Answer: 154

Small Hint:

Let the original crowd be $12k$ people, so that the adult count $\frac{5}{12} \cdot 12k = 5k$ is a whole number

Big Hint:

The new total $12k + 50$ must be divisible by $25,$ which forces $k$ to be a multiple of $25$

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 $25$ divides $12k,$ 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 $154.$

Answer: 125

Small Hint:

There are three equally likely pairings, and Carl’s semifinal opponent is Azar in exactly one of them

Big Hint:

If Carl’s semifinal opponent is Jon or Sergey, his opponent in the final is Azar with probability $\frac{3}{4}$ and Jon or Sergey with probability $\frac{1}{4}$

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 $$\begin{aligned} &\frac{3}{4}\left(\frac{3}{4} \cdot \frac{1}{3} + \frac{1}{4} \cdot \frac{3}{4}\right) \\ &= \frac{3}{4} \cdot \frac{7}{16} \\ &= \frac{21}{64}. \end{aligned}$$

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 $p + q = 29 + 96 = 125.$

Answer: 21

Small Hint:

The volume determines the height: $\frac{1}{3} \cdot 36 \cdot h = 54$

Big Hint:

The sphere’s center lies on the pyramid’s axis; set its distance to the apex equal to its distance to a base vertex, which sits $3\sqrt{2}$ from the axis

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 $m + n = 17 + 4 = 21.$

Answer: 112

Small Hint:

Change both sides to a common base: $\frac{\ln 22x}{\ln 20x} = \frac{\ln 202x}{\ln 2x}$

Big Hint:

If $\frac{p}{q} = \frac{r}{s},$ then both fractions also equal $\frac{r - p}{s - q}.$ Apply this and simplify the resulting quotient.

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: $$\begin{aligned} y &= \frac{\ln 202x - \ln 22x}{\ln 2x - \ln 20x} \\ &= \frac{\ln \frac{101}{11}}{\ln \frac{1}{10}} \\ &= -\log_{10}\frac{101}{11} \\ &= \log_{10}\frac{11}{101}. \end{aligned}$$

To check existence rather than merely use the promised $x,$ note that $y \ne 1.$ Setting $$\ln x=\frac{y\ln 20-\ln 22}{1-y}$$ makes $\ln(22x)=y\ln(20x).$ Also $$\begin{aligned} \ln(202x)-\ln(22x) &= \ln\frac{101}{11} \\ &= y\ln\frac{1}{10}, \end{aligned}$$ so $\ln(202x)=y\ln(2x)$ as well. This positive $x$ is neither excluded value (neither one satisfies the displayed linear equation), so both logarithm bases are valid. Since $\gcd(11,101)=1,$ we get $m+n=11+101=112.$

Answer: 72

Small Hint:

For labels $i \lt j \lt k,$ all three differences must be prime, and $k - i$ is the sum of the other two

Big Hint:

Two odd primes sum to an even number, so one difference must be $2:$ the differences are $2, p, p + 2$ with $p$ and $p + 2$ both prime

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 $30 + 26 + 14 + 2 = 72.$

Answer: 841

Small Hint:

The two conditions force the positive terms to sum to $\frac{1}{2}$ and the negative terms to sum to $-\frac{1}{2}$

Big Hint:

If $x_{16}$ were below $-\frac{1}{32},$ the first $16$ terms alone would sum below $-\frac{1}{2};$ bound $x_{76}$ the same way using the last $25$ terms

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 $41 + 800 = 841.$

Answer: 192

Small Hint:

The two external tangents meet at a point $P$ on the line through the centers, whose distances to the centers are in ratio $24 : 6$

Big Hint:

The third side is the common tangent at the point of tangency, perpendicular to the center line. Use $\sin\theta = \frac{24}{40}$ for the half-angle at $P.$

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

Answer: 80

Small Hint:

The integers sharing a given triple of floor values form a block of consecutive integers, so $n$ is determined exactly when its block has size $1$

Big Hint:

The block has size $1$ exactly when each of $n$ and $n + 1$ is divisible by at least one of $4, 5, 6.$ Count such $n$ in one period of $60.$

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 $60:$ $4$ dividing $n$ and $5$ dividing $n + 1$ gives $n \equiv 4, 24, 44;$ $5$ dividing $n$ and $4$ dividing $n + 1$ gives $n \equiv 15, 35, 55;$ $5$ dividing $n$ and $6$ dividing $n + 1$ gives $n \equiv 5, 35;$ and $6$ dividing $n$ and $5$ dividing $n + 1$ 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.)

Answer: 244

Small Hint:

Each choice of two $A$’s and two $B$’s produces exactly one interior crossing, so there are $\binom{7}{2}\binom{5}{2}$ crossing points

Big Hint:

Use Euler’s formula $V - E + F = 2:$ each crossing splits two segments, so the segments contribute $mn + 2\binom{m}{2}\binom{n}{2}$ edges

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

Answer: 4

Small Hint:

Expand: $\binom{\binom{n}{2}}{2}$ equals $\frac{(n+1)n(n-1)(n-2)}{8},$ a polynomial in $n$

Big Hint:

That expression is $3\binom{n+1}{4};$ sum it with the hockey stick identity

Solution:

Since $\binom{n}{2} = \frac{n(n-1)}{2}$ and $\binom{n}{2} - 1 = \frac{(n+1)(n-2)}{2},$ $$\begin{aligned} \binom{\binom{n}{2}}{2} \\ &= \frac{1}{2} \cdot \frac{n(n-1)}{2} \\ &\quad {}\cdot \frac{(n+1)(n-2)}{2} \\ &= \small \frac{(n+1)n(n-1)(n-2)}{8} \\ &= 3\binom{n+1}{4}. \end{aligned}$$

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

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

Answer: 180

Small Hint:

Reflect $B$ over the bisector from $A$ and $C$ over the bisector from $D:$ both reflections land on $\overline{AD}$

Big Hint:

Since $MB = MC,$ the midpoint $M$ is equidistant from the two reflections, which lie on $\overline{AD}$ at distances $2$ from $A$ and $3$ from $D$

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 $180.$

Answer: 23

Small Hint:

Both equations are ellipses through a common point: the first has foci $(\pm 4, 0),$ the second has foci $(20, 10)$ and $(20, 12)$

Big Hint:

$2a + 2b$ is the sum of the four focal distances of the common point; pair the foci so the triangle inequality gives segments that actually intersect

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 $$\begin{aligned} 2a + 2b \\ &= (PF_1 + PG_1) \\ &\quad {}+ (PF_2 + PG_2) \\ &\ge F_1G_1 + F_2G_2 \\ &= \sqrt{24^2 + 10^2} \\ &\quad {}+ \sqrt{16^2 + 12^2} \\ &= 26 + 20 \\ &= 46, \end{aligned}$$ 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 $16 + 7 = 23.$

Answer: 220