Solution:

Let $t$ be the number of hours since $1{:}30.$ Andy has traveled $8t$ miles north, and Betsy, who started an hour later, has traveled $12(t-1)$ miles east.

Setting the distances equal, $$8t = 12(t-1),$$ so $4t = 12$ and $t = 3.$

Three hours after $1{:}30$ is $4{:}30.$

Thus, the correct answer is E.

Solution:

The first box has $5$ lb peanuts, $2$ lb cashews, and $3$ lb almonds. Adding $x$ pounds of the second mix contributes $0.2x$ lb peanuts and $0.4x$ lb cashews.

The new peanut fraction is $40\%,$ so $$\frac{5+0.2x}{10+x}=0.4.$$ This gives $5+0.2x=4+0.4x,$ so $x=5.$

The cashews now total $2 + 0.4(5) = 4$ pounds.

Thus, the correct answer is B.

Solution:

Let $s$ be the number of students and $a$ be Ash's age. The students' ages total $12s,$ and adding Ash gives $$12s + a = 14(s+1) \implies a = 2s + 14.$$

There are $15 - s$ teachers with ages totaling $55(15-s),$ and adding Ash gives $$55(15-s) + a = 52(16-s) \implies a = 3s + 7.$$

Setting $2s+14 = 3s+7$ gives $s = 7,$ so $a = 2(7)+14 = 28.$

Thus, the correct answer is A.

Solution:

Let $T$ be the number of true statements. The statements assert $T \ge 1,$ $T \ge 2,$ $T \le 2,$ and $T \le 3,$ respectively.

Testing $T = 3$: the conditions $T\ge1,$ $T\ge2,$ $T\le3$ hold (statements one, two, four) and $T\le2$ fails (statement three). Exactly $3$ statements are true, matching $T = 3.$

No other value of $T$ is consistent, so exactly one statement is false.

Thus, the correct answer is B.

Solution:

Let the outer square have area $1.$ The nested squares have areas $1, k^2, k^4, \ldots,$ so the ring between the $n$th and $(n+1)$th squares has area $k^{2n}(1-k^2).$

The shaded rings are the alternate ones $n = 0, 2, 4, \ldots,$ with total area $$\sum_{j=0}^{\infty} k^{4j}(1-k^2) = \frac{1-k^2}{1-k^4} = \frac{1}{1+k^2}.$$

Setting $\dfrac{1}{1+k^2} = \dfrac{16}{25}$ gives $1 + k^2 = \dfrac{25}{16},$ so $k^2 = \dfrac{9}{16}$ and $k = \dfrac{3}{4}.$

Thus, the correct answer is D.

Solution:

Choosing $2$ chairs for the students and $2$ for the teachers gives $\binom{6}{2}\binom{4}{2} = 15 \cdot 6 = 90$ equally likely outcomes.

A round table has $6$ adjacent pairs of chairs. Give the students any adjacent pair; among the remaining $4$ chairs there are exactly $3$ adjacent pairs for the teachers. That is $6 \cdot 3 = 18$ favorable outcomes.

The probability is $\dfrac{18}{90} = \dfrac{1}{5}.$

Thus, the correct answer is B.

Solution:

Taking logarithms, $\log v = \log k + a\log n + b\log m.$

With $n = 5,$ this reads $\log v = (\log k + a\log 5) + b\log m,$ matching $4 + 2\log m,$ so $b = 2$ and $\log k + a\log 5 = 4.$

With $m = 25,$ it reads $\log v = (\log k + b\log 25) + a\log n,$ matching $4 + 4\log n,$ so $a = 4$ and $\log k + 2\log 25 = 4.$

Then $\log k = 4 - \log 625 = \log\dfrac{10000}{625} = \log 16,$ so $k = 16.$ Hence $k + a + b = 16 + 4 + 2 = 22.$

Thus, the correct answer is C.

Solution:

The inscribed angle $\angle BEC = 30^\circ$ subtends arc $BC = 60^\circ,$ so $\angle BAC,$ which also subtends arc $BC,$ equals $30^\circ.$ Likewise $\angle CAD = 30^\circ.$

Thus $AC$ bisects $\angle BAD = 60^\circ.$ In $\triangle ABD,$ $$BD^2 = 9^2 + 24^2 - 2(9)(24)\cos 60^\circ = 657 - 216 = 441,$$ so $BD = 21.$

Since $AF$ (along $AC$) bisects $\angle BAD,$ the Angle Bisector Theorem gives $\dfrac{BF}{FD} = \dfrac{AB}{AD} = \dfrac{9}{24} = \dfrac{3}{8}.$ Hence $BF = \dfrac{3}{11}\cdot 21 = \dfrac{63}{11}.$

Thus, the correct answer is E.

Solution:

Compute $w^2 = (2+i)^2 = 3 + 4i,$ so the points are $(2,1)$ and $(3,4).$

The line through them has slope $\dfrac{4-1}{3-2} = 3,$ giving $y = 3x - 5.$ Setting $y = 0$ yields $x = \dfrac{5}{3}.$

So $r = \dfrac{5}{3}.$

Thus, the correct answer is E.

Solution:

Let $R_1 = OA = OD$ and $R_2 = OB = OC,$ and let $\alpha = \angle AOD = \angle BOC$ (the rays coincide). The minor arc $BC$ has length $R_2\alpha = 2\pi,$ and the major arc $AD$ is the reflex arc, so $R_1(2\pi - \alpha) = 2\pi.$

Each segment $AB = CD = R_2 - R_1 = 2\pi.$

From the first two equations, $R_2 = \dfrac{2\pi}{\alpha}$ and $R_1 = \dfrac{2\pi}{2\pi - \alpha}.$ Substituting into $R_2 - R_1 = 2\pi$ and dividing by $2\pi$ gives $$\frac{1}{\alpha} - \frac{1}{2\pi - \alpha} = 1,$$ which simplifies to $\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0.$

The smaller root is $\alpha = (1+\pi) - \sqrt{1+\pi^2}.$ Then $$R_1 = \frac{2\pi}{2\pi - \alpha} = \frac{2\pi}{\pi - 1 + \sqrt{1+\pi^2}} = 1 - \pi + \sqrt{1+\pi^2},$$ after rationalizing (the denominator times $\sqrt{1+\pi^2} - (\pi-1)$ equals $2\pi$).

Thus, the correct answer is B.

Solution:

Since $B$ and $C$ both have $y = 27,$ side $BC$ is horizontal and the altitude from $A$ is the vertical line $x = 2.$

Side $AC$ has slope $\dfrac{27 - 31}{18 - 2} = -\dfrac{1}{4},$ so the altitude from $B$ has slope $4$: $$y - 27 = 4(x - 8).$$

At $x = 2,$ $y = 27 + 4(2 - 8) = 3.$ The orthocenter is $(2, 3),$ with coordinate sum $5.$

Thus, the correct answer is A.

Solution:

Each factor $kx^2 - 4x - 3$ has discriminant $16 + 12k \gt 0,$ so it has two real roots; there are $4050$ roots in all.

For the roots of $kx^2 - 4x - 3,$ the sum of reciprocals is $\dfrac{\text{sum}}{\text{product}} = \dfrac{4/k}{-3/k} = -\dfrac{4}{3},$ independent of $k.$

Summing over all $2025$ factors, $\displaystyle\sum \frac{1}{r} = 2025\left(-\frac{4}{3}\right) = -2700.$ The harmonic mean is $$\frac{4050}{-2700} = -\frac{3}{2}.$$

Thus, the correct answer is B.

Solution:

To avoid five consecutive integers, it suffices to remove two elements (for example $5$ and $10$), and no single removal breaks every run of five. Thus $N = 11.$

Choosing $11$ of $13$ elements is the same as removing $2,$ which can be done in $\binom{13}{2} = 78$ ways. The chosen set avoids five consecutive integers exactly when the two removed elements together intersect every window $\{t, t+1, t+2, t+3, t+4\}$ for $t = 1, \ldots, 9.$

This forces one removed element in $\{1,\ldots,5\},$ the other in $\{9,\ldots,13\},$ and the two within $5$ of each other. The valid removals are $\{4,9\},$ $\{5,9\},$ and $\{5,10\},$ giving $3$ of them.

The probability is $\dfrac{3}{78} = \dfrac{1}{26}.$

Thus, the correct answer is D.

Solution:

With the same eccentricity, $b = a\sqrt{1 - e^2},$ so the area $\pi a b \propto a^2.$ The area ratio $2025$ gives $\dfrac{a_1}{a_2} = \sqrt{2025} = 45,$ where $a_1, a_2$ are the semi-major axes.

Both ellipses share focus $G.$ On the large ellipse $G$ is the right focus, so its right vertex lies $a_1 - c_1$ to the right of $G.$ On the small ellipse $G$ is the left focus, so its right vertex lies $a_2 + c_2$ to the right of $G.$ Internal tangency makes these coincide: $$a_1 - c_1 = a_2 + c_2.$$

Using $c = ea,$ $a_1(1 - e) = a_2(1 + e),$ so $45(1 - e) = 1 + e,$ giving $46e = 44$ and $e = \dfrac{22}{23}.$

Thus, the correct answer is D.

Solution:

The set $\{11, 12, \ldots, 20\}$ has $10$ elements and is sum-free, since any two elements sum to at least $22 \gt 20.$

For the upper bound, let $a_1 \lt a_2 \lt \cdots \lt a_k$ be a sum-free subset. Each difference $a_k - a_i$ for $i \lt k$ cannot lie in $S,$ because $(a_k - a_i) + a_i = a_k \in S$ would violate sum-freeness.

These $k - 1$ differences are distinct, lie in $\{1, \ldots, 19\},$ and are disjoint from the $k$ elements of $S.$ So $k + (k - 1) \le 20,$ giving $k \le 10.$

Thus, the correct answer is C.

Solution:

By the Law of Cosines, $$\cos B = \frac{80^2 + 45^2 - 75^2}{2 \cdot 80 \cdot 45} = \frac{2800}{7200} = \frac{7}{18}.$$

The altitude to $AB$ is drawn from $C,$ and its foot is at distance $BC\cos B = 45 \cdot \dfrac{7}{18} = 17.5$ from $B$ along $AB.$

Along the bisector from $B,$ the component parallel to $AB$ is $BP\cos\dfrac{B}{2},$ which must reach the altitude's foot: $BP\cos\dfrac{B}{2} = 17.5.$

Since $\cos\dfrac{B}{2} = \sqrt{\dfrac{1 + 7/18}{2}} = \sqrt{\dfrac{25}{36}} = \dfrac{5}{6},$ we get $BP = \dfrac{17.5}{5/6} = 21.$

Thus, the correct answer is D.

Solution:

The sum of the roots is $-6i,$ so the centroid is $-2i.$ Substituting $z = u - 2i,$ $$(u - i)(u)(u + i) + 10 = u(u^2 + 1) + 10 = u^3 + u + 10.$$

Since $u = -2$ is a root, $u^3 + u + 10 = (u + 2)(u^2 - 2u + 5),$ giving roots $u = -2$ and $u = 1 \pm 2i.$

These are the points $(-2, 0),$ $(1, 2),$ $(1, -2).$ The base between $(1, 2)$ and $(1, -2)$ has length $4,$ at horizontal distance $3$ from $(-2, 0),$ so the area is $\dfrac{1}{2}(4)(3) = 6.$ Translation does not change the area.

Thus, the correct answer is A.

Solution:

If any variable is $0,$ say $z = 0,$ then $zx = 0 \gt y$ is impossible. So $x, y, z \in \{1, \ldots, 8\}$ are distinct positive integers.

The conditions are symmetric. For distinct values $a \lt b \lt c,$ we have $ac \gt b$ and $bc \gt a$ automatically, so the only real constraint is $ab \gt c.$ When it holds, all $6$ orderings work.

Counting $3$-subsets $\{a, b, c\}$ of $\{1, \ldots, 8\}$ with $ab \gt c$ gives $31$ sets. Multiplying by $6$ orderings yields $6 \cdot 31 = 186.$

Thus, the correct answer is C.

Solution:

By Vieta's formulas, $a + b + c = 0,$ $ab + bc + ca = k,$ and $abc = -1.$

Group the sum as $$a^2b^2(a + b) + b^2c^2(b + c) + c^2a^2(c + a).$$ Since $a + b + c = 0,$ we have $a + b = -c,$ $b + c = -a,$ $c + a = -b.$

So the sum equals $-a^2b^2 c - ab^2c^2 - a^2bc^2 = -abc(ab + bc + ca) = -(-1)(k) = k.$

Thus, the correct answer is E.

Solution:

The top is a ridge of length $7,$ centered above the base at some height $h.$ Its endpoints sit above $(3, 4)$ and $(10, 4)$ of the $13 \times 8$ base. A slant edge to a base corner has length $\sqrt{3^2 + 4^2 + h^2} = 13,$ so $h = 12.$

At height $z,$ the horizontal cross-section is a rectangle measuring $\left(13 - \dfrac{z}{2}\right)$ by $\left(8 - \dfrac{2z}{3}\right).$ At $z = 0$ its area is $104$; at $z = 6$ it is $10 \cdot 4 = 40$; at $z = 12$ the ridge has area $0.$

By the prismatoid formula, $$V = \frac{12}{6}\left(104 + 4 \cdot 40 + 0\right) = 2(264) = 528.$$

Thus, the correct answer is C.

Solution:

Summing the geometric series, the numerator is $4^a\dfrac{4^{k(m+1)} - 1}{4^k - 1}$ and the denominator is $2^a\dfrac{2^{k(m+1)} - 1}{2^k - 1}.$ Using $4^N - 1 = (2^N - 1)(2^N + 1),$ the ratio simplifies to $$2^a \cdot \frac{2^{k(m+1)} + 1}{2^k + 1} = 964.$$

Since $964 = 4 \cdot 241,$ take $a = 2,$ so $\dfrac{2^{k(m+1)} + 1}{2^k + 1} = 241.$ With $k = 4,$ $241(2^4 + 1) = 241 \cdot 17 = 4097 = 2^{12} + 1,$ so $k(m+1) = 12$ and $m = 2.$

Then $a + k + m = 2 + 4 + 2 = 8.$

Thus, the correct answer is A.

Solution:

Order the values as $x_1 \gt x_2 \gt x_3$; the joint density of the order statistics is $6$ on this region. The event is $x_1 \gt 2x_2.$

Integrating $x_3$ from $0$ to $x_2$ contributes a factor of $x_2.$ Then $$P = 6\int_0^{1/2} x_2\int_{2x_2}^{1} dx_1\, dx_2 = 6\int_0^{1/2} x_2(1 - 2x_2)\, dx_2.$$

This equals $6\left(\dfrac{1}{8} - \dfrac{1}{12}\right) = 6 \cdot \dfrac{1}{24} = \dfrac{1}{4}.$

Thus, the correct answer is E.

Solution:

The digits are distinct and drawn from $\{1, \ldots, 9\},$ and "no digit adjacent to two greater digits" means no interior digit is smaller than both neighbors.

For a fixed set of $n$ digits, build the arrangement by inserting digits from largest to smallest; each new (smaller) digit must go to one of the two ends, giving $2^{n-1}$ valid arrangements.

Summing over all nonempty digit subsets, $$\sum_{n=1}^{9}\binom{9}{n}2^{n-1} = \frac{1}{2}\sum_{n=1}^{9}\binom{9}{n}2^{n} = \frac{3^9 - 1}{2} = \frac{19682}{2} = 9841.$$

Thus, the correct answer is C.

Solution:

The centers of the $12$ outer circles lie on a circle of radius $r + 1,$ forming a regular $12$-gon. Adjacent centers are $2$ apart (both circles have radius $1$), and the central angle between them is $30^\circ.$