Small Hint:

The $14$ positions split into $7$ diametrically opposite pairs, and only the set of positions the men occupy matters.

Big Hint:

Count sets where no pair has two men: choose $5$ of the $7$ pairs to hold a man, then a position in each; compare with $\binom{14}{5}.$

Solution:

The $14$ positions split into $7$ diametrically opposite pairs. Only the set of positions occupied by the men matters, and all $\binom{14}{5} = 2002$ five-element sets are equally likely. Every man stands opposite a woman exactly when no pair contains two men, so choose which $5$ of the $7$ pairs contain a man ($\binom{7}{5} = 21$ ways) and which position of each chosen pair the man occupies ($2^5 = 32$ ways), for $21 \cdot 32 = 672$ favorable sets.

The probability is $\frac{672}{2002} = \frac{48}{143},$ so $m + n = 48 + 143 = 191.$

Small Hint:

Set $x = \log_b n$ and rewrite both equations in terms of $x$ and $b.$

Big Hint:

The first equation forces $\sqrt{x} = \frac{x}{2};$ discard the root that breaks the second equation, which reads $bx = 1 + x.$

Solution:

Let $x = \log_b n.$ The first equation says $\sqrt{x} = \log_b n^{\frac{1}{2}} = \frac{x}{2},$ so $x = \frac{x^2}{4},$ giving $x = 0$ or $x = 4.$ If $x = 0$ then $n = 1,$ and the second equation would read $0 = \log_b b = 1,$ impossible; so $x = 4.$

The second equation says $bx = \log_b b + \log_b n = 1 + x,$ so $4b = 5$ and $b = \frac{5}{4}.$ Then $$n = b^4 = \left(\frac{5}{4}\right)^4 = \frac{625}{256},$$ which is in lowest terms, so $j + k = 625 + 256 = 881.$

Small Hint:

Since no two lines are parallel, every pair of lines meets at exactly one intersection point.

Big Hint:

A point where exactly $k$ lines cross accounts for $\binom{k}{2}$ of the $\binom{40}{2}$ pairs of lines.

Solution:

Since no two of the $40$ lines are parallel, every two lines cross, giving $\binom{40}{2} = 780$ pairs of lines, and each pair meets at exactly one point. A point where exactly $k$ lines meet accounts for exactly $\binom{k}{2}$ of these pairs.

The given points account for $3\binom{3}{2} + 4\binom{4}{2}$ $+ 5\binom{5}{2} + 6\binom{6}{2}$ $= 9 + 24 + 50 + 90 = 173$ pairs of lines. Each remaining pair meets at a point where exactly $2$ lines intersect, one point per pair, so there are $780 - 173 = 607$ such points.

Small Hint:

Factor $13! = 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13$ and decide which prime exponents $m$ can have.

Big Hint:

Each exponent of $m$ must match the parity of the corresponding exponent of $13!,$ so the sum of all valid $m$ factors as a product of geometric sums.

Solution:

Since $13! = 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13,$ a valid $m = 2^x 3^y 5^z 7^w 11^u 13^v$ must leave every exponent of $\frac{13!}{m}$ even: $x \in \{0, 2, 4, 6, 8, 10\},$ $y \in \{1, 3, 5\},$ $z \in \{0, 2\},$ and $w = u = v = 1.$

The choices are independent, so the sum of all such $m$ factors as $$\begin{aligned} &(1 + 4 + \cdots + 1024) \\ &\quad {}\cdot (3 + 27 + 243) \\ &\quad {}\cdot (1 + 25) \cdot 7 \cdot 11 \cdot 13 \\ &= 1365 \cdot 273 \cdot 26 \cdot 1001. \end{aligned}$$ Since $1365 = 3 \cdot 5 \cdot 7 \cdot 13,$ $273 = 3 \cdot 7 \cdot 13,$ $26 = 2 \cdot 13,$ and $1001 = 7 \cdot 11 \cdot 13,$ the sum equals $2^1 3^2 5^1 7^3 11^1 13^4,$ and $a + b + c + d + e + f$ $= 1 + 2 + 1 + 3 + 1 + 4$ $= 12.$

Small Hint:

Put the center at the origin with $A = (R, 0),$ $B = (0, R),$ $C = (-R, 0),$ $D = (0, -R),$ and $P = (R\cos\theta, R\sin\theta).$

Big Hint:

Show $PA \cdot PC = 2R^2|\sin\theta|$ and $PB \cdot PD = 2R^2|\cos\theta|;$ a Pythagorean identity links the two products.

Solution:

Let the circle have center $O$ and radius $R,$ with $A = (R, 0),$ $B = (0, R),$ $C = (-R, 0),$ $D = (0, -R),$ and $P = (R\cos\theta, R\sin\theta).$ Then $PA^2 = 2R^2(1 - \cos\theta)$ and $PC^2 = 2R^2(1 + \cos\theta),$ so $PA \cdot PC = 2R^2|\sin\theta| = 56.$ In the same way $PB \cdot PD = 2R^2|\cos\theta| = 90.$

Squaring and adding, $4R^4 = 56^2 + 90^2 = 11236,$ so $2R^2 = \sqrt{11236} = 106.$ The square has diagonal $2R,$ hence area $\frac{(2R)^2}{2} = 2R^2 = 106.$

Small Hint:

Guessing the color with more cards remaining is optimal; let $E(r, b)$ be the expected number correct with $r$ red and $b$ black left.

Big Hint:

$E(r,b) = \frac{\max(r,b)}{r+b}$ $+ \frac{r\,E(r-1,b) + b\,E(r,b-1)}{r+b};$ build up from $E(1,1).$

Solution:

Whatever Alice guesses, the deck evolves the same way; only the immediate success probability depends on her guess, so it is optimal to guess a color with the most cards remaining. Let $E(r, b)$ be the expected number of correct guesses from a state with $r$ red and $b$ black cards left. Then $E(r, 0) = r,$ $E(0, b) = b,$ and $$\begin{aligned} E(r, b) &= \frac{\max(r, b)}{r + b} \\ &\quad {}+ \frac{r\,E(r-1,b)}{r+b} \\ &\quad {}+ \frac{b\,E(r,b-1)}{r+b}. \end{aligned}$$

By symmetry $E(r, b) = E(b, r).$ Computing upward: $E(1,1) = \frac{3}{2},$ $E(2,1) = \frac{2}{3} + \frac{2 \cdot \frac{3}{2} + 2}{3} = \frac{7}{3},$ $E(2,2) = \frac{1}{2} + \frac{7}{3} = \frac{17}{6},$ $E(3,1) = \frac{3}{4} + \frac{3 \cdot \frac{7}{3} + 3}{4} = \frac{13}{4},$ $E(3,2) = \frac{3}{5} + \frac{3 \cdot \frac{17}{6} + 2 \cdot \frac{13}{4}}{5} = \frac{18}{5},$ and finally $E(3,3) = \frac{1}{2} + \frac{18}{5} = \frac{41}{10}.$

So the expected number of correct guesses is $\frac{41}{10},$ and $m + n = 41 + 10 = 51.$

Small Hint:

The remainder mod $2$ is forced by the remainders mod $4$ and mod $6,$ and the remainder mod $3$ by the one mod $6$ — few patterns survive.

Big Hint:

Everything depends only on $n \bmod 60;$ exactly three residues work, so count each class below $1000.$

Solution:

Write $r_k$ for the remainder of $n$ modulo $k,$ and note $r_2 \equiv r_4 \pmod 2,$ $r_2 \equiv r_6 \pmod 2,$ and $r_3 \equiv r_6 \pmod 3.$ If $r_2 = 0:$ then $r_4$ is even and different from $r_2,$ so $r_4 = 2;$ then $r_6$ is even and avoids $\{0, 2\},$ so $r_6 = 4,$ which gives $r_3 = 1;$ finally $r_5$ avoids $\{0, 1, 2, 4\},$ so $r_5 = 3.$ These say $n \equiv -2$ modulo each of $2, 3, 4, 5, 6,$ i.e. $n \equiv 58 \pmod{60}.$

If $r_2 = 1:$ similarly $r_4 = 3,$ then $r_6 = 5,$ giving $r_3 = 2,$ and $r_5$ avoids $\{1, 2, 3, 5\},$ so $r_5 \in \{0, 4\}.$ The choice $r_5 = 4$ gives $n \equiv -1 \pmod{60},$ i.e. $n \equiv 59;$ the choice $r_5 = 0$ gives $n \equiv 11 \pmod{12}$ with $5$ dividing $n,$ i.e. $n \equiv 35 \pmod{60}.$

Below $1000$ there are $17$ integers congruent to $35,$ $16$ congruent to $58,$ and $16$ congruent to $59$ modulo $60,$ for a total of $17 + 16 + 16 = 49.$

Small Hint:

The distances from $P$ to the parallel lines $DA$ and $BC$ add up to the height of the rhombus, so the height is $25.$

Big Hint:

Center the incircle at the origin: $P = (\pm 12, \frac{7}{2}),$ and the distance-$5$ condition to the tangent line $AB$ becomes an equation in $\sin \angle BAD.$

Solution:

The distances from an interior point to the parallel lines $DA$ and $BC$ add up to the distance between them, the height of the rhombus. So the height is $9 + 16 = 25,$ and the incircle, tangent to both lines, has radius $\frac{25}{2}.$ Center the incircle at the origin with $DA:\ y = \frac{25}{2}$ and $BC:\ y = -\frac{25}{2}.$ Then $P$ has $y$-coordinate $\frac{25}{2} - 9 = \frac{7}{2},$ and $x^2 + \left(\frac{7}{2}\right)^2 = \left(\frac{25}{2}\right)^2$ gives $x = \pm 12.$

Let $\alpha = \angle BAD.$ Line $AB$ is tangent to the incircle and makes angle $\alpha$ with the horizontal, so (orienting the figure suitably) it is $x \sin\alpha + y \cos\alpha = -\frac{25}{2},$ and interior points satisfy $x\sin\alpha + y\cos\alpha + \frac{25}{2} \gt 0.$ The condition $\operatorname{dist}(P, AB) = 5$ reads $x \sin\alpha + \frac{7}{2}\cos\alpha + \frac{25}{2} = 5.$ For $x = 12$ the left side exceeds $\frac{25}{2},$ so $x = -12,$ and the equation becomes $24\sin\alpha - 7\cos\alpha = 15.$

Substituting $7\cos\alpha = 24\sin\alpha - 15$ into $\sin^2\alpha + \cos^2\alpha = 1$ yields $625\sin^2\alpha - 720\sin\alpha$ $+ 176 = 0,$ so $\sin\alpha = \frac{4}{5}$ or $\frac{44}{125}.$ The root $\frac{44}{125}$ makes $\cos\alpha = \frac{24\sin\alpha - 15}{7}$ negative, contradicting $\angle BAD \lt 90^\circ.$ So $\sin\alpha = \frac{4}{5},$ the side length is $\frac{25}{\sin\alpha} = \frac{125}{4},$ and the perimeter is $4 \cdot \frac{125}{4} = 125.$

Small Hint:

$c$ is irrelevant: $p(m)-p(2)=(m-2)q(m),$ where the three terms of the quadratic $q(m)$ are $m^2,$ $(a+2)m,$ and $2a+b+4.$

Big Hint:

A monic integer quadratic with one integer root has two, so its root set must be $\{2, k\}$ or a double root $k,$ with $k \ne 2.$

Solution:

Since $p(m) - p(2)$ does not involve $c,$ each valid pair $(a, b)$ contributes $41$ polynomials. Factoring, $$\begin{aligned} p(x)-p(2) &= (x-2)q(x), \\ q(x) &= x^2+(a+2)x \\ &\quad {}+(2a+b+4), \end{aligned}$$ so we need the quadratic factor $q(x)$ to have exactly one integer root different from $2.$ If $q$ has any integer root, its other root is also an integer (their sum $-(a+2)$ is an integer); if $q$ has no integer root, then no $m$ exists at all. So either $q$ has roots $2$ and $k$ with $k \ne 2,$ or a double root $k \ne 2.$

Roots $2$ and $k:$ Vieta’s formulas give $a = -4 - k$ and $b = 4k + 4.$ The constraint $-20 \le b \le 20$ forces $-6 \le k \le 4$ (and then $a$ is automatically in range), so excluding $k = 2$ leaves $10$ pairs. Double root $k:$ here $a = -2k - 2$ and $b = k^2 + 4k,$ and $b \le 20$ forces $-6 \le k \le 2,$ all valid for $a,$ so excluding $k = 2$ leaves $8$ pairs.

That is $18$ pairs $(a, b),$ hence $18 \cdot 41 = 738$ polynomials.

Small Hint:

Without the floor the sum is $\frac{1}{5}\bigl(\sum n^2 - a\sum n\bigr);$ there is an integer $a$ making it exactly $0.$

Big Hint:

For that $a,$ $U = -\frac{1}{5}\sum\bigl((n^2 - na) \bmod 5\bigr),$ and the residues repeat with period $5.$

Solution:

Ignoring the floors, $\sum_{n=1}^{2023} \frac{n^2 - na}{5}$ $= \frac{1}{5}\left(\sum n^2 - a\sum n\right)$ vanishes exactly when $a = \frac{\sum n^2}{\sum n} = \frac{2 \cdot 2023 + 1}{3} = 1349,$ an integer. For any other integer $a$ the raw sum has absolute value at least $\frac{1}{5}\sum n = \frac{2023 \cdot 1012}{5} \approx 409455,$ while taking floors changes the total by less than $2023,$ so only $a = 1349$ can put $U$ strictly between $-1000$ and $1000.$

With $a = 1349,$ each term is $\frac{n^2 - 1349n - r_n}{5}$ with $r_n = (n^2 - 1349n) \bmod 5,$ so $U = -\frac{1}{5}\sum r_n.$ Since $1349 \equiv 4 \pmod 5,$ we have $n^2 - 1349n$ $\equiv n(n+1) \pmod 5,$ whose residues for $n \equiv 0, 1, 2, 3, 4$ are $0, 2, 1, 2, 0,$ summing to $5$ per block of five. With $2023 = 5 \cdot 404 + 3,$ the leftover terms $n \equiv 1, 2, 3$ contribute $2 + 1 + 2 = 5,$ so $\sum r_n = 405 \cdot 5 = 2025.$

So $U = -\frac{2025}{5} = -405,$ which indeed lies strictly between $-1000$ and $1000,$ and $a + U = 1349 - 405 = 944.$

Small Hint:

The number of subsets of $m$ consecutive integers with no two consecutive elements is the Fibonacci number $F_{m+2}.$

Big Hint:

If the pair is $\{k, k+1\},$ the other elements must avoid $k-1$ and $k+2$ and be non-consecutive inside the two remaining blocks.

Solution:

First, the number of subsets of a block of $m$ consecutive integers containing no two consecutive elements is the Fibonacci number $F_{m+2}$ (with $F_1 = F_2 = 1$): conditioning on whether the last element is used gives the Fibonacci recursion, and the counts start $1, 2, 3, 5, \ldots.$

Suppose the unique consecutive pair is $\{k, k+1\}$ for some $1 \le k \le 9.$ The remaining elements must exclude $k - 1$ and $k + 2$ (either would create a second consecutive pair) and must contain no consecutive pair within $\{1, \ldots, k-2\}$ or within $\{k+3, \ldots, 10\},$ blocks of sizes $k - 2$ and $8 - k.$ So the count for this $k$ is $F_k \cdot F_{10-k}.$

Summing over $k = 1, \ldots, 9:$ $$\begin{aligned} \sum_{k=1}^{9} F_k F_{10-k} &= 34 + 21 + 26 \\ &\quad {}+ 24 + 25 + 24 \\ &\quad {}+ 26 + 21 + 34 \\ &= 235. \end{aligned}$$

Small Hint:

Decompose each vector from $E,$ $F,$ $D$ to $P$ into a part along the side and the distance to that side; the along-part equals the distance times $\cot\theta.$

Big Hint:

Add the three relations: the unit directions $C \to A,$ $A \to B,$ $B \to C$ sum to zero, and the three distances sum to the height (Viviani).

Solution:

Place $B = (0, 0),$ $C = (55, 0),$ $A = \left(\frac{55}{2}, \frac{55\sqrt{3}}{2}\right),$ so that $D = (7, 0),$ $E = (40, 15\sqrt{3}),$ and $F = \left(\frac{15}{2}, \frac{15\sqrt{3}}{2}\right).$ Let $\theta$ be the common angle and let $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ be the unit vectors in the directions $C \to A,$ $A \to B,$ $B \to C$ — the directions from $E$ toward $A,$ from $F$ toward $B,$ and from $D$ toward $C.$ Splitting $P - E$ into its component along $\mathbf{u}_1$ and its component perpendicular to side $CA,$ whose length is the distance $d_1$ from $P$ to line $CA,$ the angle condition gives $(P - E)\cdot\mathbf{u}_1 = d_1\cot\theta;$ similarly $(P - F)\cdot\mathbf{u}_2 = d_2\cot\theta$ and $(P - D)\cdot\mathbf{u}_3 = d_3\cot\theta.$

Now add all three relations. Since $\mathbf{u}_1 + \mathbf{u}_2 + \mathbf{u}_3 = \mathbf{0}$ (the directed sides of a triangle close up), $P$ drops out, and by Viviani’s theorem $d_1 + d_2 + d_3$ equals the height $\frac{55\sqrt{3}}{2}.$ With $\mathbf{u}_1 = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $\mathbf{u}_2 = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right),$ and $\mathbf{u}_3 = (1, 0),$ we get $E \cdot \mathbf{u}_1 = \frac{5}{2},$ $F \cdot \mathbf{u}_2 = -15,$ and $D \cdot \mathbf{u}_3 = 7,$ so $$\begin{aligned} \frac{55\sqrt{3}}{2}\,\cot\theta &= -\left(\frac{5}{2} - 15 + 7\right) \\ &= \frac{11}{2}. \end{aligned}$$