Answer: 97

Solution:

Zou wins race $1,$ so winning exactly $5$ of the $6$ races means he loses exactly one of races $2$ through $6.$ Each race after the first repeats the previous outcome with probability $\frac{2}{3}$ and switches with probability $\frac{1}{3}.$

If the loss is race $6,$ the five transitions are four repeats followed by one switch: $\left(\frac{2}{3}\right)^4 \cdot \frac{1}{3} = \frac{16}{243}.$ If the loss is race $i$ for some $2 \le i \le 5,$ there is a switch into the loss and a switch back to winning, plus three repeats: $\left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^2 = \frac{8}{243}$ for each of the $4$ positions, contributing $\frac{32}{243}.$

The total is $$\frac{16}{243} + \frac{32}{243} = \frac{48}{243} = \frac{16}{81},$$ so $m + n = 16 + 81 = 97.$

Answer: 109

Solution:

Place $B = (0, 0),$ $C = (11, 0),$ $A = (0, 3),$ $D = (11, 3).$ Solving $AF = 7$ and $CF = 9$ (consistent since $AC^2 = 11^2 + 3^2 = 130 = 7^2 + 9^2$) gives $F = \left(\frac{28}{5}, \frac{36}{5}\right),$ and since the diagonals of rectangle $AECF$ bisect each other, $E = A + C - F = \left(\frac{27}{5}, -\frac{21}{5}\right).$

Sides $AE$ and $FC$ have direction $(3, -4),$ lying on the lines $4x + 3y = 9$ and $4x + 3y = 44;$ sides $AF$ and $EC$ lie on $3x - 4y = -12$ and $3x - 4y = 33.$ Every point of $ABCD$ satisfies $-12 \le 3x - 4y \le 33,$ so the common region is just the part of the strip $0 \le y \le 3$ between the lines $4x + 3y = 9$ and $4x + 3y = 44:$ a parallelogram with vertices $A = (0, 3),$ $\left(\frac{9}{4}, 0\right),$ $C = (11, 0),$ and $\left(\frac{35}{4}, 3\right).$

Its horizontal sides have length $11 - \frac{9}{4} = \frac{35}{4}$ and the height between them is $3,$ so the area is $\frac{35}{4} \cdot 3 = \frac{105}{4},$ and $m + n = 105 + 4 = 109.$

Answer: 50

Solution:

A difference of powers of $2$ is $2^a - 2^b = 2^b(2^c - 1)$ where $c = a - b \ge 1.$ Since $2^c - 1$ is odd, the odd part of the number determines $c$ and the power of $2$ determines $b,$ so distinct pairs $(b, c)$ yield distinct integers. It suffices to count pairs with $2^b(2^c - 1) \lt 1000.$

For $c = 1, 2, \ldots, 9$ the factor $2^c - 1$ is $1, 3, 7, 15, 31, 63, 127, 255, 511,$ and the number of allowed values of $b$ is $10, 9, 8, 7, 6, 4, 3, 2, 1$ respectively (the count for $63$ drops to $4$ because $63 \cdot 16 = 1008 \gt 1000,$ while $31 \cdot 32 = 992$ still fits).

The total is $10 + 9 + 8 + 7 + 6 + 4 + 3 + 2 + 1 = 50.$

Answer: 331

Solution:

The ordered triples $(a, b, c)$ of positive integers with $a + b + c = 66$ number $\binom{65}{2} = 2080.$ Exactly one of them has all three values equal, namely $(22, 22, 22).$ Triples with exactly two values equal come from $2a + c = 66$ with $c \ne a:$ here $a$ can be $1$ through $32$ except $22,$ giving $31$ multisets, each arrangeable in $3$ ways, so $93$ ordered triples.

Hence $2080 - 1 - 93 = 1986$ ordered triples have three distinct values, and each unordered choice $a \lt b \lt c$ is counted $6$ times. The number of valid separations is $\frac{1986}{6} = 331.$

Answer: 31

Solution:

Write the terms as $a - d,$ $a,$ $a + d$ with integer $d \ge 1.$ The condition is $$(a-d)^2 + a^2 + (a+d)^2 = ad^2 \quad\Longleftrightarrow\quad 3a^2 + 2d^2 = ad^2,$$ so $d^2(a - 2) = 3a^2$ and $d^2 = \frac{3a^2}{a - 2}.$ For $d^2$ to be positive we need $a \gt 2$ (if $a = 0$ then $d = 0,$ not strictly increasing, and $0 \lt a \lt 2$ or $a \lt 0$ makes the right side negative or non-integral in the checkable cases).

Substituting $t = a - 2 \ge 1$ gives $$d^2 = \frac{3(t+2)^2}{t} = 3t + 12 + \frac{12}{t},$$ so $t \mid 12.$ Testing $t = 1, 2, 3, 4, 6, 12$ gives $d^2 = 27, 24, 25, 27, 32, 49:$ only $t = 3$ and $t = 12$ yield perfect squares.

These give $(a, d) = (5, 5)$ with sequence $0, 5, 10,$ and $(a, d) = (14, 7)$ with sequence $7, 14, 21.$ The sum of the third terms is $10 + 21 = 31.$

Answer: 192

Solution:

Let $A$ be the origin with $B = (s, 0, 0),$ $C = (0, s, 0),$ $D = (0, 0, s),$ $G = (s, s, s),$ and $P = (x, y, z).$ Expanding, $$BP^2 = AP^2 - 2sx + s^2, \quad CP^2 = AP^2 - 2sy + s^2, \quad DP^2 = AP^2 - 2sz + s^2,$$ while $GP^2 = AP^2 - 2s(x + y + z) + 3s^2.$ Therefore $$BP^2 + CP^2 + DP^2 - GP^2 = 2\,AP^2,$$ with every term involving $s$ or the coordinates of $P$ cancelling.

The given lengths yield $BP^2 = 36000,$ $CP^2 = 18000,$ $DP^2 = 28800,$ and $GP^2 = 9072,$ so $$2\,AP^2 = 36000 + 18000 + 28800 - 9072 = 73728,$$ giving $AP^2 = 36864$ and $AP = 192.$

Answer: 63

Solution:

Since each sine is at most $1,$ we need $\sin(mx) = \sin(nx) = 1,$ i.e. $mx = \frac{\pi}{2} + 2\pi a$ and $nx = \frac{\pi}{2} + 2\pi b$ for integers $a, b.$ Eliminating $x$ gives $n(4a + 1) = m(4b + 1),$ that is, $4(na - mb) = m - n.$ As $a$ and $b$ range over the integers, $na - mb$ takes exactly the multiples of $g = \gcd(m, n),$ so a solution exists if and only if $4g \mid n - m$ — equivalently, writing $m = gm'$ and $n = gn',$ if and only if $m' \equiv n' \pmod 4$ (which forces both $m'$ and $n'$ odd).

For each $g$ we count coprime pairs $m' \lt n' \le \lfloor 30/g \rfloor$ of odd numbers in the same class mod $4.$ For $g = 1:$ among $1, 3, \ldots, 29$ there are eight numbers $\equiv 1$ and seven $\equiv 3 \pmod 4,$ giving $\binom{8}{2} + \binom{7}{2} = 49$ pairs, of which the five pairs $\{3,15\},$ $\{3,27\},$ $\{9,21\},$ $\{15,27\},$ $\{5,25\}$ are not coprime, leaving $44.$ For $g = 2$ (odd numbers up to $15$): $\binom{4}{2} + \binom{4}{2} = 12$ minus the pair $\{3,15\}$ gives $11.$ For $g = 3$ (up to $10$): the pairs $\{1,5\}, \{1,9\}, \{5,9\}, \{3,7\}$ give $4.$ For $g = 4$ (up to $7$): $\{1,5\}$ and $\{3,7\}$ give $2.$ For $g = 5$ and $g = 6:$ only $\{1,5\},$ giving $1$ each. For $g \ge 7$ we would need two distinct odd numbers up to $\lfloor 30/g \rfloor \le 4$ in the same class mod $4,$ which is impossible.

The total is $44 + 11 + 4 + 2 + 1 + 1 = 63.$

Answer: 57

Solution:

Set $f(x) = \left|20|x| - x^2\right|,$ an even function; the equation says $f(x) = c + 21$ or $f(x) = c - 21.$ For $x \ge 0,$ the graph of $f$ rises from $0$ to $100$ on $[0, 10],$ falls back to $0$ at $x = 20,$ then increases without bound. So for $k$ with $0 \lt k \lt 100,$ the equation $f(x) = k$ has $3$ positive solutions, hence $6$ solutions in all; for $k = 100$ it has $4;$ for $k \gt 100$ it has $2;$ and for $k = 0$ it has $3$ (namely $0$ and $\pm 20$).

The two levels $c + 21$ and $c - 21$ are distinct, so the only way to reach $12$ solutions is $6 + 6:$ both $c - 21$ and $c + 21$ must lie strictly between $0$ and $100.$ This means $c \ge 22$ and $c \le 78,$ and every such integer works: there are $78 - 22 + 1 = 57$ values.

Answer: 567

Solution:

Since $AB \parallel CD,$ the distance $18$ from $A$ to $CD$ is the height. Put $D = (0, 0),$ $C = (c, 0),$ $A = (a, 18),$ $B = (c - a, 18),$ and let $u = AB = c - 2a \gt 0.$ The point-to-line distance formulas give $$\frac{18u}{\sqrt{324 + a^2}} = 15 \quad (\text{to } BC), \qquad \frac{18u}{\sqrt{324 + (u + a)^2}} = 10 \quad (\text{to } BD),$$ so $324 + a^2 = \left(\frac{6u}{5}\right)^2$ and $324 + (u + a)^2 = \left(\frac{9u}{5}\right)^2.$

Subtracting, $u(u + 2a) = \frac{81 - 36}{25}u^2 = \frac{9}{5}u^2,$ hence $u + 2a = \frac{9}{5}u$ and $a = \frac{2}{5}u.$ Substituting back, $324 = \frac{36u^2}{25} - \frac{4u^2}{25} = \frac{32u^2}{25},$ so $u^2 = \frac{2025}{8}$ and $u = \frac{45\sqrt{2}}{4}.$

Then $CD = c = u + 2a = \frac{9}{5}u = \frac{81\sqrt{2}}{4},$ and $$K = \frac{AB + CD}{2} \cdot 18 = 9\left(\frac{45\sqrt{2}}{4} + \frac{81\sqrt{2}}{4}\right) = \frac{567\sqrt{2}}{2},$$ so $\sqrt{2} \cdot K = 567.$

Answer: 59

Solution:

Write $a_k = \frac{m}{n}$ in lowest terms and let $d = n - m,$ so $a_j$ has the form $\frac{t}{t+1}$ exactly when $d = 1.$ One step sends $(m, n)$ to $(m + 18,\, n + 19)$ and then cancels $g = \gcd(m + 18, n + 19).$ Two facts control everything. First, $I = 19m - 18n$ satisfies $19(m + 18) - 18(n + 19) = I,$ so $I$ is unchanged by the shift and divided by $g$ upon cancellation. Second, since $I = 19(m + 18) - 18(n + 19) = (m + 18) - 18(d + 1),$ a number divides both $m + 18$ and $n + 19$ exactly when it divides both $d + 1$ and $I;$ hence $g = \gcd(d + 1,\, I),$ and after cancelling, the new difference is $\frac{d + 1}{g}.$

Initially $I = 19 \cdot 2020 - 18 \cdot 2021 = 2002 = 2 \cdot 7 \cdot 11 \cdot 13$ and $d = 1,$ so $j = 1$ works. The next step has $d + 1 = 2,$ $g = 2:$ $a_2 = \frac{2038}{2040} = \frac{1019}{1020},$ so $j = 2$ works and $I = 1001 = 7 \cdot 11 \cdot 13.$ From there $d$ climbs $1, 2, 3, \ldots$ until $d + 1$ shares a factor with $I:$ at $d + 1 = 7$ we get $a_8 = \frac{161}{162}$ (cancel $7,$ now $I = 143$); then at $d + 1 = 11,$ $a_{18} = \frac{31}{32}$ (cancel $11,$ now $I = 13$); then at $d + 1 = 13,$ $a_{30} = \frac{19}{20}$ (cancel $13,$ now $I = 1$). Each cancellation used $g = d + 1,$ so $d$ returned to $1$ at $j = 8,$ $18,$ and $30.$

Once $I = 1,$ no further cancellation is possible, so $d$ increases forever and never equals $1$ again. The valid indices are $j = 1, 2, 8, 18, 30,$ with sum $1 + 2 + 8 + 18 + 30 = 59.$

Answer: 301

Solution:

Let $P = AC \cap BD$ and let $\theta$ be the acute angle between the diagonals. Since $A$ lies on line $AC,$ its foot $A_1$ on $BD$ satisfies $PA_1 = PA\cos\theta,$ landing on the ray of $BD$ making the acute angle with ray $PA;$ the same holds for all four feet. So $A_1B_1C_1D_1$ is the image of $ABCD$ under the map that rotates each ray from $P$ onto the other diagonal (through angle $\theta$) and scales by $\cos\theta:$ corresponding triangles at $P$ are similar with ratio $\cos\theta,$ and every side of $A_1B_1C_1D_1$ is $\cos\theta$ times the corresponding side of $ABCD.$ Hence the perimeter is $(4 + 5 + 6 + 7)\cos\theta = 22\cos\theta.$

By Ptolemy, $AC \cdot BD = 4 \cdot 6 + 5 \cdot 7 = 59.$ By Brahmagupta with $s = 11,$ the area is $\sqrt{7 \cdot 6 \cdot 5 \cdot 4} = 2\sqrt{210}.$ Since the area also equals $\frac{1}{2} \, AC \cdot BD \sin\theta,$ we get $\sin\theta = \frac{4\sqrt{210}}{59},$ so $$\cos^2\theta = 1 - \frac{3360}{3481} = \frac{121}{3481}, \qquad \cos\theta = \frac{11}{59}.$$

The perimeter is $22 \cdot \frac{11}{59} = \frac{242}{59},$ which is in lowest terms, so $m + n = 242 + 59 = 301.$

Answer: 19

Solution:

Track the three gaps between consecutive frogs around the circle; they start at $(4, 4, 4)$ and always sum to $12.$ If the frogs jump by $X_1, X_2, X_3 \in \{\pm 1\},$ the gaps change by $X_2 - X_1,$ $X_3 - X_2,$ $X_1 - X_3,$ so each gap stays even and the process stops exactly when some gap becomes $0.$ Enumerating the $8$ equally likely sign choices: from $\{4,4,4\},$ the state stays with probability $\frac{2}{8}$ and moves to $\{2,4,6\}$ with probability $\frac{6}{8}.$ From $\{2,4,6\}:$ stay with probability $\frac{4}{8},$ move to $\{4,4,4\}$ or $\{2,2,8\}$ with probability $\frac{1}{8}$ each, and stop with probability $\frac{2}{8}.$ From $\{2,2,8\}:$ stay with probability $\frac{2}{8},$ move to $\{2,4,6\}$ with probability $\frac{2}{8},$ and stop with probability $\frac{4}{8}.$

Let $E_1, E_2, E_3$ be the expected remaining times from $\{4,4,4\},$ $\{2,4,6\},$ $\{2,2,8\}.$ Then $$E_1 = 1 + \tfrac{1}{4}E_1 + \tfrac{3}{4}E_2, \qquad E_2 = 1 + \tfrac{1}{2}E_2 + \tfrac{1}{8}E_1 + \tfrac{1}{8}E_3, \qquad E_3 = 1 + \tfrac{1}{4}E_3 + \tfrac{1}{4}E_2.$$ The third gives $E_3 = \frac{4}{3} + \frac{E_2}{3};$ substituting into the second yields $E_2 = 4,$ then $E_3 = \frac{8}{3}$ and $E_1 = \frac{4}{3} + E_2 = \frac{16}{3}.$

The expected number of minutes is $\frac{16}{3},$ so $m + n = 16 + 3 = 19.$

Answer: 672

Solution:

Let $O$ and $r$ be the center and radius of $\omega,$ and $O_1, O_2$ the other centers. External tangency gives $OO_1 = r + 961,$ so the power of $O$ with respect to $\omega_1$ is $OO_1^2 - 961^2 = r^2 + 2 \cdot 961r;$ similarly its power with respect to $\omega_2$ is $r^2 + 2 \cdot 625r.$ The difference is $2r(961 - 625) = 672r.$

For any point $X,$ the difference $\mathrm{pow}_{\omega_1}(X) - \mathrm{pow}_{\omega_2}(X) = (XO_1^2 - XO_2^2) - (961^2 - 625^2)$ is a linear function of $X$ that vanishes on the radical axis, which is line $AB;$ its rate of change perpendicular to $AB$ is $2 \cdot O_1O_2.$ So the difference equals $2 \cdot O_1O_2 \cdot \operatorname{dist}(O, AB).$ Meanwhile the chord $PQ$ of $\omega$ subtends a $120^\circ$ central angle, so $\operatorname{dist}(O, AB) = r\cos 60^\circ = \frac{r}{2}.$

Therefore $672r = 2 \cdot O_1O_2 \cdot \frac{r}{2} = O_1O_2 \cdot r,$ and the distance between the centers is $672.$

Answer: 125

Solution:

Note $2021 = 43 \cdot 47.$ If $a = \prod p_i^{e_i},$ then $\sigma(a^n) = \prod \sigma(p_i^{e_i n}),$ so it suffices (and is necessary, taking $a$ prime) that $\sigma(p^N) \equiv 1,$ i.e. $p + p^2 + \cdots + p^N \equiv 0 \pmod{43 \cdot 47},$ for every prime $p$ and every multiple $N$ of $n.$

Fix $q \in \{43, 47\}.$ If $q \mid p$ the sum is $0.$ If $p \equiv 1 \pmod q$ the sum is $\equiv N,$ so choosing such a prime (Dirichlet) forces $q \mid n.$ Otherwise the sum is $p \cdot \frac{p^N - 1}{p - 1}$ with $p - 1$ invertible, so we need $p^N \equiv 1 \pmod q;$ choosing $p$ to be a primitive root mod $q$ forces $q - 1 \mid n.$ Conversely, if $q(q-1) \mid n$ then for every multiple $N$ of $n$ and every prime $p,$ the sum vanishes mod $q$ in all three cases. Hence the least $n$ is $$n = \operatorname{lcm}(43 \cdot 42,\ 47 \cdot 46) = 2 \cdot 3 \cdot 7 \cdot 23 \cdot 43 \cdot 47.$$

The sum of the prime factors is $2 + 3 + 7 + 23 + 43 + 47 = 125.$

Answer: 285

Solution:

Adding the equation $x^2 - y - k = 0$ to $\frac{1}{2}$ times $2(y - 20)^2 - x - k = 0$ gives a conic through all intersection points with equal $x^2$ and $y^2$ coefficients: $$x^2 + y^2 - \tfrac{1}{2}x - 41y + 400 - \tfrac{3k}{2} = 0,$$ a circle centered at $\left(\frac{1}{4}, \frac{41}{2}\right)$ with squared radius $\frac{1}{16} + \frac{1681}{4} - 400 + \frac{3k}{2} = \frac{325}{16} + \frac{3k}{2}.$ So whenever four distinct intersection points exist, they are concyclic, and the radius is at most $21$ exactly when $\frac{325}{16} + \frac{3k}{2} \le 441,$ i.e. $k \le 280$ for integers.

Substituting $y = x^2 - k$ into the second parabola gives the quartic $f(x) = 2(x^2 - c)^2 - x - k = 0$ where $c = k + 20.$ For $1 \le k \le 4:$ if $x \le -k$ then $f(x) = 2(x^2 - c)^2 + (-x - k) \gt 0,$ and if $-k \lt x \le 0$ then $x^2 \lt 16$ while $c \ge 21,$ so $f(x) \gt 2 \cdot 25 - k \gt 0;$ thus there are no intersections with $x \le 0,$ and since $f'(x) = 8x^3 - 8cx - 1$ has exactly one positive root, $f$ has at most (and, by $f(0) \gt 0,$ $f(\sqrt{c}) \lt 0,$ exactly) two positive roots. So $k \le 4$ fails. For $k \ge 5,$ we have $k^2 \ge k + 20,$ so $f\left(-\sqrt{c}\right) = \sqrt{c} - k \le 0$ with $f$ strictly decreasing there, while $f(-\infty) = +\infty,$ $f(0) = 2c^2 - k \gt 0,$ $f\left(\sqrt{c}\right) = -\sqrt{c} - k \lt 0,$ and $f(+\infty) = +\infty:$ the sign changes produce four distinct real roots.

Hence $S = \{5, 6, \ldots, 280\},$ and the answer is $5 + 280 = 285.$