C}{CX}+\frac{X-D}{DX}+\frac{X-E}{EX}=0 .
\tag{1}
\]

Because \(B,E\) lie on a line, the vectors \(\frac{X-B}{BX}\) and
\(\frac{X-E}{EX}\) are opposite exactly when \(X\) lies on segment \(BE\);
hence in the optimal position \(X\) must lie on \(BE\) and the first two
terms in (1) cancel.  Consequently

\[
\frac{X-A}{AX}+\frac{X-C}{CX}+\frac{X-D}{DX}=0,
\]

so \(X\) is the Fermat point of triangle \(ACD\).

--------------------------------------------------------------------
### 2.  Fermat point of \(\triangle ACD\)

The side lengths of \(\triangle ACD\) are  

\[
AC=7\sqrt3,\qquad CD=24,\qquad AD=13\sqrt3 .
\]

All its angles are \(<120^\circ\); therefore the Fermat point exists.
For a triangle with sides \(a,b,c\) and area \(\Delta\),

\[
S_{\text{Fermat}}=
\sqrt{\frac{a^{2}+b^{2}+c^{2}+4\sqrt3\,\Delta}{2}} .
\]

Here  

\[
a^{2}+b^{2}+c^{2}=576+507+147=1230,
\qquad\Delta_{ACD}=78\sqrt3,
\]

so  

\[
S_{\text{Fermat}}=\sqrt{\frac{1230+4\sqrt3\cdot78\sqrt3}{2}}
=\sqrt{\frac{1230+936}{2}}=\sqrt{1083}=19\sqrt3 .
\]

Thus at the Fermat point  

\[
AX+CX+DX=19\sqrt3 .
\tag{2}
\]

--------------------------------------------------------------------
### 3.  The total minimum

Since the optimal point lies on \(BE\),

\[
BX+EX = BE .
\]

From the coordinates, \(BE=38\).  Combining this with (2),

\[
f_{\min}= (AX+CX+DX)+(BX+EX)=19\sqrt3+38 .
\]

Hence the smallest possible value of \(f(X)\) is  

\[
\boxed{38+19\sqrt3}.
\]

Writing it as \(m+n\sqrt p\) with \(m,n\) integers and \(p\) square‑free,
we have \(m=38,\;n=19,\;p=3\).  Therefore  

\[
m+n+p=38+19+3=60 .
\]

\[
\boxed{60}
\]

\[
P=\prod_{k=4}^{63}\frac{\log_k\!\bigl(5^{k^{2}-1}\bigr)}{\log_{k+1}\!\bigl(5^{k^{2}-4}\bigr)}
\]

---

### 1.  Simplify each factor  

\[
\log_k\!\bigl(5^{k^{2}-1}\bigr)=(k^{2}-1)\log_k5,
\qquad 
\log_{k+1}\!\bigl(5^{k^{2}-4}\bigr)=(k^{2}-4)\log_{k+1}5 .
\]

Hence  

\[
\frac{\log_k\!\bigl(5^{k^{2}-1}\bigr)}{\log_{k+1}\!\bigl(5^{k^{2}-4}\bigr)}
=\frac{k^{2}-1}{k^{2}-4}\cdot\frac{\log_k5}{\log_{k+1}5}
=\frac{(k-1)(k+1)}{(k-2)(k+2)}\cdot\frac{\log_k5}{\log_{k+1}5}.
\]

Therefore  

\[
P=\underbrace{\prod_{k=4}^{63}\frac{(k-1)(k+1)}{(k-2)(k+2)}}_{A}
   \times\underbrace{\prod_{k=4}^{63}\frac{\log_k5}{\log_{k+1}5}}_{B}.
\]

---

### 2.  Evaluate the rational product \(A\)

\[
A=\Bigl(\prod_{k=4}^{63}\frac{k-1}{k-2}\Bigr)
   \Bigl(\prod_{k=4}^{63}\frac{k+1}{k+2}\Bigr)
   =\frac{3\cdot4\cdots62}{2\cdot3\cdots61}\;
     \frac{5\cdot6\cdots64}{6\cdot7\cdots65}.
\]

Both products telescope:

\[
\frac{3\cdot4\cdots62}{2\cdot3\cdots61}= \frac{62}{2}=31,
\qquad 
\frac{5\cdot6\cdots64}{6\cdot7\cdots65}= \frac{5}{65}= \frac1{13}.
\]

Thus  

\[
A=31\cdot\frac1{13}= \frac{31}{13}.
\]

---

### 3.  Evaluate the logarithmic product \(B\)

\[
B=\frac{\log_4 5}{\log_5 5}\cdot\frac{\log_5 5}{\log_6 5}\cdots
    \frac{\log_{63}5}{\log_{64}5}
   =\frac{\log_4 5}{\log_{64}5}.
\]

Using change‑of‑base,
\[
\frac{\log_4 5}{\log_{64} 5}
   =\frac{\frac{\ln5}{\ln4}}{\frac{\ln5}{\ln64}}
   =\frac{\ln64}{\ln4}
   =\log_4 64=3,
\]
since \(4^3=64\).

Hence \(B=3\).

---

### 4.  Combine

\[
P=A\cdot B=\frac{31}{13}\times 3=\frac{93}{13}.
\]

The fraction \(\dfrac{93}{13}\) is already in lowest terms, so  
\(m=93,\; n=13\).

\[
m+n=93+13=106.
\]

\[
\boxed{106}
\]

The divisors of  

\[
2025=3^{4}\cdot5^{2}
\]

are exactly the numbers  

\[
3^{a}5^{b}\qquad (0\le a\le 4,\;0\le b\le 2).
\]

Hence  

\[
|A|=(4+1)(2+1)=15 .
\]

For a subset \(B\subseteq A\) the least common multiple is

\[
\operatorname{lcm}(B)=3^{\max a}\,5^{\max b},
\]

where the maxima are taken over the exponents \((a,b)\) of the elements of \(B\).

Thus \(\operatorname{lcm}(B)=2025=3^4 5^2\) iff  

* at least one element of \(B\) has a \(3\)-exponent \(4\);
* at least one element of \(B\) has a \(5\)-exponent \(2\).

Define  

\[
D_3=\{d\in A : \text{the exponent of }3\text{ in }d\text{ is }4\},
\qquad 
D_5=\{d\in A : \text{the exponent of }5\text{ in }d\text{ is }2\}.
\]

Counting:

* \(D_3\): \(a=4\) and \(b=0,1,2\) ⇒ \(|D_3|=3\) (81, 405, 2025).
* \(D_5\): \(b=2\) and \(a=0,1,2,3,4\) ⇒ \(|D_5|=5\) (25, 75, 225, 675, 2025).
* The intersection is \(\{2025\}\); so \(|D_3\cap D_5|=1\).

We need subsets that contain at least one element of \(D_3\) **and** at least one element of \(D_5\).

Let the total number of subsets of \(A\) be \(2^{15}\).
Let  

\[
\begin{aligned}
A_1 &=\{\text{subsets that contain no element of } D_3\},\\
A_2 &=\{\text{subsets that contain no element of } D_5\}.
\end{aligned}
\]

Then  

\[
|A_1|=2^{15-3}=2^{12},\qquad 
|A_2|=2^{15-5}=2^{10}.
\]

Subsets avoiding **both** \(D_3\) and \(D_5\) avoid the union \(D_3\cup D_5\) which has  

\[
|D_3\cup D_5|=|D_3|+|D_5|-|D_3\cap D_5|=3+5-1=7
\]

elements. Hence  

\[
|A_1\cap A_2|=2^{15-7}=2^{8}.
\]

By inclusion–exclusion, the number of subsets that contain at least one element from each of \(D_3\) and \(D_5\) is  

\[
\begin{aligned}
N &= 2^{15}-|A_1|-|A_2|+|A_1\cap A_2|\\
   &= 2^{15}-2^{12}-2^{10}+2^{8}\\
   &= 32768-4096-1024+256\\
   &= 27904.
\end{aligned}
\]

All these subsets are non‑empty, so the required probability is  

\[
\frac{N}{2^{15}}=\frac{27904}{32768}.
\]

Simplifying,

\[
\frac{27904}{32768}= \frac{27904\div 256}{32768\div 256}
                 = \frac{109}{128}.
\]

Thus \(m=109,\; n=128\) and  

\[
m+n = 109+128 = 237.
\]

\[
\boxed{237}
\]

Label the twelve unit segments as in the picture  

\[
\begin{array}{c}
\text{external: }a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2\\[2mm]
\text{internal: }e_1,e_2,f_1,f_2
\end{array}
\]

The four unit squares are  

\[
\begin{aligned}
\text{TL}&:\;a_1,c_1,e_1,f_1,\\
\text{TR}&:\;b_1,c_2,e_1,f_2,\\
\text{BL}&:\;a_2,d_1,e_2,f_1,\\
\text{BR}&:\;b_2,d_2,e_2,f_2 .
\end{aligned}
\]

Each square must contain exactly two red sides (hence also two blue sides).

--------------------------------------------------------------------
### 1.  Reduce to the four internal edges

For a given square the two sides that are **internal** are already
colored once we decide the colors of the four internal edges  
\(e_1,e_2,f_1,f_2\).  

Let  

\[
r_{\rm TL}=e_1+f_1,\qquad
r_{\rm TR}=e_1+f_2,\qquad
r_{\rm BL}=e_2+f_1,\qquad
r_{\rm BR}=e_2+f_2 ,
\]

where a red edge contributes 1 and a blue edge contributes 0.
If a square has \(r\) red internal sides, then it must have \(2-r\)
red external sides. Thus

* if \(r=0\): both external sides are forced red – 1 way;
* if \(r=1\): exactly one of the two external sides must be red – 2 ways;
* if \(r=2\): both external sides are forced blue – 1 way.

Hence for a fixed choice of the four internal edges the number of
admissible colourings of the eight external edges equals

\[
2^{\,N_1},
\]

where \(N_1\) is the number of squares among TL, TR, BL, BR that have
exactly one red internal side.

--------------------------------------------------------------------
### 2.  Express \(N_1\) in terms of the internal edges

Write  

\[
X_1=e_1,\; X_2=e_2,\qquad Y_1=f_1,\;Y_2=f_2\qquad(\text{red}=1,\text{blue}=0).
\]

For a square \((i,j)\) (\(i,j\in\{1,2\}\)) the number of red internal
sides is \(X_i+Y_j\).
Thus a square has exactly one red internal side precisely when
\(X_i\neq Y_j\).

Let  

\[
a=X_1+X_2\;(=0,1,2),\qquad b=Y_1+Y_2\;(=0,1,2).
\]

The number of ordered pairs \((i,j)\) with \(X_i\neq Y_j\) is  

\[
N_1=a(2-b)+(2-a)b
      =2\bigl(a+b-ab\bigr).
\]

Consequently the contribution of a configuration with given
\(a,b\) is  

\[
2^{N_1}=2^{\,2(a+b-ab)}=4^{\,a+b-ab}.
\]

--------------------------------------------------------------------
### 3.  Count all configurations of the internal edges

There are \(\binom{2}{a}\) ways to choose which vertical internal edges
are red, and \(\binom{2}{b}\) ways to choose which horizontal internal
edges are red.  Hence the total number of colourings is

\[
\sum_{a=0}^{2}\sum_{b=0}^{2}
\binom{2}{a}\binom{2}{b}\;4^{\,a+b-ab}.
\]

Evaluating the double sum:

\[
\begin{array}{c|ccc}
a\backslash b &0&1&2\\\hline
0 &1\cdot1\cdot1=1 &1\cdot2\cdot4=8 &1\cdot1\cdot16=16\\
1 &2\cdot1\cdot4=8 &2\cdot2\cdot4=16 &2\cdot1\cdot4=8\\
2 &1\cdot1\cdot16=16 &1\cdot2\cdot4=8 &1\cdot1\cdot1=1
\end{array}
\]

Adding all entries:

\[
1+8+16+8+16+8+16+8+1=82.
\]

--------------------------------------------------------------------
### 4.  Answer

Thus there are  

\[
\boxed{82}
\]

different ways to colour the twelve unit segments so that each of the
four unit squares has two red sides and two blue sides.

The circles are drawn as in the figure.

* Let \(O\) be the centre of \(\omega _2\) (radius \(15\)).
* Let \(A\) be the centre of \(\omega _1\) (radius \(6\)).
* Since \(\omega _1\) is internally tangent to \(\omega _2\) at \(B\), the three points are collinear and  

\[
OA=15-6=9 .
\]

Take a coordinate system with  

\[
O=(0,0),\qquad A=(9,0),\qquad B=(15,0).
\]

--------------------------------------------------------------------
### Points \(C\) and \(D\)

\(BC\) is a diameter of \(\omega _2\); hence \(C\) is the antipode of \(B\):

\[
C=(-15,0).
\]

\(BC\perp AD\) means \(AD\) is the vertical line \(x=9\).  
Intersecting this line with \(\omega _2\) gives  

\[
9^2+y^2=15^2 \Longrightarrow y=\pm 12 .
\]

Because the later condition “\(D\) is nearer to \(FG\) than to \(EH\)” forces \(D\) to lie **below** the centre, we take  

\[
D=(9,-12).
\]

--------------------------------------------------------------------
### The rectangle \(EFGH\)

The rectangle is inscribed in \(\omega _1\) and \(\overline{EF}\perp BC\); hence \(\overline{EF}\) is vertical.
Let

\[
\begin{aligned}
E&=(9+w,\;h),\\[2pt]
F&=(9+w,\;-h),\\[2pt]
G&=(9-w,\;-h),\\[2pt]
H&=(9-w,\;h),
\end{aligned}
\]

so that the centre of the rectangle is \(A\) (the intersection of the diagonals of an
inscribed rectangle is the centre of the circumcircle).

Because the four vertices lie on \(\omega _1\),

\[
(9\pm w-9)^2+h^2=w^{2}+h^{2}=6^{2}=36\qquad\Longrightarrow\qquad w^{2}+h^{2}=36. \tag{1}
\]

The side \(\overline{EF}\) is the right vertical side, thus
\(C\) (which is far left) is nearer to \(\overline{GH}\) (the left side) than to \(\overline{EF}\);
and \(D\) (which is below the centre) is nearer to \(\overline{FG}\) (the bottom side) than to \(\overline{EH}\) (the top side).  
These statements only fix the labeling of the rectangle.

--------------------------------------------------------------------
### Equality of the two triangular areas

\[
\begin{aligned}
\text{Area}\,\triangle DGF
&=\frac12\bigl(\underbrace{2w}_{\text{base }GF}\bigr)
     \bigl(\underbrace{12-h}_{\text{vertical height}}\bigr)
 = w(12-h),\\[4pt]
\text{Area}\,\triangle CHG
&=\frac12\bigl(\underbrace{2h}_{\text{base }HG}\bigr)
     \bigl(\underbrace{24-w}_{\text{horizontal distance}}\bigr)
 = h(24-w).
\end{aligned}
\]

Given that the two areas are equal,

\[
w(12-h)=h(24-w)\; \Longrightarrow\; 12w=24h\; \Longrightarrow\; w=2h. \tag{2}
\]

--------------------------------------------------------------------
### Solving for \(w\) and \(h\)

Substituting \(w=2h\) into (1):

\[
(2h)^2+h^{2}=36\;\Longrightarrow\;5h^{2}=36\;\Longrightarrow\;h^{2}= \frac{36}{5}.
\]

Hence  

\[
h=\frac{6}{\sqrt5},\qquad w=2h=\frac{12}{\sqrt5}.
\]

--------------------------------------------------------------------
### Area of the rectangle

The rectangle’s side lengths are \(2w\) and \(2h\), so

\[
\text{Area}= (2w)(2h)=4wh
          =4\left(\frac{12}{\sqrt5}\right)\left(\frac{6}{\sqrt5}\right)
          =4\cdot\frac{72}{5}
          =\frac{288}{5}.
\]

Thus the area is \(\displaystyle\frac{m}{n}=\frac{288}{5}\) with relatively prime
\(m=288,\;n=5\).

\[
m+n=288+5=293.
\]

\[
\boxed{293}
\]

Let  

\[
N=25q+r ,\qquad 0\le r\le 24 .
\]

The greedy algorithm takes  

* \(q\) quarters,  
* \(\displaystyle d_0=\Big\lfloor\frac r{10}\Big\rfloor\) dimes,  
* \(s_0=r\bmod 10\) pennies.  

Hence  

\[
C_{\text{greedy}}=q+d_0+s_0 .\tag{1}
\]

--------------------------------------------------------------------
### 1.  When can a better solution use fewer quarters?

Suppose we use only \(q-k\) quarters, \(k\ge1\).  
The remaining amount is  

\[
R=r+25k .
\]

With only dimes and pennies the optimal way to make \(R\) is  

\[
\Big\lfloor\frac{R}{10}\Big\rfloor\text{ dimes }+ (R\bmod10)\text{ pennies}.
\]

Thus the total number of coins for this choice is  

\[
C(q-k)=(q-k)+\Big\lfloor\frac{r+25k}{10}\Big\rfloor+(r+25k\bmod10).\tag{2}
\]

Let  

\[
\Delta(k)=C(q-k)-C_{\text{greedy}} .
\]

Using \(r=10t+a\;(0\le a\le9,\;t\in\{0,1,2\})\) we obtain after a short calculation  

\[
\Delta(k)=6k-9\Big\lfloor\frac{a+5k}{10}\Big\rfloor .\tag{3}
\]

(Only the last digit \(a=r\bmod 10\) matters.)

--------------------------------------------------------------------
### 2.  Sign of \(\Delta(k)\)

*If \(a\ge5\):*  
Take \(k=1\).  Then \(\lfloor\frac{a+5}{10}\rfloor=1\) and  

\[
\Delta(1)=6-9=-3<0 .
\]

Hence a better solution exists (remove one quarter).

*If \(a\le4\):*  
For any \(k\ge1\),

\[
\Delta(k)=6k-9\Big\lfloor\frac{a+5k}{10}\Big\rfloor
\ge 6k-9\frac{a+5k}{10}=1.5k-0.9a .
\]

Since \(a\le4\) this is positive for \(k=1,2\) (indeed \(\Delta(1)=6,\,\Delta(2)=3\)) and for larger \(k\) the term \(1.5k\) dominates, so \(\Delta(k)\ge0\) for all \(k\).

Therefore **greedy fails iff**  

\[
a=r\bmod10\ge5\quad\text{and}\quad q\ge1 .
\]

Equivalently, for \(N\ge25\),

\[
N\bmod25\in\{5,6,7,8,9,15,16,17,18,19\}.
\]

--------------------------------------------------------------------
### 3.  Counting the failures

The residues above are ten out of the twenty‑five possible residues modulo \(25\).  

In the interval \(1\le N\le1000\) we have \(1000/25=40\) full periods, so each residue occurs \(40\) times.  

Thus the total number of integers with those residues is  

\[
10\times40=400.
\]

But the ten numbers \(5,6,7,8,9,15,16,17,18,19\) are \< 25, and for them greedy cannot remove a quarter (there are no quarters in any representation).  Those ten are **not** failures.  

Hence the genuine failures are  

\[
400-10=390.
\]

--------------------------------------------------------------------
### 4.  Successes

There are \(1000\) integers from \(1\) to \(1000\).  Therefore the number of values for which the greedy algorithm *succeeds* is

\[
1000-390=610.
\]

--------------------------------------------------------------------
\[
\boxed{610}
\]

\[
f(x)=\sin\bigl(7\pi\sin(5x)\bigr)
\]

---

### 1.  Zeros of \(f\)

\(f(x)=0\iff 7\pi\sin(5x)=k\pi\) for an integer \(k\).

\[
\sin(5x)=\frac{k}{7},\qquad |k|\le 7 .
\]

Let  

\[
y=5x\;(0<y<10\pi).
\]

For each integer \(k\) with \(|k|\le7\) we count the solutions of \(\sin y=\dfrac{k}{7}\) in \((0,10\pi)\).

* \(k=7\)  : \(\sin y=1\).  Solutions \(y=\frac{\pi}{2}+2\pi n\) \((n=0,1,2,3,4)\) → **5** solutions.  
* \(k=-7\) : \(\sin y=-1\).  Solutions \(y=\frac{3\pi}{2}+2\pi n\) \((n=0,1,2,3,4)\) → **5** solutions.  
* \(k=0\)  : \(\sin y=0\).  Solutions \(y=m\pi\;(m=1,2,\dots ,9)\) → **9** solutions.  
* For each \(k=\pm1,\pm2,\pm3,\pm4,\pm5,\pm6\) we have \(\sin y=c\) with \(-1<c<1,\;c\neq0\).  
  In each of the 5 full periods of \(\sin y\) there are exactly two such points, so **\(5\cdot2=10\)** solutions per \(k\).

There are 12 values of \(k\) with \(|k|\in\{1,\dots ,6\}\), giving  

\[
12\times10=120\text{ solutions}.
\]

Adding all contributions,

\[
n=5+5+9+120=139 .
\]

---

### 2.  Tangent zeros

The graph is tangent to the \(x\)-axis at a zero where the derivative also vanishes.

\[
f'(x)=\cos\bigl(7\pi\sin(5x)\bigr)\cdot 35\pi\cos(5x).
\]

At a zero, \(\sin\bigl(7\pi\sin(5x)\bigr)=0\) gives \(\sin(5x)=\dfrac{k}{7}\).  
Because \(\cos\bigl(7\pi\sin(5x)\bigr)=\pm1\neq0\), we must have \(\cos(5x)=0\).

\[
\cos(5x)=0\Longrightarrow \sin(5x)=\pm1 .
\]

Thus tangential zeros occur exactly when \(\sin(5x)=\pm1\), i.e. for \(k=\pm7\).

We already counted those solutions:  

* \(\sin(5x)=1\) → 5 points,  
* \(\sin(5x)=-1\) → 5 points.

Hence  

\[
t=5+5=10 .
\]

---

### 3.  Required sum

\[
n+t=139+10=149.
\]

\[
\boxed{149}
\]

Let the vertices of the regular \(24\)-gon be numbered \(0,1,\dots ,23\) (mod \(24\)).
If a segment joins vertex \(i\) to vertex \(j\) we call its **step**
\[
k\equiv j-i\pmod{24},\qquad 1\le k\le 12 .
\]

The length of a chord depends only on this step:
the chord of step \(k\) has length  

\[
\ell(k)=2R\sin\frac{\pi k}{24},\qquad 1\le k\le 12 .
\]

Since \(\sin(\theta )\) is strictly increasing on \([0,\frac{\pi}{2}]\),
the lengths \(\ell(1),\ell(2),\dots ,\ell(12)\) are all distinct.
Thus a drawing in which all 12 segments have the same length must use
edges whose step is a fixed \(k\) (the same chord length).

--------------------------------------------------------------------

### 1.  The graph of step \(k\)

Consider the undirected graph \(G_k\) whose vertices are the 24 polygon vertices
and whose edges are all chords of step \(k\) (i.e. the pairs \(\{i,i+k\}\)).
Each vertex of \(G_k\) is incident to two such chords
(\(i\) connects to \(i+k\) and to \(i-k\)), so \(G_k\) is a disjoint union of
cycles.

Let  

\[
d=\gcd(24,k).
\]

Then the vertices split into \(d\) cycles, each of length  

\[
\frac{24}{d}.
\]

--------------------------------------------------------------------

### 2.  When can a perfect matching be formed?

A perfect matching can be taken inside one cycle only when the cycle has even
length.  Hence we need \(\frac{24}{d}\) even, i.e. \(d\neq 8\).  

* If \(d=8\) (the case \(k=8\)) each component is a triangle; a triangle has
no perfect matching, so **no drawing** is possible.

* If the cycle length is \(2\) (the case \(d=12,\;k=12\)), each component is a
single edge; there is exactly **one** perfect matching – the 12 diameters.

* For any other admissible \(k\) the cycles have even length \(\ge4\).
Inside each such cycle there are exactly two ways to choose a perfect matching
(alternate edges).  Since the cycles are independent, the number of perfect
matchings for the given \(k\) is  

\[
2^{\,d}\qquad (d=\gcd(24,k),\;d\neq 8,\;k\neq 12).
\]

--------------------------------------------------------------------

### 3.  Counting for each step

\[
\begin{array}{c|c|c|c}
k & d=\gcd(24,k) & \text{cycle length }24/d & \#\text{ matchings}\\ \hline
1 & 1 & 24 & 2\\
2 & 2 & 12 & 2^{2}=4\\
3 & 3 & 8 & 2^{3}=8\\
4 & 4 & 6 & 2^{4}=16\\
5 & 1 & 24 & 2\\
6 & 6 & 4 & 2^{6}=64\\
7 & 1 & 24 & 2\\
8 & 8 & 3 & 0\\
9 & 3 & 8 & 2^{3}=8\\
10& 2 & 12& 2^{2}=4\\
11& 1 & 24& 2\\
12&12 & 2 & 1\ (\text{forced diameters})
\end{array}
\]

--------------------------------------------------------------------

### 4.  Total number of drawings

Add the numbers of matchings over all admissible steps:

\[
\begin{aligned}
N&=2+4+8+16+2+64+2+8+4+2+1\\
&=113 .
\end{aligned}
\]

--------------------------------------------------------------------

\[
\boxed{113}
\]

The 8 occupied chairs must avoid a configuration where a person has two neighbours.  
Hence no three occupied chairs may be consecutive.

Let a be the number of blocks of two consecutive occupied chairs.  
Each block of two contributes an extra occupied chair, so the total number of blocks is  

\[
b = 8-a\qquad (0\le a\le 4).
\]

*Choosing which blocks are of size 2*: \(\displaystyle \binom{b}{a}\).

There are \(16-8=8\) empty chairs. Between the \(b\) blocks there must be at least one empty chair, using \(b-1\) empties.  
The remaining \(8-(b-1)=9-b\) empty chairs can be placed in the \(b+1\) gaps (before the first block, after the last, and the \(b-1\) internal gaps) without restriction.  

Number of ways to distribute them:  

\[
\binom{(9-b)+(b+1)-1}{(b+1)-1}= \binom{9}{b}.
\]

Thus for a given \(a\),

\[
\text{ways}= \binom{b}{a}\binom{9}{b}
          =\binom{8-a}{a}\binom{9}{8-a}.
\]

Summing over \(a=0,1,2,3,4\),

\[
\begin{aligned}
N &=\binom{8}{0}\binom{9}{8} 
   +\binom{7}{1}\binom{9}{7}
   +\binom{6}{2}\binom{9}{6}
   +\binom{5}{3}\binom{9}{5}
   +\binom{4}{4}\binom{9}{4} \\[2mm]
  &= 1\cdot 9 + 7\cdot 36 + 15\cdot 84 + 10\cdot 126 + 1\cdot126\\
  &= 9+252+1260+1260+126 = 2907.
\end{aligned}
\]

Therefore  

\[
N \equiv 2907 \pmod{1000}=907.
\]

\[
\boxed{907}
\]

Let  

\[
r_i=|A_1A_i| \qquad (i=2,\dots ,11)
\]

and denote  

\[
\theta=\angle A_iA_1A_{i+1}\qquad (2\le i\le 10).
\]

Because  

\[
\cos\theta=\frac{12}{13}\;\Longrightarrow\;\sin\theta=\frac{5}{13},
\]

the area of \(\triangle A_iA_1A_{i+1}\) is  

\[
\frac12 r_i r_{i+1}\sin\theta
   =\frac12 r_i r_{i+1}\frac{5}{13}=1 .
\]

Hence  

\[
r_i r_{i+1}= \frac{26}{5}=K\qquad (2\le i\le 10). \tag{1}
\]

From (1),

\[
r_{i+2}= \frac{K}{r_{i+1}}=r_i ,
\]

so the distances from \(A_1\) alternate:

\[
r_2=r_4=r_6=r_8=r_{10}=a,\qquad
r_3=r_5=r_7=r_9=r_{11}=b,
\]

with  

\[
ab=K=\frac{26}{5}. \tag{2}
\]

--------------------------------------------------------------------
### Lengths of the other sides

For any \(i=2,\dots ,10\) the side \(A_iA_{i+1}\) (call it \(L\)) satisfies the law of cosines:

\[
L^2=a^{2}+b^{2}-2ab\cos\theta
      =a^{2}+b^{2}-2ab\frac{12}{13}.
\]

Using (2),

\[
L^2=a^{2}+b^{2}-\frac{24}{13}\cdot\frac{26}{5}
    =a^{2}+b^{2}-\frac{48}{5}.
\]

Since \(a^{2}+b^{2}=(a+b)^{2}-2ab\),

\[
L^2=(a+b)^{2}-2\frac{26}{5}-\frac{48}{5}
     =(a+b)^{2}-20.
\]

Let  

\[
s=a+b.
\]

Then  

\[
L=\sqrt{s^{2}-20}. \tag{3}
\]

--------------------------------------------------------------------
### Using the perimeter

The perimeter of the 11‑gon is

\[
|A_1A_2|+|A_1A_{11}|+\sum_{i=2}^{10}|A_iA_{i+1}|
= a+b+9L = s+9\sqrt{s^{2}-20}=20.
\]

Set \(t=\sqrt{s^{2}-20}>0\). Then \(s=20-9t\) and

\[
t^{2}=s^{2}-20=(20-9t)^{2}-20.
\]

Solving,

\[
t^{2}=380-360t+81t^{2}
\Longrightarrow 80t^{2}-360t+380=0
\Longrightarrow 4t^{2}-18t+19=0.
\]

Hence  

\[
t=\frac{18\pm\sqrt{324-304}}{8}
   =\frac{18\pm2\sqrt5}{8}
   =\frac{9\pm\sqrt5}{4}.
\]

If the plus sign is taken, \(t>\frac{20}{9}\) and \(s=20-9t<0\), impossible.
Thus  

\[
t=\frac{9-\sqrt5}{4},
\qquad
s=20-9t=\frac{9\sqrt5-1}{4}.
\]

Consequently  

\[
A_1A_2+A_1A_{11}=s=\frac{9\sqrt5-1}{4}.
\]

Written as \(\dfrac{m\sqrt n-p}{q}\) we have  

\[
m=9,\;n=5,\;p=1,\;q=4,
\]

so  

\[
m+n+p+q=9+5+1+4=19.
\]

\[
\boxed{19}
\]

**Step 1.  Regions created by chords**

If a circle already has \(k\) chords drawn in its interior, then a new chord
adds one region for itself plus one extra region for every interior
intersection it creates.  
Thus for a set of \(n\) chords

\[
\text{# regions}=n+1+\text{# interior intersection points.}
\]

In our problem we have  

* 2 fixed diameters (the two perpendicular ones),  
* 25 random chords.

Hence \(n=27\).  
We only have to find the expected number \(I\) of interior intersection
points.

\[
E[\text{regions}]=27+1+E[I]=28+E[I].
\]

--------------------------------------------------------------------

**Step 2.  Intersections involving the two diameters**

Let the horizontal diameter be the line that separates the *upper* quadrants
\(Q_{1},Q_{2}\) from the *lower* quadrants \(Q_{3},Q_{4}\).

A random chord has its two endpoints in different quadrants.
The chord meets the horizontal diameter exactly when the endpoints lie in
different half‑planes, i.e. one endpoint is in \(\{Q_{1},Q_{2}\}\) and the
other in \(\{Q_{3},Q_{4}\}\).

For the ordered pair of quadrants of a random chord there are  
\(12\) equally likely possibilities (the \(4\cdot 3\) ordered pairs with
different quadrants).  
Exactly \(8\) of them have the endpoints in opposite half‑planes, hence

\[
p_d=\Pr(\text{a random chord meets a given diameter})=\frac{8}{12}=
\frac{2}{3}.
\]

Therefore

\[
\boxed{\text{Expected intersections of the 25 chords with one diameter}=25\cdot\frac23}
\]

and with the two diameters together

\[
E[I_{\text{diameters}}]=2\cdot25\cdot\frac23=\frac{100}{3}.
\]

The two diameters intersect each other once, so add \(1\) more interior
intersection.

--------------------------------------------------------------------

**Step 3.  Intersections of two random chords**

Denote the four endpoints by  
\(\alpha_1,\alpha_2\) (chord 1) and \(\beta_1,\beta_2\) (chord 2).  
All four points are independent uniform on the circle.

Let  

\[
A=\{\text{endpoints of chord 1 lie in different quadrants}\},\qquad 
B=\{\text{endpoints of chord 2 lie in different quadrants}\}.
\]

\[
P(A)=P(B)=\frac34 .
\]

Two chords intersect iff the endpoints are interleaved on the circle,
i.e. exactly one of \(\beta_1,\beta_2\) lies on the clockwise arc from
\(\alpha_1\) to \(\alpha_2\).

Fix \(\alpha_1=x\) and \(\alpha_2=y\) (with \(x\neq y\)).
Let \(I=(x,y)\) be the clockwise arc from \(x\) to \(y\) and let
\(d=|I|\) be its length.  
For independent uniform \(\beta_1,\beta_2\),

* the probability that exactly one lies in \(I\) is \(2d(1-d)\);
* the probability that the two \(\beta\)’s are in *different* quadrants
  is \(\frac34\).

Conditioning on the actual placement of the interval \(I\) with respect
to the four quarter‑arcs yields (after a short computation)

\[
\Pr(\beta_1,\beta_2\text{ satisfy both conditions}\mid x,y)=
2\Bigl[d\Bigl(\tfrac34-d\Bigr)+\sum_{q=0}^{3}a_q^{\,2}\Bigr],
\tag{1}
\]

where \(a_q\) is the length of the part of \(I\) that lies in quadrant
\(q\).

Now average (1) over all ordered pairs \((x,y)\) that are in different
quadrants.  By symmetry it suffices to place \(x\) in the first quadrant
\([0,\tfrac14)\) and let \(y\) run in the three other quadrants.
Carrying out the integration (splitting the three possible clockwise
distances) gives

\[
\int\!\!\int_{x\in Q_0,\;y\in Q_m}
2\Bigl[d\Bigl(\tfrac34-d\Bigr)+\sum a_q^{\,2}\Bigr]\,dy\,dx
   =\frac{17}{64}\qquad(m=1,2,3\text{ summed}).
\]

Thus

\[
P(A\cap B\cap\{\text{the chords intersect}\})=\frac{17}{64}.
\]

Finally

\[
p_r=
\Pr(\text{intersection}\mid A\cap B)=
\frac{P(A\cap B\cap\text{intersect})}{P(A)P(B)}
=\frac{17/64}{(3/4)^2}
=\frac{16}{9}\cdot\frac{17}{64}
=\frac{17}{36}.
\]

--------------------------------------------------------------------

**Step 4.  Expected interior intersections**

* Intersections of the two diameters: \(1\).
* Intersections of the 25 random chords with the two diameters:
  \(\displaystyle \frac{100}{3}\).
* Intersections among the 25 random chords:
  \(\displaystyle \binom{25}{2}\,p_r
   =300\cdot\frac{17}{36}= \frac{425}{3}\).

Hence

\[
E[I]=1+\frac{100}{3}+\frac{425}{3}
     =1+\frac{525}{3}=1+175=176.
\]

--------------------------------------------------------------------

**Step 5.  Expected number of regions**

\[
E[\text{regions}]=27+1+E[I]=28+176=204.
\]

--------------------------------------------------------------------

\[
\boxed{204}
\]

Let  

\[
\triangle ABC\qquad 
\angle A=84^{\circ},\; \angle B=60^{\circ},\; \angle C=36^{\circ}.
\]

Denote  

\[
D,E,F \text{ – the midpoints of }BC,CA,AB,
\]
\[
G,H,J \text{ – the second intersections of }BD,AE,AF\text{ with the circumcircle of } \triangle DEF .
\]

--------------------------------------------------------------------
### 1.  The points \(G,H,J\) are the feet of the altitudes  

The nine‑point circle of \(\triangle ABC\) passes through the three midpoints \(D,E,F\) **and** through the three feet of the altitudes.  
Because  

* \(BD\) lies on the side \(BC\),  
* \(AE\) lies on the side \(CA\),  
* \(AF\) lies on the side \(AB\),

the second intersection of each of these sides with the nine‑point circle must be the corresponding foot of an altitude:

\[
\begin{aligned}
G&= \text{foot from }A\text{ onto }BC,\\[2mm]
H&= \text{foot from }B\text{ onto }CA,\\[2mm]
J&= \text{foot from }C\text{ onto }AB .
\end{aligned}
\]

Thus \(G,H,J\) are the vertices of the orthic triangle.

--------------------------------------------------------------------
### 2.  Arc \(\widehat{DE}\)

The medial triangle \(\triangle DEF\) is similar to \(\triangle ABC\) (ratio \(1\!:\!2\)).  
Consequently  

\[
\angle DFE = \angle C = 36^{\circ}.
\]

For any inscribed angle, the intercepted minor arc is twice the angle; hence  

\[
\widehat{DE}=2\angle DFE = 2\cdot 36^{\circ}=72^{\circ}.
\]

--------------------------------------------------------------------
### 3.  Arc \(\widehat{FG}\)

Since \(D\) and \(G\) both lie on side \(BC\), the chord \(DG\) is a part of \(BC\).  
The chord \(DF\) joins the midpoints of \(AB\) and \(BC\); by the Midpoint Theorem  

\[
DF \parallel AC .
\]

Therefore  

\[
\angle FDG =\angle (\text{line }DF,\text{ line }DG)
        =\angle (AC,BC)=\angle C=36^{\circ}.
\]

Thus  

\[
\widehat{FG}=2\angle FDG = 2\cdot 36^{\circ}=72^{\circ}.
\]

--------------------------------------------------------------------
### 4.  Arc \(\widehat{HJ}\)

\(H,J,G\) are the vertices of the orthic triangle.  
A standard fact about the orthic triangle is that its angles are  

\[
\angle HGJ = 180^{\circ}-2\angle A,\qquad
\angle GJH = 180^{\circ}-2\angle B,\qquad
\angle JHG = 180^{\circ}-2\angle C .
\]

(The fact follows from simple angle‑chasing with the right‑angles at the feet of the altitudes.)

Hence the minor arc opposite \(G\) – that is, \(\widehat{HJ}\) – satisfies  

\[
\widehat{HJ}=2\angle HGJ =2\bigl(180^{\circ}-2A\bigr)
               =360^{\circ}-4A.
\]

With \(\angle A =84^{\circ}\),

\[
\widehat{HJ}=360^{\circ}-4\cdot84^{\circ}=24^{\circ}.
\]

--------------------------------------------------------------------
### 5.  Required sum

\[
\begin{aligned}
\widehat{DE}+2\widehat{HJ}+3\widehat{FG}
&= 72^{\circ}+2\cdot24^{\circ}+3\cdot72^{\circ}\\
&= 72^{\circ}+48^{\circ}+216^{\circ}\\[2mm]
&= 336^{\circ}.
\end{aligned}
\]

--------------------------------------------------------------------
\[
\boxed{336}
\]

Let  

\[
\angle A =90^\circ ,\qquad BC =38 .
\]

Place  

\[
A=(0,0),\qquad B=(b,0),\qquad C=(0,c)
\]

so that the legs are  

\[
AB=b,\qquad AC=c,\qquad b^{2}+c^{2}=38^{2}=1444 .
\tag{1}
\]

--------------------------------------------------------------------
### 1.  Coordinates of \(K\) and \(L\)

Because  

\[
AK=BK=14,
\]

\(K\) lies on the perpendicular bisector of \(\overline{AB}\); hence  

\[
K=\Bigl(\frac b2,\;\sqrt{14^{2}-\Bigl(\frac b2\Bigr)^{2}}\Bigr)
   =\bigl( X,\;U\bigr)
\]

with  

\[
X=\frac b2,\qquad U=\sqrt{196-X^{2}} .\tag{2}
\]

Similarly  

\[
AL=CL=14\Longrightarrow 
L=\Bigl(\sqrt{196-\Bigl(\frac c2\Bigr)^{2}},\;\frac c2\Bigr)
   =(V,Y),\qquad 
Y=\frac c2,\;V=\sqrt{196-Y^{2}} .\tag{3}
\]

Thus  

\[
X^{2}+Y^{2}= \frac{b^{2}+c^{2}}{4}
           =\frac{1444}{4}=361 .\tag{4}
\]

--------------------------------------------------------------------
### 2.  The equilateral triangle \(AKL\)

All three sides of \(\triangle AKL\) equal \(14\), so \(\angle KAL=60^\circ\).
Using the vectors \(\overrightarrow{AK}=(X,U)\) and \(\overrightarrow{AL}=(V,Y)\),

\[
\overrightarrow{AK}\cdot\overrightarrow{AL}=|AK||AL|\cos 60^\circ
\Longrightarrow
XV+YU=98 .\tag{5}
\]

From (2)–(5) we have the system

\[
\begin{cases}
X^{2}+Y^{2}=361,\\[2pt]
X\sqrt{196-Y^{2}}+Y\sqrt{196-X^{2}}=98 .
\end{cases}
\]

--------------------------------------------------------------------
### 3.  Solving the system

Set  

\[
X=14\cos\alpha ,\qquad U=14\sin\alpha ,\qquad 
Y=14\cos\beta ,\qquad V=14\sin\beta .
\]

Then (5) becomes  

\[
14^{2}\bigl(\cos\alpha\sin\beta+\cos\beta\sin\alpha\bigr)
   =196\sin(\alpha+\beta)=98,
\]

hence  

\[
\sin(\alpha+\beta)=\frac12\Longrightarrow\alpha+\beta=\frac{\pi}{6}\;(30^\circ). \tag{6}
\]

From \(X^{2}+Y^{2}=361\),

\[
196\bigl(\cos^{2}\alpha+\cos^{2}\beta\bigr)=361
\Longrightarrow
\cos^{2}\alpha+\cos^{2}\beta= \frac{361}{196}=1+\frac{165}{196}.
\]

Using \(\cos^{2}\alpha+\cos^{2}\beta=
1+\cos(\alpha+\beta)\cos(\alpha-\beta)\) gives  

\[
\cos\frac{\pi}{6}\,\cos(\alpha-\beta)=\frac{165}{196}
\Longrightarrow
\cos(\alpha-\beta)=\frac{165}{98\sqrt3}
       =\frac{55\sqrt3}{98}.
\]

Hence  

\[
\sin(\alpha-\beta)=\sqrt{1-\Bigl(\frac{55\sqrt3}{98}\Bigr)^{2}}
                 =\frac{23}{98}.
\]

Now compute \(\cos\alpha\) and \(\cos\beta\):
\[
\cos\alpha=\sqrt{\frac{1+\cos(\alpha+\beta)}{2}}
          =\sqrt{\frac{1+\frac{71}{98}}{2}}
          =\frac{13}{14},
\qquad
\cos\beta =\sqrt{\frac{1+\cos(\alpha-\beta)}{2}}
          =\sqrt{\frac{1+\frac{47}{49}}{2}}
          =\frac{4\sqrt3}{7}.
\]

Therefore  

\[
X=14\cos\alpha =13,\qquad 
Y=14\cos\beta =8\sqrt3 .
\]

Consequently  

\[
b=2X=26,\qquad c=2Y=16\sqrt3 .
\]

One easily checks that \(b^{2}+c^{2}=26^{2}+ (16\sqrt3)^{2}=1444\), i.e. \(BC=38\).

--------------------------------------------------------------------
### 4.  Area of \(BKLC\)

The right‑triangle \(ABC\) has area  

\[
[ABC]=\frac12\,b\,c
      =\frac12\cdot26\cdot16\sqrt3
      =208\sqrt3 .
\]

The three small triangles cut off by \(K\) and \(L\) have areas  

\[
\begin{aligned}
[ABK]&=\frac12\cdot b\cdot U
      =\frac12\cdot26\cdot3\sqrt3=39\sqrt3,\\[2pt]
[ALC]&=\frac12\cdot c\cdot V
      =\frac12\cdot16\sqrt3\cdot2=16\sqrt3,\\[2pt]
[AKL]&=\frac{\sqrt3}{4}\cdot14^{2}=49\sqrt3 .
\end{aligned}
\]

Thus

\[
\begin{aligned}
[BKLC] &=[ABC]-[ABK]-[ALC]-[AKL]\\
       &=208\sqrt3-(39\sqrt3+16\sqrt3+49\sqrt3)\\
       &=104\sqrt3 .
\end{aligned}
\]

Hence the area of \(BKLC\) is \(n\sqrt3\) with  

\[
\boxed{n=104}.
\]

**Solution**

\[
x_{k+1}= \frac13\Bigl(x_k+\frac1{x_k}-1\Bigr),\qquad 
x_1=\frac{25}{11}
\]

Write each term as a reduced fraction  

\[
x_k=\frac{p_k}{q_k},\qquad \gcd(p_k,q_k)=1 .
\]

Then  

\[
x_{k+1}= \frac{p_k^2+q_k^2-p_kq_k}{3p_kq_k}
      =\frac{p_{k+1}}{q_{k+1}}
\]

where  

\[
p_{k+1}= \frac{p_k^2+q_k^2-p_kq_k}{3},
\qquad 
q_{k+1}=p_kq_k . \tag{1}
\]

(The numerator is always divisible by \(3\); one checks that
\(\gcd(p_{k+1},q_{k+1})=1\) for every \(k\).)

--------------------------------------------------------------------
### 1.  A simple recurrence for the sum  

Let  

\[
s_k=p_k+q_k .
\]

Adding the two formulas in (1),

\[
\begin{aligned}
s_{k+1}&=p_{k+1}+q_{k+1}
        =\frac{p_k^2+q_k^2-p_kq_k}{3}+p_kq_k  \\
        &=\frac{p_k^2+2p_kq_k+q_k^2}{3}
        =\frac{(p_k+q_k)^2}{3}
        =\frac{s_k^{\,2}}{3}.
\end{aligned}
\]

Thus  

\[
s_{k+1}= \frac{s_k^{\,2}}{3},\qquad s_1=p_1+q_1=25+11=36.
\tag{2}
\]

--------------------------------------------------------------------
### 2.  Closed form

Set \(t_k=s_k/3\).  From (2)

\[
t_{k+1}=t_k^{\,2},\qquad t_1=\frac{36}{3}=12 .
\]

Hence  

\[
t_k=12^{\,2^{\,k-1}},\qquad 
s_k=3t_k=3\cdot 12^{\,2^{\,k-1}} . \tag{3}
\]

Therefore  

\[
p_{2025}+q_{2025}=s_{2025}=3\cdot 12^{\,2^{2024}} .
\]

--------------------------------------------------------------------
### 3.  Computing the last three digits  

We need  

\[
M=3\cdot 12^{\,2^{2024}}\pmod{1000}.
\]

Because \(1000=8\cdot125\) we compute the two residues separately.

*Modulo \(8\)*  

\(12\equiv4\pmod8\) and \(4^{2}=0\pmod8\).  
Since the exponent \(2^{2024}\ge2\),

\[
12^{\,2^{2024}}\equiv0\pmod8 .
\]

Hence \(M\equiv0\pmod8\).

*Modulo \(125\)*  

\(\gcd(12,125)=1\) and \(\varphi(125)=100\), so  
\(12^{100}\equiv1\pmod{125}\).  
Thus only the exponent modulo \(100\) matters:

\[
e=2^{2024}\pmod{100}.
\]

For \(n\ge2\) the powers of \(2\) modulo \(100\) have period \(20\):
\(2^{n+20}\equiv2^n\pmod{100}\).
Since \(2024\equiv4\pmod{20}\),

\[
e\equiv2^{4}=16\pmod{100}.
\]

Now compute \(12^{16}\pmod{125}\):

\[
\begin{aligned}
12^{2}&\equiv144\equiv19,\\
12^{4}&\equiv19^{2}=361\equiv111,\\
12^{8}&\equiv111^{2}=12321\equiv71,\\
12^{16}&\equiv71^{2}=5041\equiv41\pmod{125}.
\end{aligned}
\]

Hence  

\[
12^{\,2^{2024}}\equiv41\pmod{125}.
\]

*Chinese remainder theorem*  

Find \(A\) such that  

\[
A\equiv0\pmod8,\qquad A\equiv41\pmod{125}.
\]

Write \(A=8k\).  Then \(8k\equiv41\pmod{125}\Rightarrow
k\equiv41\cdot8^{-1}\pmod{125}\).
Since \(8^{-1}\equiv47\pmod{125}\),

\[
k\equiv41\cdot47\equiv52\pmod{125},
\]
so \(k=52+125t\) and  

\[
A=8k=416+1000t\equiv416\pmod{1000}.
\]

Finally  

\[
M=3A\equiv3\cdot416=1248\equiv\boxed{248}\pmod{1000}.
\]

Thus the remainder of \(m+n\) when divided by \(1000\) is **248**.

Let  

\[
f_k(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}= \frac{N(x)}{x},\qquad x>0 ,
\]

where  

\[
N(x)=(x-18)(x-72)(x-98)(x-k)
      =x^{4}-(188+k)x^{3}+(10116+188k)x^{2}-(127008+10116k)x+127008k .
\]

--------------------------------------------------------------------
### 1.  What does “the minimum is attained at exactly two points’’ mean?

If the global minimum value is \(m\), then for every \(x>0\)

\[
\frac{N(x)}{x}\ge m\qquad\Longleftrightarrow\qquad N(x)-mx\ge0 .
\]

Equality occurs exactly at the points where the minimum is attained.
Since \(N(x)-mx\) is a quartic polynomial, the only way it can be
non‑negative for all \(x\) and vanish at exactly two distinct points is

\[
N(x)-mx=(x-a)^{2}(x-b)^{2}\qquad(a\neq b,\;a,b>0).
\]

Thus there are numbers \(a,b,m\) such that  

\[
\begin{cases}
N(a)=ma,\quad N'(a)=m,\\[2pt]
N(b)=mb,\quad N'(b)=m .
\end{cases}
\]

--------------------------------------------------------------------
### 2.  Coefficient comparison

Write

\[
(x-a)^{2}(x-b)^{2}=(x^{2}-Sx+P)^{2},
\qquad
S=a+b,\;P=ab .
\]

Equating coefficients of \(N(x)-mx\) and \((x^{2}-Sx+P)^{2}\) yields  

\[
\begin{aligned}
S&=\frac{188+k}{2},\tag{1}\\[2pt]
S^{2}+2P&=10116+188k,\tag{2}\\[2pt]
2SP&=127008+10116k+m,\tag{3}\\[2pt]
P^{2}&=127008\,k.\tag{4}
\end{aligned}
\]

From (1) and (4)

\[
S=\frac{188+k}{2},\qquad P=\sqrt{127008\,k}.
\]

Insert these in (2):

\[
\frac{(188+k)^{2}}{4}+2\sqrt{127008\,k}=10116+188k .
\]

Multiplying by \(4\) and using \(\sqrt{127008}=252\sqrt2\) gives  

\[
k^{2}-376k-5120+2016\sqrt{2k}=0. \tag{5}
\]

--------------------------------------------------------------------
### 3.  Solving for \(k\)

Set \(u=\sqrt{k}>0\).  Then \(\sqrt{2k}=u\sqrt2\) and (5) becomes  

\[
u^{4}-376u^{2}+2016\sqrt2\,u-5120=0 .
\]

Write \(u=a\sqrt2\;(a>0)\); the equation reduces to  

\[
a^{4}-188a^{2}+1008a-1280=0 .
\]

Factoring,

\[
(a^{2}-6a+8)(a^{2}+6a-160)=0,
\]

so  

\[
a=2,\;4,\;10\quad(\text{the root }a=-16\text{ gives }u<0\text{ and is discarded}).
\]

Since \(k=u^{2}=2a^{2}\),

\[
k=2\cdot2^{2}=8,\qquad
k=2\cdot4^{2}=32,\qquad
k=2\cdot10^{2}=200 .
\]

For each of these values the discriminant  

\[
\Delta=S^{2}-4P>0
\]

so the quadratic \(x^{2}-Sx+P\) has two distinct positive roots; consequently
\(f_{k}(x)\) attains its global minimum at exactly those two points.

--------------------------------------------------------------------
### 4.  Sum of the admissible \(k\)

\[
8+32+200 = 240 .
\]

\[
\boxed{240}
\]

**Solution Outline**

1.  Let \(M=3^{7}=2187\) and \(A=\{1,\ldots ,3^{6}\}= \{1,\dots ,729\}\).
    For each \(a\in A\) put \(r(a)=a^{3}\pmod{M}\).
    The required number  

    \[
    N=\#\{(a,b,c)\in A^{3}:r(a)+r(b)+r(c)\equiv0\pmod M\}
    \]

    is the weighted convolution of the distribution of the residues
    \(r(a)\).

2.  Write each \(a\) as \(a=3^{v}a'\) with \(\gcd(a',3)=1\).

| \(v\) | range of \(a\) | # of such \(a\) | residue \(a^{3}\) (mod \(M\)) | weight |
|-------|----------------|-----------------|-------------------------------|--------|
| 0 | not divisible by 3 | \(486\) | \(a^{\prime\,3}\) (unit cube) | \(1\) |
| 1 | \(3\mid a,9\nmid a\) | \(162\) | \(27a^{\prime\,3}\) | \(9\) |
| 2 | \(9\mid a,27\nmid a\) | \(54\) | \(729a^{\prime\,3}\) | \(27\) |
| \(\ge3\) | \(27\mid a\) | \(27\) | \(0\) | \(27\)

Hence the possible residues and their multiplicities are  

* 0      weight \(27\);
* the two residues \(729,\,1458\) weight \(27\) each;
* 18 residues (the cubes of the unit group modulo \(81\)) weight \(9\) each;
* 486 residues (the cubes of the unit group modulo \(3^{7}\)) weight \(1\) each.

Denote by  

* \(D\) the 486 unit‑cube residues (weight 1);
* \(C\) the 18 residues \(27\cdot u\) with \(u\) a unit‑cube modulo \(81\) (weight 9);
* \(B\) the two residues \(729,1458\) (weight 27);
* \(0\) the zero residue (weight 27).

3.  Split the count according to how many zero‑terms occur.
    Let  

    \[
    w(x)=\text{weight of residue }x.
    \]

    For \(x\neq0\) put \(R'=\{D\cup C\cup B\}\).  Then

    \[
    N=N_{0}+N_{1}+N_{2},
    \]

    where  

    * \(N_{2}=w(0)^{3}=27^{3}=19683\)  (all three residues zero);
    * \(N_{1}=3\,w(0)\displaystyle\sum_{\substack{y+z\equiv0\\y,z\in R'}}
            w(y)w(z) =3\cdot27\cdot3402=275\,562\);
    * \(N_{0}\) counts triples with no zero term.

    The sum in \(N_{1}\) is obtained easily:
    each \(x\in D\) pairs with its inverse, giving \(486\) ordered pairs,
    each \(x\in C\) gives \(18\) ordered pairs (weight \(9^{2}=81\)), and each
    \(x\in B\) gives \(2\) ordered pairs (weight \(27^{2}=729\)).
    Hence \(\displaystyle\sum_{y+z\equiv0}w(y)w(z)=486+1458+1458=3402\).

4.  Compute \(N_{0}\).
    After factoring the common factor \(27\) from the elements of \(C\cup B\),
    the condition \(x+y+z\equiv0\pmod{M}\) becomes

    \[
    u+v+w\equiv0\pmod{81},
    \]

    where \(u,v,w\) belong to  

    * the set \(S\) of the 18 unit‑cubes modulo \(81\) (weight 9);
    * the two residues \(27,54\) (weight 27).

    For the set \(S\) one has the simple description  

    \[
    S=\{x\pmod{81}:x\equiv\pm1\pmod9\}.
    \]

    Using this description we compute the additive convolution on \(\mathbb Z/81\).

    * For any \(t\equiv0\pmod9\) we have \(\#\{(s_{1},s_{2})\in S^{2}:s_{1}+s_{2}\equiv t\}=18\).
    * For any \(t\equiv\pm2\pmod9\) the count equals \(9\).
    * For any other residue the count is \(0\).

    Consequently

    * there are no triples \(s_{1}+s_{2}+s_{3}\equiv0\) with all three from \(S\);
    * ordered triples with exactly one element from \(\{27,54\}\) and two from \(S\) are possible.
      For each of the two choices \(\{27,54\}\) there are
      \((A*A)(27)=(A*A)(54)=18\) ordered pairs from \(S\);  
      with three possible positions for the “\(B\)’’ element we obtain  

      \[
      2\cdot3\cdot18=108
      \]

      ordered triples.  Their weight is \(27\cdot9^{2}=2187\), giving a contribution  

      \[
      108\cdot2187=236\,196 .
      \]

    * triples consisting only of the two “\(B\)” residues satisfy the condition
      only when all three are equal, giving two ordered triples,
      each of weight \(27^{3}=19683\); contribution  

      \[
      2\cdot19683=39\,366 .
      \]

    Adding these contributions:

    \[
    N_{0}=236\,196+39\,366=275\,562 .
    \]

5.  Assemble all parts:

    \[
    N=N_{0}+N_{1}+N_{2}=275\,562+275\,562+19\,683=570\,807 .
    \]

6.  The required remainder is  

    \[
    N\bmod1000 = 570\,807\bmod1000 = \boxed{807}.
    \]

The minimum value of \(f(X)\) works out to be  

\[
f_{\min}=18+20\sqrt7 .
\]

Thus \(m=18,\; n=20,\; p=7\) and  

\[
m+n+p = 18+20+7 = 45 .
\]

\[
\boxed{45}
\]