Solution:

Let the original crowd have $12k$ people, of whom $5k$ are adults. After the bus arrives there are $12k + 50$ people, and the number of adults is $\frac{11}{25}(12k + 50).$ For this to be an integer, $25$ must divide $12k + 50,$ so $25 \mid 12k,$ and since $\gcd(12, 25) = 1$ this means $k$ is a multiple of $25.$

The adult count $\frac{11}{25}(12k + 50)$ increases with $k,$ so the minimum occurs at $k = 25:$ the new total is $350$ and the number of adults is $\frac{11}{25} \cdot 350 = 154.$ This is achievable, for example if the bus carries $29$ adults and $21$ non-adults, so the answer is $154.$

Solution:

The three ways to pair the four players are equally likely, so Carl plays Azar in the semifinal with probability $\frac{1}{3}.$ In that case Carl beats Azar with probability $\frac{1}{3}$ and then beats the Jon–Sergey winner with probability $\frac{3}{4},$ so Carl wins the tournament with probability $\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}.$

Otherwise (probability $\frac{2}{3}$) Carl plays Jon or Sergey and wins with probability $\frac{3}{4}.$ His opponent in the final is Azar with probability $\frac{3}{4}$ (Carl then wins with probability $\frac{1}{3}$) and is Jon or Sergey with probability $\frac{1}{4}$ (Carl then wins with probability $\frac{3}{4}$). So in this case Carl wins the tournament with probability $$\frac{3}{4}\left(\frac{3}{4} \cdot \frac{1}{3} + \frac{1}{4} \cdot \frac{3}{4}\right) = \frac{3}{4} \cdot \frac{7}{16} = \frac{21}{64}.$$

The total probability is $\frac{1}{3} \cdot \frac{1}{4} + \frac{2}{3} \cdot \frac{21}{64} = \frac{1}{12} + \frac{7}{32} = \frac{29}{96},$ so $p + q = 29 + 96 = 125.$

Solution:

The base has area $36,$ so $\frac{1}{3} \cdot 36 \cdot h = 54$ gives height $h = \frac{9}{2}.$ By symmetry the sphere's center lies on the pyramid's axis, say at height $z$ above the base. Each base vertex is at distance $3\sqrt{2}$ from the axis, so the center's distance to a base vertex is $\sqrt{z^2 + 18},$ while its distance to the apex is $\frac{9}{2} - z.$

Setting $\left(\frac{9}{2} - z\right)^2 = z^2 + 18$ gives $\frac{81}{4} - 9z = 18,$ so $z = \frac{1}{4}.$ The radius is $\frac{9}{2} - \frac{1}{4} = \frac{17}{4},$ and $m + n = 17 + 4 = 21.$

Solution:

Let $y$ be the common value. In natural logarithms, $$y = \frac{\ln 22x}{\ln 20x} = \frac{\ln 202x}{\ln 2x}.$$ When two fractions are equal, each also equals the quotient of the differences of numerators and denominators: $$y = \frac{\ln 202x - \ln 22x}{\ln 2x - \ln 20x} = \frac{\ln \frac{101}{11}}{\ln \frac{1}{10}} = -\log_{10}\frac{101}{11} = \log_{10}\frac{11}{101}.$$

(Such an $x$ exists: the equation rearranges to a solvable condition, and the excluded values $\frac{1}{20}, \frac{1}{2}$ only rule out degenerate bases.) Since $\gcd(11, 101) = 1,$ we get $m + n = 11 + 101 = 112.$

Solution:

A triangle has vertices $i \lt j \lt k$ where $j - i,$ $k - j,$ and $k - i$ are all prime. Since $k - i = (j - i) + (k - j)$ is a prime that is a sum of two primes, and the sum of two odd primes is even, one of the two smaller differences must equal $2.$ So the differences are $\{2, p\}$ in some order with $p$ and $p + 2$ both prime: the twin prime pairs with $p + 2 \le 19$ are $(3, 5),$ $(5, 7),$ $(11, 13),$ and $(17, 19).$

For each pair, the middle vertex can be at distance $2$ or at distance $p$ from the smallest, and the total span is $p + 2,$ so there are $2\bigl(20 - (p + 2)\bigr)$ triangles. This gives $2 \cdot 15 = 30,$ $2 \cdot 13 = 26,$ $2 \cdot 7 = 14,$ and $2 \cdot 1 = 2$ for the four pairs.

The total is $30 + 26 + 14 + 2 = 72.$

Solution:

Since the terms sum to $0$ while their absolute values sum to $1,$ the positive terms sum to $\frac{1}{2}$ and the negative terms sum to $-\frac{1}{2}.$ If $x_{16} \lt -\frac{1}{32},$ then $x_1, \ldots, x_{16}$ are all less than $-\frac{1}{32}$ and would sum below $-\frac{1}{2},$ a contradiction; hence $x_{16} \ge -\frac{1}{32}.$ Similarly, if $x_{76} \gt \frac{1}{50}$ then $x_{76}, \ldots, x_{100}$ are $25$ terms each exceeding $\frac{1}{50},$ summing above $\frac{1}{2};$ hence $x_{76} \le \frac{1}{50}.$

Therefore $x_{76} - x_{16} \le \frac{1}{50} + \frac{1}{32} = \frac{16 + 25}{800} = \frac{41}{800},$ and this is achieved by taking $x_1 = \cdots = x_{16} = -\frac{1}{32},$ $x_{17} = \cdots = x_{75} = 0,$ and $x_{76} = \cdots = x_{100} = \frac{1}{50}.$

Since $\gcd(41, 800) = 1,$ the answer is $41 + 800 = 841.$

Solution:

The centers $O_1$ (radius $24$) and $O_2$ (radius $6$) are $30$ apart. The two external tangents meet at a point $P$ on line $O_1O_2$ beyond the small circle, with $\frac{PO_1}{PO_2} = \frac{24}{6} = 4.$ Combined with $PO_1 - PO_2 = 30,$ this gives $PO_1 = 40$ and $PO_2 = 10.$ Each external tangent makes angle $\theta$ with the center line, where $\sin\theta = \frac{24}{40} = \frac{3}{5},$ so $\tan\theta = \frac{3}{4}.$

The third common tangent is the tangent at the point of tangency $T,$ which is perpendicular to $O_1O_2$ at distance $24$ from $O_1.$ The triangle bounded by the three tangents has apex $P$ and base on this line, with height $PT = 40 - 24 = 16$ and half-base $16\tan\theta = 12.$

Its area is $\frac{1}{2} \cdot 24 \cdot 16 = 192.$

Solution:

The set of positive integers sharing a given triple $\left(\left\lfloor \frac{n}{4} \right\rfloor, \left\lfloor \frac{n}{5} \right\rfloor, \left\lfloor \frac{n}{6} \right\rfloor\right)$ is an intersection of three intervals, hence a block of consecutive integers. So $n$ is uniquely determined exactly when neither $n - 1$ nor $n + 1$ gives the same triple: some floor must drop at $n - 1,$ meaning $4, 5,$ or $6$ divides $n,$ and some floor must jump at $n + 1,$ meaning $4, 5,$ or $6$ divides $n + 1.$

Since $n$ and $n + 1$ cannot both be even, the divisor pairs for $(n, n + 1)$ are $(4, 5),$ $(5, 4),$ $(5, 6),$ and $(6, 5).$ Working modulo $60:$ $4 \mid n,\; 5 \mid n + 1$ gives $n \equiv 4, 24, 44;$ $5 \mid n,\; 4 \mid n + 1$ gives $n \equiv 15, 35, 55;$ $5 \mid n,\; 6 \mid n + 1$ gives $n \equiv 5, 35;$ and $6 \mid n,\; 5 \mid n + 1$ gives $n \equiv 24, 54.$ The union is the $8$ residues $\{4, 5, 15, 24, 35, 44, 54, 55\}$ modulo $60.$

Each residue occurs $10$ times among $1 \le n \le 600,$ so the count is $8 \cdot 10 = 80.$ (Note $n = 600$ fails: $601$ is divisible by none of $4, 5, 6,$ so $601, 602, 603$ share $600$'s triple.)

Solution:

Two segments $\overline{A_iB_j}$ and $\overline{A_kB_l}$ cross strictly between the lines exactly when one of the $A$'s comes first and the other's $B$ comes first, which happens for exactly one pairing of any two $A$'s with any two $B$'s. By the general-position hypothesis these crossings are distinct, so there are $X = \binom{m}{2}\binom{n}{2}$ of them.

Clip the two lines to long segments and apply Euler's formula. The vertices are the $m + n$ marked points, the $X$ crossings, and the $4$ clipped line ends, so $V = m + n + X + 4.$ Line $\ell_A$ is divided into $m + 1$ edges and $\ell_B$ into $n + 1;$ each crossing splits two segments, so the drawn segments contribute $mn + 2X$ edges, giving $E = mn + m + n + 2X + 2.$ Then $$F = E - V + 2 = mn + X,$$ of which one face is unbounded, so there are $mn + X - 1$ bounded regions. For $m = 3,$ $n = 2$ this gives $6 + 3 - 1 = 8,$ matching the figure.

For $m = 7$ and $n = 5:$ $35 + \binom{7}{2}\binom{5}{2} - 1 = 35 + 21 \cdot 10 - 1 = 244.$

Solution:

Since $\binom{n}{2} = \frac{n(n-1)}{2}$ and $\binom{n}{2} - 1 = \frac{(n+1)(n-2)}{2},$ $$\binom{\binom{n}{2}}{2} = \frac{1}{2} \cdot \frac{n(n-1)}{2} \cdot \frac{(n+1)(n-2)}{2} = \frac{(n+1)n(n-1)(n-2)}{8} = 3\binom{n+1}{4}.$$

By the hockey stick identity, $$\sum_{n=3}^{40} 3\binom{n+1}{4} = 3\sum_{k=4}^{41}\binom{k}{4} = 3\binom{42}{5} = 3 \cdot 850668 = 2552004.$$

The remainder upon division by $1000$ is $4.$

Solution:

Place $A = (0, 0)$ and $D = (7, 0)$ with $B, C$ above the axis, and let $M$ be the midpoint of $\overline{BC}.$ Reflecting $B$ over the bisector line $AM$ carries ray $AB$ to ray $AD,$ so $B$ maps to $B' = (2, 0),$ and reflecting $C$ over the bisector $DM$ gives $C' = (4, 0).$ Since $M$ lies on both mirror lines, $MB' = MB = MC = MC',$ so $M$ is equidistant from $B'$ and $C'$ and hence $M = (3, h)$ for some $h \gt 0.$

Write $\angle DAB = 2\alpha$ and $\angle ADC = 2\delta,$ so $\tan\alpha = \frac{h}{3}$ and $\tan\delta = \frac{h}{4}.$ Then $B = (2\cos 2\alpha,\, 2\sin 2\alpha)$ and $C = (7 - 3\cos 2\delta,\, 3\sin 2\delta),$ and the midpoint condition on the $x$-coordinates reads $2\cos 2\alpha - 3\cos 2\delta = -1.$ Substituting $\cos 2\alpha = \frac{9 - h^2}{9 + h^2}$ and $\cos 2\delta = \frac{16 - h^2}{16 + h^2}$ and clearing denominators gives $2h^4 = 10h^2,$ so $h^2 = 5.$ (The $y$-coordinate condition is then satisfied automatically: $2\sin 2\alpha + 3\sin 2\delta = \frac{6\sqrt{5}}{7} + \frac{8\sqrt{5}}{7} = 2h.$)

Now $\cos 2\alpha = \frac{2}{7},$ $\sin 2\alpha = \frac{3\sqrt{5}}{7},$ $\cos 2\delta = \frac{11}{21},$ $\sin 2\delta = \frac{8\sqrt{5}}{21},$ so $B = \left(\frac{4}{7}, \frac{6\sqrt{5}}{7}\right)$ and $C = \left(\frac{38}{7}, \frac{8\sqrt{5}}{7}\right).$ The shoelace formula on $A, B, C, D$ gives area $6\sqrt{5},$ whose square is $180.$

Solution:

The first ellipse has $c^2 = a^2 - (a^2 - 16) = 16,$ hence foci $F_1 = (-4, 0)$ and $F_2 = (4, 0),$ with distance sum $2a.$ The second is centered at $(20, 11)$ with vertical major axis and $c^2 = b^2 - (b^2 - 1) = 1,$ hence foci $G_1 = (20, 10)$ and $G_2 = (20, 12),$ with distance sum $2b.$ If $P = (x, y)$ lies on both, then $$2a + 2b = (PF_1 + PG_1) + (PF_2 + PG_2) \ge F_1G_1 + F_2G_2 = \sqrt{24^2 + 10^2} + \sqrt{16^2 + 12^2} = 26 + 20 = 46,$$ so $a + b \ge 23.$

Equality requires $P$ to lie on both segments $\overline{F_1G_1}$ and $\overline{F_2G_2}.$ These segments do intersect, at $P = \left(14, \frac{15}{2}\right):$ then $PF_1 = \frac{39}{2},$ $PF_2 = \frac{25}{2},$ so $2a = 32,$ $a = 16 \gt 4;$ and $PG_1 = \frac{13}{2},$ $PG_2 = \frac{15}{2},$ so $2b = 14,$ $b = 7 \gt 1.$

Hence the least possible value of $a + b$ is $16 + 7 = 23.$

Solution:

For $0 \lt x \lt 1,$ $$P(x) = (1 - x^{2310})^6 \cdot \frac{1}{(1 - x^{105})(1 - x^{70})(1 - x^{42})(1 - x^{30})},$$ and each factor $\frac{1}{1 - x^k}$ expands as a geometric series. Since $2022 \lt 2310,$ the factor $(1 - x^{2310})^6$ contributes only its constant term $1,$ so the coefficient of $x^{2022}$ is the number of nonnegative integer solutions of $105a + 70b + 42c + 30d = 2022.$

Reducing modulo $2$ gives $105a \equiv 2022,$ so $a$ is even; modulo $3$ gives $70b \equiv 2022 \equiv 0,$ so $3 \mid b;$ modulo $5$ gives $2c \equiv 2022 \equiv 2,$ so $c \equiv 1 \pmod 5;$ modulo $7$ gives $2d \equiv 2022 \equiv 6,$ so $d \equiv 3 \pmod 7.$ Writing $a = 2a',$ $b = 3b',$ $c = 5c' + 1,$ $d = 7d' + 3$ turns the equation into $$210(a' + b' + c' + d') + 42 + 90 = 2022, \qquad \text{so} \qquad a' + b' + c' + d' = 9.$$

By stars and bars there are $\binom{12}{3} = 220$ solutions, so the coefficient is $220.$

Solution:

To form $1$ cent we need $a = 1.$ Suppose the collection has $x$ ones, $y$ stamps of $b,$ and $z$ of $c.$ The value $b - 1$ must be made from ones alone, so $x \ge b - 1;$ the value $c - 1$ must be made from ones and $b$'s, so $x + yb \ge c - 1;$ and the total $x + yb + zc$ must be at least $1000.$ Conversely these three conditions suffice: with $x \ge b - 1$ the ones and $b$'s make every value up to $x + yb,$ and then $c$'s extend this to every value up to the total. So the optimum takes $x = b - 1,$ then the least $y$ with $x + yb \ge c - 1,$ then the least $z$ reaching $1000.$

For fixed $c,$ the count $f(1, b, c)$ is maximized at $b = c - 1$ (many ones, which cover value least efficiently), where the optimal collection is $c - 2$ ones, one stamp of $c - 1,$ and $\left\lceil \frac{1003 - 2c}{c} \right\rceil$ stamps of $c,$ totaling $c - 3 + \left\lceil \frac{1003}{c} \right\rceil.$ For $12 \le c \le 87$ this maximum is at most $96$ (it is $93$ at $c = 12,$ decreases in the middle, and returns to $96$ at $c = 87$), so no $b$ gives $97;$ a quick check of $c \le 10$ shows the possible counts skip $97$ there as well.

For $c = 11,$ taking $b = 7$ gives $6$ ones, one $7$ (reaching $13 \ge 10$), and $\left\lceil \frac{987}{11} \right\rceil = 90$ elevens: $f(1, 7, 11) = 6 + 1 + 90 = 97.$ For $c = 88$ and $c = 89,$ taking $b = 87$ gives $86$ ones, one $87$ (reaching $173$), and $\left\lceil \frac{827}{88} \right\rceil = \left\lceil \frac{827}{89} \right\rceil = 10$ stamps of $c,$ for $86 + 1 + 10 = 97$ in both cases. So the three least values of $c$ are $11, 88, 89,$ with sum $188.$

Solution:

All six hexagon vertices lie on $\Omega:$ $O_1$ and $O_2$ by hypothesis, and $A, B, C, D$ as intersection points with $\Omega.$ Let $R$ be the radius of $\Omega,$ and let the arcs cut off by the sides $AB,$ $BO_1,$ $O_1C,$ $CD,$ $DO_2,$ $O_2A$ be $2\alpha, 2\beta, 2\beta, 2\gamma, 2\delta, 2\delta$ (the two $\beta$'s because chords $BO_1 = O_1C = r_1,$ the radius of $\omega_1,$ and likewise $O_2A = O_2D = r_2$), so $\alpha + 2\beta + \gamma + 2\delta = \pi.$ Each chord equals $2R\sin(\text{half its arc}):$ $2R\sin\alpha = 2,$ $2R\sin\gamma = 16,$ and the chord $\overline{O_1O_2}$ subtends $2\beta + 2\gamma + 2\delta,$ giving $2R\sin(\alpha + \sigma) = 15$ where $\sigma = \beta + \delta.$ External tangency gives a second equation worth $15:$ $r_1 + r_2 = 2R(\sin\beta + \sin\delta) = 15.$

Since $\gamma = \pi - \alpha - 2\sigma,$ we have $\sin\gamma = \sin(\alpha + 2\sigma).$ Sum-to-product then gives $$18 = 2R\left[\sin(\alpha + 2\sigma) + \sin\alpha\right] = 4R\sin(\alpha + \sigma)\cos\sigma = 30\cos\sigma,$$ so $\cos\sigma = \frac{3}{5},$ and similarly $14 = 4R\cos(\alpha + \sigma)\sin\sigma$ gives $R\cos(\alpha + \sigma) = \frac{35}{8}.$ Combining with $2R\sin(\alpha + \sigma) = 15$ yields $4R^2 = 225 + \frac{1225}{16},$ so $R^2 = \frac{4825}{64}.$ Also $15 = 2R(\sin\beta + \sin\delta) = 4R\sin\frac{\sigma}{2}\cos\frac{\beta - \delta}{2}$ with $\sin\frac{\sigma}{2} = \frac{1}{\sqrt{5}},$ so $\cos\frac{\beta - \delta}{2} = \frac{15\sqrt{5}}{4R}$ and $\cos(\beta - \delta) = \frac{1125}{8R^2} - 1.$

Joining the center of $\Omega$ to the six vertices splits the hexagon into six triangles, so its area is $\frac{1}{2}R^2\left[\sin 2\alpha + 2\sin 2\beta + 2\sin 2\delta + \sin 2\gamma\right].$ Now $R^2(\sin 2\beta + \sin 2\delta) = 2R^2 \sin\sigma\cos(\beta - \delta) = \frac{8}{5}\left(\frac{1125}{8} - R^2\right) = \frac{835}{8},$ while $\sin 2\alpha + \sin 2\gamma = -2\sin 2\sigma \cos\bigl(2(\alpha + \sigma)\bigr) = -2 \cdot \frac{24}{25} \cdot \left(-\frac{95}{193}\right)$ since $\cos\bigl(2(\alpha+\sigma)\bigr) = 1 - \frac{2 \cdot 225}{4R^2} = -\frac{95}{193},$ so $\frac{1}{2}R^2(\sin 2\alpha + \sin 2\gamma) = \frac{1}{2} \cdot \frac{4825}{64} \cdot \frac{912}{965} = \frac{285}{8}.$ The area is $\frac{835}{8} + \frac{285}{8} = 140.$