Solution:

The legs of right triangle $ABE$ are $AB = 10$ and $BE.$ From $\tfrac12\cdot 10\cdot BE = 40,$ we get $BE = 8.$

Thus, the correct answer is E.

Solution:

The team can only score twice as many runs as its opponent when its own score is even. Those games have scores $2, 4, 6, 8, 10,$ so their opponents scored $1 + 2 + 3 + 4 + 5 = 15.$

The other five games had scores $1, 3, 5, 7, 9$ and were one-run losses, so their opponents scored $2 + 4 + 6 + 8 + 10 = 30.$ The total is $15 + 30 = 45.$

Thus, the correct answer is C.

Solution:

Six tenths of the flowers are pink and four tenths are red. Since two thirds of the pink flowers are carnations, pink carnations make up $\tfrac{2}{3}\cdot\tfrac{6}{10} = \tfrac{4}{10}$ of the flowers.

Since three fourths of the red flowers are carnations, red carnations make up $\tfrac{3}{4}\cdot\tfrac{4}{10} = \tfrac{3}{10}$ of the flowers. Together the carnations are $\tfrac{4}{10} + \tfrac{3}{10} = \tfrac{7}{10} = 70\%.$

Thus, the correct answer is E.

Solution:

Factoring $2^{2012}$ from each term gives $$\dfrac{2^{2012}(2^2 + 1)}{2^{2012}(2^2 - 1)} = \dfrac{4 + 1}{4 - 1} = \dfrac{5}{3}.$$

Thus, the correct answer is C.

Solution:

The total spent was $105 + 125 + 175 = 405,$ so each fair share is $\tfrac13\cdot 405 = 135$ dollars.

Then $t = 135 - 105 = 30$ and $d = 135 - 125 = 10,$ so $t - d = 30 - 10 = 20.$

Thus, the correct answer is B.

Solution:

If Shenille attempted $x$ three-point shots and $30 - x$ two-point shots, she scored $$0.2\cdot 3\cdot x + 0.3\cdot 2\cdot(30 - x) = 0.6x + 0.6(30 - x) = 0.6\cdot 30 = 18$$ points.

Thus, the correct answer is B.

Solution:

Since $S_9 = S_7 + S_8,$ we get $S_8 = 110 - 42 = 68.$ Then $S_6 = S_8 - S_7 = 68 - 42 = 26,$ $S_5 = S_7 - S_6 = 42 - 26 = 16,$ and $S_4 = S_6 - S_5 = 26 - 16 = 10.$

Thus, the correct answer is C.

Solution:

Multiplying by $xy$ gives $x^2 y + 2y = xy^2 + 2x,$ so $$x^2 y - xy^2 - 2x + 2y = (x - y)(xy - 2) = 0.$$

Since $x \ne y,$ it follows that $xy = 2.$

Thus, the correct answer is D.

Solution:

Because $EF \parallel AB,$ triangle $FEC$ is similar to triangle $ABC,$ which is isosceles, so $FE = FC.$

Half the perimeter of parallelogram $ADEF$ is $AF + FE = AF + FC = AC = 28.$ The entire perimeter is $56.$

Thus, the correct answer is C.

Solution:

If $\dfrac{1}{n} = 0.\overline{ab},$ then $\dfrac{99}{n} = \overline{ab},$ a two-digit number. The positive divisors of $99$ are $1, 3, 9, 11, 33, 99.$

Only $n = 11, 33, 99$ make $\dfrac{99}{n}$ equal to $09, 03, 01,$ which have two different digits. The requested sum is $11 + 33 + 99 = 143.$

Thus, the correct answer is D.

Solution:

Let $x = DE$ and $y = FG.$ The parallel cuts make the small regions equilateral or isosceles trapezoids, so the perimeters are $$\triangle ADE: 3x, \quad DFGE: 3y - x, \quad FBCG: 3 - y.$$

Setting them equal, $3x = 3y - x$ gives $4x = 3y,$ and $3x = 3 - y.$ Solving yields $x = \tfrac{9}{13}$ and $y = \tfrac{12}{13},$ so $DE + FG = \tfrac{21}{13}.$

Thus, the correct answer is C.

Solution:

If the angles are $\alpha - \delta, \alpha, \alpha + \delta,$ their sum $3\alpha = 180^\circ$ gives $\alpha = 60^\circ,$ so one angle is $60^\circ.$

If $x$ is opposite the $60^\circ$ angle, the Law of Cosines gives $$x^2 = 4^2 + 5^2 - 2\cdot 4\cdot 5\cos 60^\circ = 21,$$ so $x = \sqrt{21}.$

If $5$ is opposite the $60^\circ$ angle, then $25 = x^2 - 4x + 16,$ whose positive solution is $x = 2 + \sqrt{13}.$ If $4$ is opposite, then $16 = x^2 - 5x + 25$ has no real solution.

The sum of the possible values is $2 + \sqrt{13} + \sqrt{21},$ so $a + b + c = 2 + 13 + 21 = 36.$

Thus, the correct answer is A.

Solution:

By the shoelace formula, the area of $ABCD$ is $\tfrac{15}{2}.$ Let the line meet $\overline{CD}$ at $G.$ Triangle $ADG$ must have area $\tfrac{15}{4}.$

Since $AD = 4$ lies on the $x$-axis, $\tfrac12\cdot 4\cdot y_G = \tfrac{15}{4}$ gives $y_G = \tfrac{15}{8}.$ Line $CD$ is $y = -3(x - 4),$ so $x_G = \tfrac{27}{8}.$

Then $p + q + r + s = 27 + 8 + 15 + 8 = 58.$

Thus, the correct answer is B.

Solution:

Because the logarithms are in arithmetic progression, $162, x, y, z, 1250$ is a geometric sequence. Its common ratio $r$ satisfies $162 r^4 = 1250,$ so $r^4 = \tfrac{625}{81}$ and $r = \tfrac53.$

Therefore $x = 162\cdot\tfrac53 = 270.$

Thus, the correct answer is B.

Solution:

If the two parents share a store, there are $4$ choices for it, and each child must go to one of the other three stores: $4\cdot 3^3 = 108$ ways.

If the parents go to different stores, there are $4\cdot 3 = 12$ choices, and each child must go to one of the two remaining stores: $12\cdot 2^3 = 96$ ways.

The total is $108 + 96 = 204.$

Thus, the correct answer is D.

Solution:

Let $a, b, c$ be the numbers of rocks in the piles. From $\dfrac{40a + 50b}{a + b} = 43,$ we get $7b = 3a,$ so $a = 7k$ and $b = 3k.$

Let $\mu_{BC}$ be the mean of $B$ and $C.$ Using the $A, C$ mean $44$ to express $\mu_C = \dfrac{28k + 44c}{c},$ we find $\mu_{BC} = \dfrac{178k + 44c}{3k + c},$ so $(\mu_{BC} - 44)c = (178 - 3\mu_{BC})k.$

Since $B$ is heavier than $A,$ the mean of $B$ and $C$ exceeds $44,$ forcing $178 - 3\mu_{BC} \gt 0,$ i.e. $\mu_{BC} \lt \tfrac{178}{3} = 59\tfrac13.$ The value $59$ is attainable, so the greatest integer mean is $59.$

Thus, the correct answer is E.

Solution:

For $1 \le k \le 11,$ the number of coins before the $k$th pirate takes a share is $\dfrac{12}{12 - k}$ times the number afterward. So if $n$ coins are left for the $12$th pirate, the initial count is $$\dfrac{12^{11}\, n}{11!} = \dfrac{2^{14}\cdot 3^{7}\, n}{5^2\cdot 7\cdot 11}.$$

The smallest $n$ making this a positive integer is $5^2\cdot 7\cdot 11 = 1925,$ and one checks each earlier pirate then receives a whole number of coins. The $12$th pirate receives $1925$ coins.

Thus, the correct answer is D.

Solution:

Each small center is $2$ from the center $O,$ and the small spheres have radius $1,$ so the large sphere has radius $3.$ Let the eighth sphere have radius $r$ and center $G$ at distance $x$ from $O;$ then $x + r = 3.$

Since $G$ is equidistant from two opposite hexagon vertices, $GO$ is perpendicular to the line to a vertex, and the Pythagorean Theorem gives $$(r + 1)^2 = 2^2 + x^2 = 4 + (3 - r)^2.$$

This simplifies to $2r + 1 = 13 - 6r,$ so $r = \tfrac32.$

Thus, the correct answer is B.

Solution:

By the Power of a Point Theorem, $BC\cdot CX = AC^2 - AB^2$ where $AB$ is the radius. Thus $BC\cdot CX = 97^2 - 86^2 = 2013.$

Since $BC = BX + CX$ and $CX$ are integers, they are complementary factors of $2013 = 3\cdot 11\cdot 61.$ As $CX \lt BC \lt AB + AC = 183,$ the only possibility is $CX = 33$ and $BC = 61.$

Thus, the correct answer is D.

Solution:

Reading the elements modulo $19,$ the relation $a \succ b$ holds exactly when $0 \lt (a - b) \bmod 19 \le 9.$

There are $19$ choices for $x.$ Once $x$ is fixed, take $y = x + i$ for some $1 \le i \le 9.$ Then $z$ must satisfy $x + 10 \le z \le x + 9 + i,$ giving $i$ choices.

The total is $19(1 + 2 + \cdots + 9) = 19\cdot 45 = 855.$

Thus, the correct answer is B.

Solution:

Let $A_n = \log(n + \log((n-1) + \cdots + \log(3 + \log 2)\cdots)).$ One checks $0 \lt A_n \lt 1$ for $2 \le n \le 9,$ then $1 \lt A_n \lt 2$ for $10 \le n \le 98,$ then $2 \lt A_n \lt 3$ for $99 \le n \le 997,$ and $3 \lt A_n \lt 4$ for $998 \le n \le 9996.$

Hence $3 \lt A_{2012} \lt 4,$ so $2016 \lt 2013 + A_{2012} \lt 2017$ and therefore $\log 2016 \lt A \lt \log 2017.$

Thus, the correct answer is A.

Solution:

Let $m = \dfrac{n}{11}.$ A $4$-digit $m$ leads to a contradiction, so $m$ is a $5$-digit palindrome $\overline{abcba}.$

Writing $n = 11m,$ there are no carries exactly when $a + b \le 9$ and $b + c \le 9,$ and only then is $n$ a palindrome. The number of valid $m$ is $$\sum_{b=0}^{9}(10 - b)(9 - b) = 330.$$

There are $9\cdot 10^2 = 900$ six-digit palindromes, so the probability is $\dfrac{330}{900} = \dfrac{11}{30}.$

Thus, the correct answer is E.

Solution:

Let $A', B', C', D'$ be the images of the vertices under the rotation. The swept region decomposes into four circular sectors and four triangles.

Since $AP = \sqrt{2}$ and $PC = AC - AP = \sqrt{6},$ the sectors at $A$ and $C$ have areas $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}.$ One finds $PB = 2,$ so the two $60^\circ$ sectors along $BC$ each have area $\tfrac{2\pi}{3}.$ The four triangles together contribute $(\sqrt{3} - 1) + (3 - \sqrt{3}) = 2.$

The total area is $$\dfrac{\pi}{2} + \dfrac{3\pi}{2} + 2\cdot\dfrac{2\pi}{3} + 2 = \dfrac{10\pi + 6}{3},$$ so $a + b + c = 10 + 6 + 3 = 19.$

Thus, the correct answer is C.

Solution:

Inscribe the $12$-gon in a unit circle. The segment lengths are $d_k = 2\sin(15k^\circ)$ for $1 \le k \le 6,$ with $12$ segments of each length $d_1, \ldots, d_5$ and $6$ of length $d_6.$

Comparing sums, the forbidden index triples $(a, b, c)$ with $d_a \le d_b \le d_c$ and $d_c \ge d_a + d_b$ are $$(1,1,3),(1,1,4),(1,1,5),(1,1,6),(1,2,4),(1,2,5),(1,2,6),(1,3,5),(1,3,6),(2,2,6).$$

Counting the corresponding segment selections and dividing by $\binom{66}{3}$ gives a failure probability of $\dfrac{63}{286},$ so the answer is $1 - \dfrac{63}{286} = \dfrac{223}{286}.$

Thus, the correct answer is E.

Solution:

On the upper half-plane $H,$ if $f(z_1) = f(z_2)$ then $(z_1 - z_2)(z_1 + z_2 + i) = 0;$ since $\operatorname{Im}(z_1), \operatorname{Im}(z_2) \gt 0,$ the factor $z_1 + z_2 + i \ne 0,$ so $f$ is one-to-one on $H.$

The image is $f(H) = \{w : \operatorname{Re}(w) \lt (\operatorname{Im}(w))^2 + 1\}.$ Thus we count $w = a + ib$ with $a, b \in \mathbb{Z},$ $|a|, |b| \le 10,$ and $a \lt b^2 + 1:$ $$|S| = 21^2 - \sum_{b=-3}^{3}(10 - b^2) = 441 - 42 = 399.$$

Thus, the correct answer is A.