Answer: 336

Solution:

The geometric series sums to $S(r) = \frac{12}{1 - r}.$ Therefore $$S(a)S(-a) = \frac{12}{1 - a} \cdot \frac{12}{1 + a} = \frac{144}{1 - a^2} = 2016,$$ so $\frac{1}{1 - a^2} = 14.$

Adding the two sums over a common denominator, $$S(a) + S(-a) = \frac{12}{1 - a} + \frac{12}{1 + a} = \frac{24}{1 - a^2} = 24 \cdot 14 = 336.$$

Answer: 71

Solution:

Since $1 + 2 + \cdots + 6 = 21,$ each die rolls $k$ with probability $\frac{k}{21}.$ A total of $7$ arises from the pairs $(k, 7-k)$ for $k = 1, \ldots, 6,$ so its probability is $$\frac{1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1}{21^2} = \frac{56}{441} = \frac{8}{63}.$$

Thus $m + n = 8 + 63 = 71.$

Answer: 810

Solution:

Each vertex of the upper pentagon is adjacent to the top vertex, two upper-pentagon neighbors, and two vertices of the lower pentagon; each vertex of the lower pentagon is adjacent to two upper vertices, two lower-pentagon neighbors, and the bottom vertex. So a downward-or-horizontal path with no repeated vertex must descend to the upper pentagon, circle part of it in one direction, drop to the lower pentagon, circle part of it in one direction, and end at the bottom.

There are $5$ choices for the first step down. On the upper pentagon the path can take $0, 1, 2, 3,$ or $4$ horizontal steps, in either of two directions (a reversal would repeat a vertex), for $1 + 2 \cdot 4 = 9$ options. Then there are $2$ edges down to the lower pentagon, again $9$ horizontal options there, and $1$ final step down.

The total is $5 \cdot 9 \cdot 2 \cdot 9 = 810.$

Answer: 108

Solution:

The three vertices adjacent to $A$ are its two neighbors $B$ and $C$ in the same hexagonal base and the vertex $D$ directly above $A,$ with $DA = h$ perpendicular to the base. The face in the base is $ABC,$ and the face avoiding $A$ is $BCD;$ they meet along $\overline{BC}.$

Let $E$ be the midpoint of $\overline{BC}.$ Since $AB = AC = 12$ and $\angle BAC = 120^\circ$ (the interior angle of a regular hexagon), $\overline{AE} \perp \overline{BC}$ and $AE = 12\cos 60^\circ = 6.$ Because $\overline{DA}$ is perpendicular to the base, $\overline{DE} \perp \overline{BC}$ as well, so the dihedral angle is $\angle DEA = 60^\circ.$

In right triangle $DAE,$ $h = AE\tan 60^\circ = 6\sqrt{3},$ so $h^2 = 108.$

Answer: 53

Solution:

Say Anh finished on day $k.$ Summing the arithmetic progressions of pages and of minutes, $$\frac{k(2n + k - 1)}{2} = 374 \qquad \text{and} \qquad \frac{k(2t + k - 1)}{2} = 319,$$ so $k(2n + k - 1) = 748$ and $k(2t + k - 1) = 638.$

Subtracting, $2k(n - t) = 110,$ so $k(n - t) = 55.$ Thus $k$ divides both $55$ and $\gcd(748, 638) = 22,$ so $k \mid 11.$ Since the story spans more than one day, $k = 11.$

Then $2n + 10 = \frac{748}{11} = 68$ gives $n = 29,$ and $2t + 10 = \frac{638}{11} = 58$ gives $t = 24.$ Hence $n + t = 29 + 24 = 53.$

Answer: 13

Solution:

The incenter $I$ lies on the bisector $\overline{CL},$ between $C$ and $L.$ In triangle $ACI,$ the exterior angle at $I$ gives $\angle DIA = \angle IAC + \angle ICA.$ On the other hand, $\angle DAB = \angle DCB$ (both subtend arc $DB$) and $\angle DCB = \angle ICA$ (the bisector), so $$\angle DAI = \angle DAB + \angle BAI = \angle ICA + \angle IAC = \angle DIA.$$ Hence triangle $DAI$ is isosceles with $DA = DI = DL + LI = 5.$

Triangles $DAL$ and $DCA$ have a common angle at $D,$ and $\angle DAL = \angle DAB = \angle DCB = \angle DCA,$ so they are similar. Therefore $\frac{DA}{DC} = \frac{DL}{DA},$ giving $DC = \frac{DA^2}{DL} = \frac{25}{3}.$

Finally $IC = DC - DI = \frac{25}{3} - 5 = \frac{10}{3},$ so $p + q = 10 + 3 = 13.$

Answer: 103

Solution:

If $ab + 2016 \ge 0,$ the first term is real, so the number is real exactly when $\sqrt{|a + b|} = 0,$ that is $b = -a.$ Then $ab + 2016 = 2016 - a^2 \ge 0$ forces $|a| \le 44,$ and the denominator $ab + 100 = 100 - a^2$ rules out $a = \pm 10.$ That gives $89 - 2 = 87$ pairs.

If $ab + 2016 \lt 0,$ then $\sqrt{ab + 2016} = i\sqrt{-ab - 2016},$ so the whole number is $\frac{\sqrt{-ab - 2016} - \sqrt{|a + b|}}{ab + 100}\,i,$ which is real exactly when $-ab - 2016 = |a + b|.$ Note $a + b = 0$ is impossible here since $a^2 = 2016$ has no integer solution. For $a + b \gt 0$ the equation becomes $ab + a + b + 2016 = 0,$ that is $(a + 1)(b + 1) = -2015,$ and for $a + b \lt 0$ it becomes $(a - 1)(b - 1) = -2015.$

Since $2015 = 5 \cdot 13 \cdot 31$ has $8$ positive divisors, $(a+1)(b+1) = -2015$ has $16$ ordered integer solutions, and $a + b \gt 0$ holds exactly when the positive factor is the larger in absolute value: $8$ solutions. Symmetrically the other case gives $8$ more. In all of these $ab + 100 = -1916 - |a + b| \ne 0.$ The total is $87 + 8 + 8 = 103.$

Answer: 162

Solution:

By place value, $s(p) = 100(a_1 + a_4 + a_7) + 10(a_2 + a_5 + a_8) + (a_3 + a_6 + a_9),$ and all nine digits sum to $45.$ The units digit of $s(p)$ is $0$ exactly when $a_3 + a_6 + a_9 = 10$ or $20.$ Writing $X = a_1 + a_4 + a_7,$ if the units column sums to $10$ then $s(p) = 100X + 10(35 - X) + 10 = 90X + 360 \ge 900,$ while if it sums to $20$ then $s(p) = 90X + 270 \ge 90 \cdot 6 + 270 = 810.$ So $m = 810,$ achieved exactly when $\{a_1, a_4, a_7\} = \{1, 2, 3\}$ and the units digits sum to $20.$

The remaining digits $\{4, 5, 6, 7, 8, 9\}$ must split so the units triple sums to $20:$ the possibilities are $\{4, 7, 9\},$ $\{5, 6, 9\},$ and $\{5, 7, 8\}.$ Each of the $3$ splits allows $3! \cdot 3! \cdot 3! = 216$ arrangements of the three columns, so $n = 3 \cdot 216 = 648.$

Therefore $|m - n| = |810 - 648| = 162.$

Answer: 744

Solution:

Let $\beta = \angle BAQ$ and $\gamma = \angle CAS,$ so $\beta + \gamma = 90^\circ - A.$ From the right triangles $AQB$ and $ASC,$ the sides of the rectangle are $AQ = 40\cos\beta$ and $AS = 31\cos\gamma,$ so its area is $$40 \cdot 31 \cos\beta\cos\gamma = 620\bigl(\cos(\beta - \gamma) + \cos(\beta + \gamma)\bigr) = 620\bigl(\cos(\beta - \gamma) + \sin A\bigr),$$ using the product-to-sum identity and $\cos(90^\circ - A) = \sin A.$

This is maximized when $\beta = \gamma,$ which the constraint allows, giving area $620\left(1 + \frac{1}{5}\right) = 744.$

Answer: 504

Solution:

Write the common ratio of $a_1, a_2, a_3$ as $\frac{b}{a}$ in lowest terms, with $b \gt a \ge 1$ since the sequence increases. Because $a_3 = a_1 \left(\frac{b}{a}\right)^2$ is an integer and $\gcd(a, b) = 1,$ we get $a^2 \mid a_1;$ set $c = \frac{a_1}{a^2}.$ Then $a_1 = ca^2,$ $a_2 = cab,$ $a_3 = cb^2,$ and the arithmetic condition gives $a_4 = 2cb^2 - cab = cb(2b - a).$ Continuing, induction shows for every $k$ that $$a_{2k+1} = c\,\bigl(kb - (k-1)a\bigr)^2, \qquad a_{2k+2} = c\,\bigl(kb - (k-1)a\bigr) \bigl((k+1)b - ka\bigr).$$

In particular $a_{13} = c\,(6b - 5a)^2 = 2016 = 2^5 \cdot 3^2 \cdot 7.$ Let $N = 6b - 5a;$ then $N^2 \mid 2016,$ so $N \le 12.$ But $N = a + 6(b - a) \ge a + 6 \ge 7,$ and the only value in range with $N^2 \mid 2016$ is $N = 12,$ giving $c = \frac{2016}{144} = 14.$ From $6(b - a) = 12 - a$ we need $6 \mid a$ with $a \le 6,$ so $a = 6$ and $b = 7,$ which are coprime.

Therefore $a_1 = ca^2 = 14 \cdot 36 = 504.$ (Indeed the sequence begins $504, 588, 686, 784, 896, \ldots$ and reaches $a_{13} = 14 \cdot 12^2 = 2016.$)

Answer: 109

Solution:

Setting $x = 1$ in the identity gives $0 = 3P(1),$ so $P(1) = 0.$ Setting $x = 0$ gives $-P(1) = 2P(0),$ so $P(0) = 0,$ and setting $x = -2$ gives $-3P(-1) = 0,$ so $P(-1) = 0.$ Hence $P(x) = x(x - 1)(x + 1)L(x)$ for some polynomial $L.$

Substituting back, $(x - 1)\,(x + 1)x(x + 2)L(x + 1) = (x + 2)\,x(x - 1)(x + 1)L(x),$ so $L(x + 1) = L(x)$ for all real $x,$ which forces $L$ to be a constant $c.$ The normalization $\left(P(2)\right)^2 = P(3)$ reads $(6c)^2 = 24c,$ so $c = \frac{2}{3}.$

Then $$P\left(\tfrac{7}{2}\right) = \frac{2}{3} \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{9}{2} = \frac{105}{4},$$ and $m + n = 105 + 4 = 109.$

Answer: 132

Solution:

Let $e(m) = m^2 - m + 11.$ Since $m^2 - m$ is always even, $e(m)$ is odd. Checking all residues shows $m^2 - m + 11$ is never $0$ modulo $3,$ $5,$ or $7$ either, so every prime factor of $e(m)$ is at least $11.$ A product of four such primes is at least $11^4 = 14641,$ and the two smallest candidates are $11^4$ and $11^3 \cdot 13 = 17303.$

For $e(m) = 14641,$ the discriminant of $m^2 - m - 14630 = 0$ is $58521,$ which lies strictly between $241^2 = 58081$ and $242^2 = 58564,$ so there is no integer solution. For $e(m) = 17303:$ since $e(m) = m(m - 1) + 11$ must be divisible by $11,$ either $m = 11k$ or $m = 11k + 1.$ Trying $m = 11k$ gives $11k^2 - k + 1 = 1573,$ that is $k(11k - 1) = 1572,$ which $k = 12$ satisfies: $12 \cdot 131 = 1572.$

Since $e$ is increasing for $m \ge 1,$ every smaller $m$ has $e(m) \lt 17303,$ and the only four-prime value below that, $11^4,$ is unattainable. Hence the least $m$ is $11 \cdot 12 = 132,$ where $e(132) = 17303 = 11^3 \cdot 13.$

Answer: 273

Solution:

Horizontal jumps change nothing that matters, so let $T(y)$ be the expected number of jumps to reach the river from height $y.$ Then $T(24) = 0;$ for $1 \le y \le 23$ each jump goes up, down, or sideways with probabilities $\frac{1}{4}, \frac{1}{4}, \frac{1}{2},$ so $$T(y) = 1 + \tfrac{1}{4}T(y + 1) + \tfrac{1}{4}T(y - 1) + \tfrac{1}{2}T(y),$$ which simplifies to $2T(y) = 4 + T(y - 1) + T(y + 1).$ At the fence the three equally likely moves give $T(0) = 1 + \frac{2}{3}T(0) + \frac{1}{3}T(1),$ that is $T(0) = 3 + T(1).$

Summing $2T(y) = 4 + T(y - 1) + T(y + 1)$ over $y = 1, \ldots, 23$ telescopes to $T(1) + T(23) = 92 + T(0) + T(24).$ Substituting $T(0) = 3 + T(1)$ and $T(24) = 0$ yields $T(23) = 95.$

Now run the recurrence downward as $T(y - 1) = 2T(y) - T(y + 1) - 4:$ from $T(24) = 0$ and $T(23) = 95,$ we get $T(22) = 2 \cdot 95 - 0 - 4 = 186$ and $T(21) = 2 \cdot 186 - 95 - 4 = 273.$ Freddy starts at height $21,$ so the answer is $273.$

Answer: 574

Solution:

Since $\gcd(1001, 429) = 143,$ the segment passes through the lattice points $(7k, 3k)$ for $k = 0, \ldots, 143$ and consists of $143$ translated copies of the segment from $(0, 0)$ to $(7, 3).$ The line is $y = \frac{3}{7}x.$ It meets the square centered at $(m, n)$ exactly when its height passes within $\frac{1}{10}$ of $n$ for some $x$ within $\frac{1}{10}$ of $m,$ that is when $\left|\frac{3m}{7} - n\right| \le \frac{1}{10} + \frac{3}{7} \cdot \frac{1}{10} = \frac{1}{7},$ or equivalently $|3m - 7n| \le 1.$

For $0 \le m \le 7$ the solutions are $(0, 0)$ and $(7, 3)$ with $3m - 7n = 0,$ and $(2, 1)$ and $(5, 2)$ with $3m - 7n = \mp 1.$ In the first two the line passes through the center, so it meets the circle as well. In the other two, equality means the line passes exactly through a corner of the square (for $(2, 1),$ the corner $(2.1, 0.9)$), while its distance to the center is $\frac{1}{\sqrt{3^2 + 7^2}} = \frac{1}{\sqrt{58}} \gt \frac{1}{10},$ so it misses the circle. Thus each copy of the segment meets $4$ squares and $2$ circles.

The $142$ interior lattice points $(7k, 3k)$ are each shared by two consecutive copies, so $m = 4 \cdot 143 - 142 = 430$ and $n = 2 \cdot 143 - 142 = 144,$ giving $m + n = 574.$

Answer: 270

Solution:

Line $AD$ is the radical axis of $\omega$ and $\omega_1,$ line $BC$ that of $\omega$ and $\omega_2,$ and line $XY$ that of $\omega_1$ and $\omega_2,$ so the three lines meet at the radical center $Z.$ (They cannot be parallel: that would force a symmetric configuration with $XC = XD.$) Let $M = XY \cap AB.$ The power of $M$ with respect to each circle gives $MA^2 = MX \cdot MY = MB^2,$ so $M$ is the midpoint of $\overline{AB},$ with $X$ between $M$ and $Y.$

Since $ADYX$ is cyclic, $\angle XAZ = \angle XYD,$ and since $BCYX$ is cyclic, $\angle XBZ = \angle XYC;$ as $C,$ $Y,$ $D$ are collinear these add to $180^\circ,$ so $ZAXB$ is cyclic. The tangent-chord angle at $B$ gives $\angle XYB = \angle ABX = \angle AZX,$ so $BY \parallel ZA,$ and symmetrically $AY \parallel ZB.$ Hence $AYBZ$ is a parallelogram, and since $M$ is the midpoint of diagonal $\overline{AB},$ it is also the midpoint of $\overline{ZY}:$ therefore $XZ = XM + MZ = MX + MY.$ Moreover $\angle XCZ = \angle XYB = \angle XZD$ and (by the tangent-chord angle at $A$) $\angle XZC = \angle XAB = \angle XYA = \angle XDZ,$ so triangles $XZC$ and $XDZ$ are similar, giving $XZ^2 = XC \cdot XD.$

Putting it together, $$AB^2 = 4MA^2 = 4\,MX \cdot MY = (MX + MY)^2 - (MY - MX)^2 = XZ^2 - XY^2 = XC \cdot XD - XY^2,$$ which equals $67 \cdot 37 - 47^2 = 2479 - 2209 = 270.$