Solution:

Place the line on a number line with $A = 0.$ Then $C = 26,$ $E = 26 + 31 = 57,$ $F = 73,$ $D = 73 - 33 = 40,$ and $B = 40 - 22 = 18.$

Write $G = (x, y).$ From $CG = 40$ and $DG = 30,$ $$(x - 26)^2 + y^2 = 1600, \qquad (x - 40)^2 + y^2 = 900.$$ Subtracting gives $14(2x - 66) = 700,$ so $x = 58,$ and then $y^2 = 1600 - 32^2 = 576,$ so $G$ is at height $24$ above the line.

Since $B$ and $E$ both lie on the line, $BE = 57 - 18 = 39$ is a base with height $24,$ so the area is $\frac{1}{2} \cdot 39 \cdot 24 = 468.$

Solution:

Work modulo $n + 2,$ where $n \equiv -2.$ Then $$3(n + 3)(n^2 + 9) \equiv 3 \cdot 1 \cdot (4 + 9) = 39 \pmod{n + 2},$$ so $n + 2$ divides $3(n+3)(n^2+9)$ exactly when $n + 2$ divides $39.$

The divisors of $39$ that are at least $3$ are $3,$ $13,$ and $39,$ giving $n = 1,$ $11,$ and $37.$ The sum is $1 + 11 + 37 = 49.$

Solution:

The $12$ segments split into the $4$ interior segments forming the central cross and $8$ boundary segments, and each unit square has exactly two interior sides (its two cross arms) and two boundary sides. Color the cross first. A square that already has $j$ red interior sides needs $2 - j$ red boundary sides, which can be chosen in $\binom{2}{2-j}$ ways: $1$ way if $j = 0$ or $j = 2,$ and $2$ ways if $j = 1.$

Group the $2^4 = 16$ cross colorings by the set of red arms. If all four arms have the same color ($2$ colorings), every square has $j = 0$ or $j = 2,$ contributing $1$ each: total $2.$ If exactly one arm is red or exactly one is blue ($8$ colorings), the two squares touching the odd arm have $j = 1$ and the others do not, contributing $2 \cdot 2 = 4$ each: total $32.$ If two adjacent arms are red ($4$ colorings), the squares have $j = 2, 1, 1, 0,$ contributing $4$ each: total $16.$ If two opposite arms are red ($2$ colorings), all four squares have $j = 1,$ contributing $2^4 = 16$ each: total $32.$

The number of colorings is $2 + 32 + 16 + 32 = 82.$

Solution:

By the change-of-base formula, $\log_k (5^{k^2-1}) = \frac{(k^2 - 1)\log 5}{\log k},$ so each factor of the product equals $$\frac{(k^2-1)/\log k}{(k^2-4)/\log(k+1)} = \frac{(k-1)(k+1)}{(k-2)(k+2)} \cdot \frac{\log(k+1)}{\log k}.$$

All three pieces telescope over $k = 4, \ldots, 63:$ $$\prod_{k=4}^{63} \frac{k-1}{k-2} = \frac{62}{2} = 31, \qquad \prod_{k=4}^{63} \frac{k+1}{k+2} = \frac{5}{65} = \frac{1}{13}, \qquad \prod_{k=4}^{63} \frac{\log(k+1)}{\log k} = \frac{\log 64}{\log 4} = 3.$$

The product is $31 \cdot \frac{1}{13} \cdot 3 = \frac{93}{13},$ which is in lowest terms, so $m + n = 93 + 13 = 106.$

Solution:

The medial triangle $DEF$ has sides parallel to those of $ABC,$ so $\angle FDE = 84^\circ,$ $\angle DEF = 60^\circ,$ and $\angle DFE = 36^\circ.$ Its circumcircle is the nine-point circle, whose second intersections with the sides of $ABC$ are the feet of the altitudes: $G$ is the foot from $A,$ $H$ the foot from $B,$ and $J$ the foot from $C.$ By the inscribed angle theorem, $\widehat{DE} = 2\angle DFE = 72^\circ.$

For $\widehat{FG}:$ since $\overline{DF} \parallel \overline{CA}$ and $G$ lies on ray $DB,$ the angle $\angle FDG$ equals the angle between lines $CA$ and $CB,$ which is $36^\circ,$ so $\widehat{FG} = 2 \cdot 36^\circ = 72^\circ.$ For $\widehat{HJ}:$ because $\angle BJC = \angle BHC = 90^\circ,$ both $H$ and $J$ lie on the circle with diameter $\overline{BC}$ centered at $D,$ so $DJ = DB$ and $DH = DC.$ Isosceles triangle $BDJ$ gives $\angle JDB = 180^\circ - 2 \cdot 60^\circ = 60^\circ,$ and isosceles triangle $CDH$ gives $\angle HDC = 180^\circ - 2 \cdot 36^\circ = 108^\circ.$ Hence $\angle JDH = 180^\circ - 60^\circ - 108^\circ = 12^\circ$ and $\widehat{HJ} = 24^\circ.$

Therefore $\widehat{DE} + 2 \cdot \widehat{HJ} + 3 \cdot \widehat{FG} = 72 + 48 + 216 = 336.$

Solution:

Center $\omega_2$ at the origin with $B = (15, 0).$ Internal tangency at $B$ puts $A = (9, 0),$ and $C = (-15, 0).$ Since $\overline{AD} \perp \overline{BC}$ and $D$ is on $\omega_2,$ we get $D = (9, 12)$ (taking $D$ above the line). Because $\overline{EF} \perp \overline{BC},$ the rectangle has vertical sides, so its vertices are $(9 \pm a, \pm b)$ with $a^2 + b^2 = 36.$ The conditions on $C$ and $D$ make $\overline{GH}$ the left side and $\overline{FG}$ the top side: $F = (9 + a, b),$ $G = (9 - a, b),$ $H = (9 - a, -b),$ $E = (9 + a, -b).$

Triangle $DGF$ has base $GF = 2a$ and height $12 - b,$ so its area is $a(12 - b).$ Triangle $CHG$ has base $GH = 2b$ and height $(9 - a) - (-15) = 24 - a,$ so its area is $b(24 - a).$ Setting these equal, $12a - ab = 24b - ab,$ so $a = 2b,$ and then $a^2 + b^2 = 5b^2 = 36.$

The area of the rectangle is $2a \cdot 2b = 8b^2 = \frac{288}{5},$ so $m + n = 288 + 5 = 293.$

Solution:

Since $2025 = 3^4 \cdot 5^2,$ the set $A$ has $5 \cdot 3 = 15$ elements, and there are $2^{15}$ subsets. A subset has least common multiple $2025$ exactly when it contains at least one divisor divisible by $3^4 = 81$ and at least one divisible by $5^2 = 25$ (such a subset is automatically nonempty). There are $12$ divisors not divisible by $81,$ $10$ not divisible by $25,$ and $8$ divisible by neither.

By inclusion-exclusion, the number of good subsets is $$2^{15} - 2^{12} - 2^{10} + 2^8 = 32768 - 4096 - 1024 + 256 = 27904.$$ Since $27904 = 2^8 \cdot 109,$ the probability is $\frac{27904}{32768} = \frac{109}{128},$ and $m + n = 109 + 128 = 237.$

Solution:

In any optimal collection there are at most $9$ pennies (ten pennies could become a dime) and at most $4$ dimes (five dimes could become two quarters), so its dimes and pennies are worth at most $49$ cents. Hence an optimal collection uses either $q = \lfloor N/25 \rfloor$ quarters, like greedy, or $q - 1$ quarters. For an amount $v$ made only of dimes and pennies, the best count is $f(v) = \lfloor v/10 \rfloor + (v \bmod 10),$ which is what greedy does on the remainder.

Let $r = N \bmod 25.$ Greedy uses $q + f(r)$ coins, and the only rival uses $(q - 1) + f(r + 25)$ coins (possible when $q \ge 1$), so greedy fails exactly when $f(r + 25) \le f(r).$ Tabulating: for $r = 0, \ldots, 4,$ $f(r+25) = r + 7 \gt f(r) = r;$ for $r = 5, \ldots, 9,$ $f(r+25) = r - 2 \le r;$ for $r = 10, \ldots, 14,$ $f(r+25) = r - 2 \gt r - 9;$ for $r = 15, \ldots, 19,$ $f(r+25) = r - 11 \le r - 9;$ for $r = 20, \ldots, 24,$ $f(r+25) = r - 11 \gt r - 18.$ So greedy fails exactly when $N \ge 25$ and $r \in \{5, \ldots, 9\} \cup \{15, \ldots, 19\}.$

Each residue class mod $25$ contains $40$ values of $N$ in $1, \ldots, 1000,$ so these $10$ residues give $400$ values, of which the $10$ values less than $25$ do not count (there $q = 0$). Greedy fails for $390$ values and succeeds for $1000 - 390 = 610.$

Solution:

$f(x) = 0$ exactly when $7\pi \sin(5x)$ is a multiple of $\pi,$ that is, $\sin(5x) = \frac{k}{7}$ for an integer $k$ with $|k| \le 7.$ As $x$ runs over $(0, 2\pi),$ the quantity $5x$ runs over $(0, 10\pi),$ five full periods. For $k = 0,$ the solutions are $5x = \pi, 2\pi, \ldots, 9\pi:$ $9$ values. For each of the $12$ values $k = \pm 1, \ldots, \pm 6,$ each period contributes $2$ solutions: $10$ values each. For $k = \pm 7,$ we need $\sin(5x) = \pm 1,$ which happens $5$ times each. So $n = 9 + 120 + 10 = 139.$

The graph is tangent to the $x$-axis at a zero exactly when $f'(x) = 35\pi \cos(7\pi \sin(5x)) \cos(5x) = 0$ there. At any zero, $\cos(7\pi \sin(5x)) = \cos(k\pi) = \pm 1 \ne 0,$ so tangency requires $\cos(5x) = 0,$ which means $\sin(5x) = \pm 1:$ exactly the $10$ zeros with $k = \pm 7$ (there $\sin(5x)$ has an extremum, so $f$ touches without crossing). Thus $t = 10$ and $n + t = 149.$

Solution:

A person sits next to two others exactly when three consecutive chairs are all occupied, so we count $8$-element subsets of the $16$ chairs with no three consecutive chairs chosen. The occupied chairs then form maximal blocks of size $1$ or $2.$ If there are $m$ blocks, then $8 - m$ of them are pairs and $2m - 8$ are singles, so $4 \le m \le 8,$ and the pair positions can be chosen in $\binom{m}{8-m}$ ways. The $8$ empty chairs create $9$ gaps (including the ends), and the $m$ blocks occupy $m$ distinct gaps: $\binom{9}{m}$ ways.

Therefore $$N = \sum_{m=4}^{8} \binom{m}{8-m}\binom{9}{m} = 1 \cdot 126 + 10 \cdot 126 + 15 \cdot 84 + 7 \cdot 36 + 1 \cdot 9 = 2907.$$

The remainder when $N = 2907$ is divided by $1000$ is $907.$

Solution:

Two chords of a circle through equally spaced points have equal length exactly when they skip the same number of vertices, so all $12$ segments join pairs of vertices exactly $k$ apart for one common $k \in \{1, \ldots, 12\}.$ For fixed $k,$ form the graph on the $24$ vertices joining each $i$ to $i \pm k \pmod{24}:$ we need a perfect matching in this graph. For $k \lt 12$ the graph is a disjoint union of $\gcd(24, k)$ cycles of length $24/\gcd(24, k),$ while for $k = 12$ it is $12$ disjoint diameters.

A cycle of even length has exactly $2$ perfect matchings (alternate edges), and a cycle of odd length has none. So each $k \lt 12$ with even cycle length contributes $2^{\gcd(24, k)}:$ $k = 1, 5, 7, 11$ give $2$ each; $k = 2, 10$ give $4$ each; $k = 3, 9$ give $8$ each; $k = 4$ gives $16;$ $k = 6$ gives $64.$ For $k = 8$ the cycles have odd length $3,$ giving $0.$ For $k = 12$ the matching is forced: $1$ way.

The total is $4 \cdot 2 + 2 \cdot 4 + 2 \cdot 8 + 16 + 64 + 0 + 1 = 113.$

Solution:

Let $r_i = A_1A_i$ for $2 \le i \le 11,$ and let $\theta$ be the common angle, with $\cos\theta = \frac{12}{13}$ and $\sin\theta = \frac{5}{13}.$ Each area condition says $\frac{1}{2} r_i r_{i+1} \cdot \frac{5}{13} = 1,$ so $r_i r_{i+1} = \frac{26}{5}$ for $i = 2, \ldots, 10.$ Consecutive products being equal forces the $r_i$ to alternate between two values $a = r_2 = r_4 = \cdots$ and $b = r_3 = r_5 = \cdots,$ with $ab = \frac{26}{5};$ in particular $r_{11} = b.$

By the law of cosines, every side $A_iA_{i+1}$ with $2 \le i \le 10$ has the same length $s,$ where $$s^2 = a^2 + b^2 - 2ab \cdot \tfrac{12}{13} = (a + b)^2 - 2ab - \tfrac{48}{5} = (a+b)^2 - 20.$$ Writing $u = a + b,$ the perimeter condition is $9\sqrt{u^2 - 20} + u = 20.$ Squaring $9\sqrt{u^2 - 20} = 20 - u$ gives $81u^2 - 1620 = 400 - 40u + u^2,$ which simplifies to $4u^2 + 2u - 101 = 0,$ so $u = \frac{-1 + 9\sqrt{5}}{4}$ (the positive root; then $20 - u \gt 0$ as required).

Thus $A_1A_2 + A_1A_{11} = a + b = \frac{9\sqrt{5} - 1}{4},$ with $5$ squarefree and no prime dividing all of $9,$ $1,$ $4.$ The answer is $9 + 5 + 1 + 4 = 19.$

Solution:

Let $y_k = \frac{2x_k - 1}{x_k + 1}.$ From the recurrence, $2x_{k+1} - 1 = \frac{(2x_k - 1)(x_k - 2)}{3x_k}$ and $x_{k+1} + 1 = \frac{(x_k + 1)^2}{3x_k},$ so $$y_{k+1} = \frac{(2x_k - 1)(x_k - 2)}{(x_k + 1)^2} = y_k(y_k - 1) = y_k^2 - y_k,$$ since $y_k - 1 = \frac{x_k - 2}{x_k + 1}.$ Here $y_1 = \frac{39/11}{36/11} = \frac{13}{12}.$ By induction $y_k = \frac{c_k}{12^{2^{k-1}}}$ where $c_1 = 13$ and $c_{k+1} = c_k\bigl(c_k - 12^{2^{k-1}}\bigr);$ since $12^{2^{k-1}}$ is divisible by $6,$ every $c_k$ stays coprime to $6.$

Inverting the substitution, $x_k = \frac{1 + y_k}{2 - y_k} = \frac{d + c}{2d - c}$ with $d = 12^{2^{k-1}}$ and $c = c_k.$ All $x_k$ are positive (for $x \gt 0,$ $x + \frac{1}{x} - 1 \ge 1$), so $y_k = \frac{2x_k - 1}{x_k + 1} \in (-1, 2),$ making both $d + c$ and $2d - c$ positive. Any common divisor of $d + c$ and $2d - c$ divides their combinations $3d$ and $3c;$ as $\gcd(c, d) = 1,$ it divides $3,$ but $3 \mid d$ and $3 \nmid c,$ so $3 \nmid d + c.$ Hence the fraction is in lowest terms and $m + n = 3d = 3 \cdot 12^{2^{2024}}.$

Modulo $8,$ $12^{2^{2024}} \equiv 0.$ Modulo $125,$ the multiplicative order of $12$ divides $\lambda(125) = 100,$ and $2^{2024} \equiv 16 \pmod{100}$ (it is $0$ mod $4,$ and $2^{20} \equiv 1$ mod $25$ with $2024 \equiv 4$ mod $20$), so $12^{2^{2024}} \equiv 12^{16} \equiv 41 \pmod{125}.$ The Chinese remainder theorem gives $12^{2^{2024}} \equiv 416 \pmod{1000},$ so $m + n \equiv 3 \cdot 416 = 1248 \equiv 248 \pmod{1000}.$

Solution:

Since $AK = AL = KL = 14,$ triangle $AKL$ is equilateral and $\angle KAL = 60^\circ.$ Let $\alpha = \angle BAK$ and $\beta = \angle LAC,$ so $\alpha + \beta = 30^\circ.$ Because $AK = KB,$ point $K$ lies on the perpendicular bisector of $\overline{AB},$ so $AB = 2 \cdot 14\cos\alpha = 28\cos\alpha;$ similarly $AC = 28\cos\beta.$ Then $AB^2 + AC^2 = 38^2$ gives $\cos^2\alpha + \cos^2\beta = \frac{361}{196},$ i.e. $\cos 2\alpha + \cos 2\beta = \frac{165}{98}.$ By sum-to-product, $2\cos(\alpha+\beta)\cos(\alpha-\beta) = \sqrt{3}\cos(\alpha - \beta) = \frac{165}{98},$ so $\cos(\alpha - \beta) = \frac{55\sqrt{3}}{98}.$

Decompose $[BKLC] = [ABC] - [ABK] - [ACL] - [AKL].$ First, $$[ABC] = \tfrac{1}{2} AB \cdot AC = 392\cos\alpha\cos\beta = 196\bigl(\cos(\alpha - \beta) + \cos(\alpha + \beta)\bigr) = 110\sqrt{3} + 98\sqrt{3} = 208\sqrt{3}.$$ Next, $K$ has height $14\sin\alpha$ over $\overline{AB},$ so $[ABK] = \frac{1}{2} \cdot 28\cos\alpha \cdot 14\sin\alpha = 98\sin 2\alpha,$ and likewise $[ACL] = 98\sin 2\beta;$ their sum is $196\sin(\alpha + \beta)\cos(\alpha - \beta) = 98 \cdot \frac{55\sqrt{3}}{98} = 55\sqrt{3}.$ Finally $[AKL] = \frac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}.$

Therefore $[BKLC] = 208\sqrt{3} - 55\sqrt{3} - 49\sqrt{3} = 104\sqrt{3},$ so $n = 104.$

Solution:

For $x \gt 0,$ $f(x) \to +\infty$ both as $x \to 0^+$ (the numerator tends to $18 \cdot 72 \cdot 98 \cdot k \gt 0$) and as $x \to \infty,$ so $f$ attains a global minimum value $c$ on $(0, \infty).$ It is attained at exactly two points precisely when $f(x) - c \ge 0$ with two distinct positive double roots, i.e. $$(x - 18)(x - 72)(x - 98)(x - k) - cx = (x^2 - Sx + P)^2$$ where the roots of $x^2 - Sx + P$ are positive and distinct (so $S, P \gt 0$).

Matching coefficients of $x^3,$ $x^2,$ and the constant (the $x$-coefficient just determines $c$): $$2S = 188 + k, \qquad S^2 + 2P = 10116 + 188k, \qquad P^2 = 18 \cdot 72 \cdot 98 \cdot k = 127008k.$$ Substitute $k = 2t^2$ with $t \gt 0:$ then $S = 94 + t^2$ and $P = 504t.$ The middle equation becomes $(94 + t^2)^2 + 1008t = 10116 + 376t^2,$ i.e. $$t^4 - 188t^2 + 1008t - 1280 = 0,$$ which factors as $(t - 2)(t - 4)(t + 16)(t - 10) = 0.$