Solution:

There are $2^8 = 256$ subsets in all. The ones to exclude are those contained in $\{1,2,3,4,5\}$ (there are $2^5 = 32$) or contained in $\{4,5,6,7,8\}$ (another $32$). A subset of both is exactly a subset of the intersection $\{4, 5\},$ and there are $2^2 = 4$ of those.

By inclusion-exclusion, $32 + 32 - 4 = 60$ subsets fail, so $256 - 60 = 196$ subsets have the required property.

Solution:

To be champion, $T_4$ must first beat $T_1,$ which happens with probability $\frac{4}{4+1} = \frac{4}{5}.$ The other semifinal sends $T_2$ to the final with probability $\frac{2}{2+3} = \frac{2}{5}$ and $T_3$ with probability $\frac{3}{5};$ in the final, $T_4$ beats $T_2$ with probability $\frac{4}{4+2} = \frac{2}{3}$ and beats $T_3$ with probability $\frac{4}{4+3} = \frac{4}{7}.$

The probability that $T_4$ is champion is therefore $$\frac{4}{5}\left(\frac{2}{5} \cdot \frac{2}{3} + \frac{3}{5} \cdot \frac{4}{7}\right) = \frac{4}{5} \cdot \frac{64}{105} = \frac{256}{525}.$$ Since $256 = 2^8$ and $525 = 3 \cdot 5^2 \cdot 7,$ this is in lowest terms, and $p + q = 256 + 525 = 781.$

Solution:

The points closer to $B$ than to $A$ lie to the right of the perpendicular bisector of $\overline{AB},$ the line $x = 6.$ The points closer to $B$ than to $C$ lie below the perpendicular bisector of $\overline{BC},$ which passes through the midpoint $(10, 5)$ with slope $\frac{2}{5}$ (the negative reciprocal of the slope $-\frac{5}{2}$ of $BC$): the line $y = \frac{2}{5}x + 1.$

Inside the triangle, the favorable region is the quadrilateral with vertices $(6, 0),$ $B(12, 0),$ the midpoint $(10, 5)$ of $\overline{BC},$ and $\left(6, \frac{17}{5}\right),$ where the two bisectors meet. Splitting it along the segment from $(6, 0)$ to $(10, 5),$ its area is $$\frac{1}{2} \cdot \frac{17}{5} \cdot 4 + \frac{1}{2} \cdot 6 \cdot 5 = \frac{34}{5} + 15 = \frac{109}{5}.$$

The triangle has area $\frac{1}{2} \cdot 12 \cdot 10 = 60,$ so the probability is $\frac{109/5}{60} = \frac{109}{300},$ and $p + q = 109 + 300 = 409.$

Solution:

A positive integer has no $0$ in base three exactly when every digit is $1$ or $2.$ For $k = 1, 2, \ldots, 6$ there are $2^k$ such $k$-digit numbers, and all of them are at most $222222_3 = 728 \lt 2017.$

Since $2017 = 2202201_3,$ a seven-digit string of $1$s and $2$s is at most $2017$ exactly when it begins with $11,$ $12,$ or $21:$ any string beginning $22$ already beats $2202201_3$ at the third digit, since its digits are nonzero. That gives $3 \cdot 2^5 = 96$ seven-digit numbers.

The total is $2 + 4 + 8 + 16 + 32 + 64 + 96 = 222.$

Solution:

Let the set be $\{a, b, c, d\}$ with total $s = a + b + c + d.$ The six pairwise sums come in three complementary pairs: $$(a+b) + (c+d) = (a+c) + (b+d) = (a+d) + (b+c) = s.$$ No two of the pairings of $189,$ $320,$ $287,$ $234$ into two pairs give equal totals ($509 \ne 521,$ $476 \ne 554,$ $423 \ne 607$), so $x$ and $y$ are not paired with each other; each is paired with a given sum, and the remaining two given sums are paired together. Adding all six values, $x + y = 3s - (189 + 320 + 287 + 234) = 3s - 1030,$ where $s$ is the sum of two of the given numbers.

The largest choice is $s = 320 + 287 = 607,$ giving $x + y = 3 \cdot 607 - 1030 = 791.$ This is attained by the set $\{51.5,\ 137.5,\ 182.5,\ 235.5\},$ whose pairwise sums are $189,$ $234,$ $287,$ $320,$ $373,$ and $418,$ with $373 + 418 = 791.$

Solution:

Suppose $n^2 + 85n + 2017 = m^2$ for a positive integer $m.$ Multiplying by $4$ and completing the square gives $(2n + 85)^2 + 843 = 4m^2,$ so $$(2m - 2n - 85)(2m + 2n + 85) = 843 = 3 \cdot 281,$$ where $281$ is prime. Both factors are positive with the second one larger, so either $2m - 2n - 85 = 1$ and $2m + 2n + 85 = 843,$ or $2m - 2n - 85 = 3$ and $2m + 2n + 85 = 281.$

The first system gives $m = 211$ and $n = 168,$ and indeed $168^2 + 85 \cdot 168 + 2017 = 44521 = 211^2.$ The second gives $m = 71$ and $n = 27,$ with $27^2 + 85 \cdot 27 + 2017 = 5041 = 71^2.$

The requested sum is $168 + 27 = 195.$

Solution:

The equation requires $x + 2 \gt 0$ and $kx \gt 0,$ and under those restrictions it is equivalent to $kx = (x + 2)^2,$ that is, $x^2 + (4 - k)x + 4 = 0.$

For $k \lt 0$ the restrictions force $-2 \lt x \lt 0.$ On this interval $kx$ decreases from $-2k \gt 0$ to $0$ while $(x + 2)^2$ increases from $0$ to $4,$ so the graphs cross exactly once. Hence every one of the $500$ negative values of $k$ works, while $k = 0$ makes $\log(kx)$ undefined.

For $k \gt 0$ the restrictions force $x \gt 0.$ The quadratic has root product $4,$ so any real roots have the same sign, and the discriminant $(4 - k)^2 - 16 = k(k - 8)$ is negative for $0 \lt k \lt 8.$ When $k \gt 8$ there are two distinct positive roots (root sum $k - 4 \gt 0$), giving two solutions; only $k = 8$ gives exactly one solution, the double root $x = 2.$ In total $500 + 1 = 501$ values of $k$ work.

Solution:

Multiplying by $6! = 720,$ the sum is an integer exactly when $$720 \mid n^6 + 6n^5 + 30n^4 + 120n^3 + 360n^2.$$ If $n$ were odd, every term except $n^6$ would be even, making the total odd. If $3 \nmid n,$ then modulo $3$ every term except $n^6$ vanishes while $n^6 \equiv 1 \pmod 3.$ So $n$ must be a multiple of $6.$

Write $n = 6k.$ Then $30n^4 = 720 \cdot 54k^4,$ $120n^3 = 720 \cdot 36k^3,$ and $360n^2 = 720 \cdot 18k^2$ are all divisible by $720,$ while $n^6 + 6n^5 = 6^6 k^5(k + 1).$ Since $6^6 = 2^6 3^6$ supplies the factors $2^4$ and $3^2$ of $720 = 2^4 \cdot 3^2 \cdot 5,$ the condition reduces to $5 \mid k(k + 1),$ that is, $k \equiv 0$ or $4 \pmod 5.$

For $n = 6k \lt 2017$ we need $1 \le k \le 336.$ That range contains $67$ multiples of $5$ and $67$ values with $k \equiv 4 \pmod 5,$ so there are $67 + 67 = 134$ such $n.$

Solution:

Since the eight cards cover all seven numbers and all seven colors, exactly one number and exactly one color appear twice. Sharon can discard a card exactly when a single card carries both the repeated number and the repeated color: that card is then the unique discardable one, while if no card carries both, removing any card loses a number or a color.

Hands of the first type consist of a rainbow set of seven cards — one of each number and each color, which is one of $7!$ permutation patterns — plus any of the remaining $42$ cards, and every such hand arises exactly once this way: $7! \cdot 42 = 211680$ hands. For the second type, choose the repeated number ($7$ ways) and the two colors of its cards ($\binom{7}{2} = 21$ ways); the repeated color must be one of the other $5$ colors, and the numbers of its two cards come from the remaining $6$ numbers ($\binom{6}{2} = 15$ ways); finally match the last four numbers to the last four colors ($4! = 24$ ways). That is $7 \cdot 21 \cdot 5 \cdot 15 \cdot 24 = 264600$ hands.

The probability is $\frac{211680}{211680 + 264600} = \frac{4}{9},$ so $p + q = 4 + 9 = 13.$

Solution:

Place $A = (0, 0),$ $B = (84, 0),$ $C = (84, 42),$ $D = (0, 42),$ so $M = (0, 21)$ and $N = (28, 0).$ Line $CM$ is $y = \frac{x}{4} + 21$ and line $DN$ is $y = 42 - \frac{3x}{2},$ which meet at $O = (12, 24).$

By the Shoelace Formula, quadrilateral $BCON$ has area $2184,$ so each half must have area $1092.$ Triangle $BCO$ alone has area $\frac{1}{2} \cdot 42 \cdot 72 = 1512 \gt 1092$ (base $\overline{BC},$ and $O$ is at horizontal distance $72$ from it), so the bisecting segment ends at a point $P$ on $\overline{CO}.$ For $[BPC] = 1092,$ the distance $d$ from $P$ to line $BC$ must satisfy $\frac{1}{2} \cdot 42 \cdot d = 1092,$ so $d = 52,$ giving $P$ the $x$-coordinate $84 - 52 = 32.$ Since $P$ lies on line $CO,$ which is $y = \frac{x}{4} + 21,$ we get $P = (32, 29).$

Triangle $CDP$ has base $CD = 84$ on the line $y = 42$ and height $42 - 29 = 13,$ so its area is $\frac{1}{2} \cdot 84 \cdot 13 = 546.$

Solution:

The assignment works if and only if no town has all four roads inbound or all four outbound. One direction is clear: an all-inbound town cannot be left, and an all-outbound town cannot be reached. Conversely, suppose every town has an inbound and an outbound road, yet town $B$ cannot be reached from town $A.$ Let $S$ be the set of towns reachable from $A$ (including $A$) and $T$ the set of towns from which $B$ is reachable (including $B$). These sets are disjoint, every outbound road of a town in $S$ stays inside $S,$ and every inbound road of a town in $T$ comes from inside $T.$ Since $A$ has an outbound road, $|S| \ge 2,$ and similarly $|T| \ge 2;$ as $|S| + |T| \le 5,$ one of the two sets has exactly two towns. If $S = \{A, X\},$ the outbound roads of $A$ and of $X$ must both stay inside $S,$ forcing the single road between them to point both ways — a contradiction (and $|T| = 2$ is symmetric).

Now count assignments with a bad town among the $2^{10} = 1024$ total. Choosing a town to be all-outbound ($5$ ways) and orienting the remaining $\binom{4}{2} = 6$ roads freely gives $5 \cdot 2^6 = 320$ assignments, and there can be at most one all-outbound town. Similarly $320$ assignments have an all-inbound town. Assignments with both are counted twice: choose the all-outbound town ($5$), the all-inbound town ($4$), and the other $3$ roads freely, $5 \cdot 4 \cdot 2^3 = 160.$ So $320 + 320 - 160 = 480$ assignments fail.

The number that work is $1024 - 480 = 544.$

Solution:

Work in the complex plane with $C_0$ centered at $O_0 = 0$ and $A_0 = 1,$ and let $O_n$ be the center of $C_n.$ Inductively, $A_n = O_n + r^n i^n:$ this holds for $n = 0,$ and since $C_{n+1}$ is internally tangent to $C_n$ at $A_n,$ its center is $O_{n+1} = A_n - r^{n+1} i^n;$ then $A_n$ sits in direction $i^n$ from $O_{n+1},$ so rotating $90^\circ$ counterclockwise gives $A_{n+1} = O_{n+1} + r^{n+1} i^{n+1}.$

Therefore $O_{n+1} - O_n = (r^n - r^{n+1})\, i^n = (1 - r)(ir)^n.$ The circles are nested, and their radii shrink to $0,$ so the common point $B$ is the limit of the centers: $$B = (1 - r)\sum_{n=0}^{\infty} (ir)^n = \frac{1 - r}{1 - ir},$$ at distance $\frac{1 - r}{|1 - ir|} = \frac{1 - r}{\sqrt{1 + r^2}}$ from the origin.

For $r = \frac{11}{60}$ this equals $\frac{49/60}{\sqrt{3721}/60} = \frac{49}{61},$ so $m + n = 49 + 61 = 110.$

Solution:

Count isosceles triangles by apex. For a vertex $P$ of a regular $n$-gon, the isosceles triangles whose two equal sides meet at $P$ have their other two vertices symmetric about the diameter through $P,$ giving $\lfloor (n-1)/2 \rfloor$ such pairs. Summing over all $n$ vertices counts each non-equilateral isosceles triangle once (it has one apex) and each equilateral triangle three times; equilateral triangles exist exactly when $3 \mid n,$ and then there are $n/3$ of them. Hence $f(n) = n \lfloor (n-1)/2 \rfloor,$ minus $2n/3$ when $3 \mid n.$

Writing $n = 6k + j$ and computing $f(n+1) - f(n)$ in each residue class gives $13k$ for $j = 0,$ $3k$ for $j = 1,$ $5k + 1$ for $j = 2,$ $7k + 3$ for $j = 3,$ $9k + 6$ for $j = 4,$ and $-(k + 2)$ for $j = 5.$ Setting each equal to $78:$ $13k = 78$ gives $k = 6,$ $n = 36;$ $3k = 78$ gives $k = 26,$ $n = 157;$ $9k + 6 = 78$ gives $k = 8,$ $n = 52;$ and the other three cases have no positive integer solutions.

The sum of all such $n$ is $36 + 157 + 52 = 245.$

Solution:

Take a primitive direction vector $(a, b, c)$ for the line. Any nonzero component of absolute value $2$ or more limits the line to at most $5$ grid points, so every component is $0$ or $\pm 1.$ Lines parallel to a coordinate axis contain $10$ points, never $8.$ If exactly one component is $0,$ the line lies in one of the $30$ planes parallel to a face of the cube ($3$ orientations, $10$ positions), and within that $10 \times 10$ grid it is a diagonal of slope $\pm 1$ shifted off center; the shift by $2$ in either direction from each of the two main diagonals gives exactly $8$ points. That is $4$ lines per plane, and each lies in only one of the $30$ planes: $4 \cdot 30 = 120$ lines.

Otherwise the direction is one of the four space-diagonal directions $(1, \pm 1, \pm 1)$ up to sign; by symmetry, count lines parallel to $(1, 1, 1)$ and multiply by $4.$ Such a line $(d + t,\ e + t,\ f + t)$ meets the grid in $10 - (\max(d,e,f) - \min(d,e,f))$ points, so exactly $8$ points means $\max - \min = 2.$ Normalizing the base point so that $\min(d, e, f) = 1,$ we need $(d, e, f)$ with entries in $\{1, 2, 3\}$ using both $1$ and $3:$ there are $27 - 8 - 8 + 1 = 12$ of them, hence $12$ lines per direction and $4 \cdot 12 = 48$ in all.

The total is $120 + 48 = 168.$

Solution:

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{CD}.$ The medians from $C$ and from $D$ to $\overline{AB}$ are equal, since triangles $ABC$ and $BAD$ are congruent by $SSS;$ by the median length formula, $4MD^2 = 2 \cdot 28^2 + 2 \cdot 44^2 - 52^2 = 2736,$ so $MC^2 = MD^2 = 684.$ Likewise $NA = NB.$ Then $MN,$ as a median of the isosceles triangles $MCD$ and $NAB,$ is perpendicular to both $\overline{AB}$ and $\overline{CD},$ so the $180^\circ$ rotation about line $MN$ swaps $A \leftrightarrow B$ and $C \leftrightarrow D.$ Also $MN^2 = MD^2 - ND^2 = 684 - 26^2 = 8.$

For any point $X,$ let $X'$ be its image under this rotation, and let $Q$ be the midpoint of $\overline{XX'},$ which lies on line $MN.$ Then $BX = AX'$ and $CX = DX',$ so $$f(X) = (AX + AX') + (DX + DX') \ge 2AQ + 2DQ = f(Q),$$ because a median of a triangle is at most half the sum of the two adjacent sides. So it suffices to minimize $f(Q) = 2(AQ + DQ)$ over points $Q$ on line $MN.$

Rotate $D$ about line $MN$ into the plane of $A$ and line $MN,$ on the opposite side of $MN$ from $A,$ landing at $D'$ with $ND' = ND = 26.$ For $Q$ on line $MN,$ $AQ + DQ = AQ + D'Q \ge AD',$ with equality where segment $\overline{AD'}$ crosses $MN.$ Since $AM \perp MN$ and $D'N \perp MN$ with $AM = 26$ and $ND' = 26,$ $$AD'^2 = (AM + ND')^2 + MN^2 = 52^2 + 8 = 2712 = 4 \cdot 678.$$ Hence the minimum of $f$ is $2AD' = 4\sqrt{678},$ and since $678 = 2 \cdot 3 \cdot 113$ is squarefree, $m + n = 4 + 678 = 682.$