Solution:

The volume of water is the same before and after, so $\dfrac{5}{6}V_1=\dfrac{3}{4}V_2.$

Then $$\dfrac{V_1}{V_2}=\dfrac{3/4}{5/6}=\dfrac{3}{4}\cdot\dfrac{6}{5}=\dfrac{9}{10}.$$

Thus, D is the correct answer.

Solution:

A counterexample needs $n$ not prime (so the hypothesis holds) and $n-2$ not prime (so the conclusion fails).

Among the choices, $27$ is not prime and $27-2=25$ is not prime. The primes $11$ and $19$ fail the hypothesis, and $15-2=13,$ $21-2=19$ are prime.

Thus, E is the correct answer.

Solution:

Each point maps by $(x,y)\to(-x,-y):$ indeed $(-2,1)\to(2,-1)$ and $(-1,4)\to(1,-4).$

The map $(x,y)\to(-x,-y)$ is a $180^\circ$ rotation about the origin.

Thus, E is the correct answer.

Solution:

Factor the left side: $$(n+1)!+(n+2)!=(n+1)!\,[1+(n+2)]=(n+1)!\,(n+3).$$

Dividing both sides by $n!$ and using $(n+1)!=(n+1)\,n!$ gives $$(n+1)(n+3)=440.$$

So $n^2+4n-437=0,$ which factors as $(n-19)(n+23)=0,$ giving $n=19.$ Its digit sum is $1+9=10.$

Thus, C is the correct answer.

Solution:

Let $M$ be Casper's money in cents. Since he can exactly buy $12,$ $14,$ or $15$ whole-cent pieces, $M$ is a multiple of $\operatorname{lcm}(12,14,15)=420.$

Purple candy costs $20$ cents, so $n=\dfrac{M}{20}.$ The smallest $M$ is $420,$ giving $n=\dfrac{420}{20}=21.$

Thus, B is the correct answer.

Solution:

The perimeter condition gives $CA+CB=50-10=40,$ so $C$ lies on an ellipse with foci $A,B$ and major axis $2a=40.$ Thus $a=20$ and $c=5,$ so the semi-minor axis is $$b=\sqrt{a^2-c^2}=\sqrt{375}\approx19.36.$$

For area $100$ with base $AB=10,$ the height from $C$ must be $\dfrac{2\cdot100}{10}=20.$ But the greatest possible height on the ellipse is $b\approx19.36\lt20,$ so no such $C$ exists.

Thus, A is the correct answer.

Solution:

The mean is $\dfrac{4+6+8+17+x}{5}=\dfrac{35+x}{5}.$

If $x\le6,$ the median is $6,$ so $\dfrac{35+x}{5}=6$ gives $x=-5,$ which is consistent.

If $6\lt x\lt8,$ the median is $x,$ so $\dfrac{35+x}{5}=x$ gives $x=8.75,$ not in range. If $x\ge8,$ the median is $8,$ so $\dfrac{35+x}{5}=8$ gives $x=5,$ not in range.

The only solution is $x=-5,$ so the sum is $-5.$

Thus, A is the correct answer.

Solution:

Since $f(1-x)=(1-x)^2x^2=f(x),$ we have $f\!\left(\tfrac{k}{2019}\right)=f\!\left(\tfrac{2019-k}{2019}\right).$

In the sum, the term with index $k$ has sign $(-1)^{k+1},$ while the term with index $2019-k$ equals it in value but has sign $(-1)^{2019-k+1}=(-1)^{k},$ the opposite.

Every term cancels with its partner, so the total is $0.$

Thus, A is the correct answer.

Solution:

Let $t=\log_2 x.$ Then $\log_4 x=\dfrac{t}{2},$ and the sides are $t,\ \dfrac{t}{2},\ 3.$

The triangle inequalities give $t+\dfrac{t}{2}\gt3$ (so $t\gt2$) and $\dfrac{t}{2}+3\gt t$ (so $t\lt6$); the third inequality is automatic.

Thus $2\lt\log_2 x\lt6,$ i.e. $4\lt x\lt64.$ The integers $5,6,\ldots,63$ number $59.$

Thus, B is the correct answer.

Solution:

A route uses $13$ roads as an open trail from $A$ to $L,$ so on the used roads exactly $A$ and $L$ have odd degree and every other city has even degree.

In the full map the corner cities $A$ and $L$ already have even degree $2,$ and six edge-cities have odd degree $3.$ Removing $4$ roads must flip the parity of $A,$ $L,$ and those six cities, and of no others. This forces the four removed roads to pair up those eight cities in the only possible way, so the set of $13$ used roads is uniquely determined.

Counting the Eulerian trails from $A$ to $L$ on that graph gives exactly $4$ routes.

Thus, E is the correct answer.

Solution:

Two edges determine a plane exactly when they are coplanar, that is, parallel or intersecting.

The $12$ edges split into $3$ directions of $4$ parallel edges, giving $3\binom{4}{2}=18$ parallel pairs. Edges sharing a vertex give $8\binom{3}{2}=24$ intersecting pairs.

The total is $18+24=42.$

Thus, D is the correct answer.

Solution:

Triangle $ABC$ has perimeter $1+1+\sqrt2=2+\sqrt2$ and $AC=\sqrt2.$ In $\triangle ACD$ let $CD=d,$ so $AD=\sqrt{2+d^2}$ and equal perimeters give $$\sqrt2+d+\sqrt{2+d^2}=2+\sqrt2.$$

Then $\sqrt{2+d^2}=2-d,$ so $2+d^2=4-4d+d^2,$ giving $d=\dfrac12$ and $AD=\dfrac32.$

Since $\angle BAC=45^\circ,$ writing $\theta=\angle CAD$ gives $2\angle BAD=90^\circ+2\theta,$ so $\sin(2\angle BAD)=\cos 2\theta.$ With $\tan\theta=\dfrac{CD}{AC}=\dfrac{1}{2\sqrt2},$ we get $$\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\tfrac18}{1+\tfrac18}=\dfrac{7}{9}.$$

Thus, D is the correct answer.

Solution:

The probability the balls land in the same bin is $$\sum_{k=1}^\infty \left(2^{-k}\right)^2=\sum_{k=1}^\infty 4^{-k}=\dfrac{1/4}{1-1/4}=\dfrac13.$$

By symmetry, the red ball being higher and the green ball being higher are equally likely, so each has probability $$\dfrac{1-\tfrac13}{2}=\dfrac13.$$

Thus, C is the correct answer.

Solution:

Since $100{,}000=2^5\cdot5^5,$ every divisor is $2^a5^b$ with $0\le a,b\le5.$ A product of two divisors is $2^x5^y$ with $0\le x,y\le10,$ and every such pair $(x,y)$ is attainable, giving $11\cdot11=121$ values.

We need two \emph{distinct} divisors. A value $2^x5^y$ is forced to be a divisor times itself only when both $x$ and $y$ have a unique split, which happens exactly when $x,y\in\{0,10\}.$ Those $4$ corner values ($1,\ 2^{10},\ 5^{10},\ 2^{10}5^{10}$) cannot use two distinct divisors.

The count is $121-4=117.$

Thus, C is the correct answer.

Solution:

Put $A=(0,0),\ B=(2,0),\ C=(4,0),\ D=(6,0),$ so the semicircles are centered at $(1,0),(3,0),(5,0)$ and their tops are $E=(1,1),\ F=(3,1),\ G=(5,1).$ The circle has center $F=(3,1)$ and radius $2,$ so it passes through $E$ and $G,$ and has area $4\pi.$

The middle semicircle $BFC$ lies entirely inside the circle, removing area $\dfrac{\pi}{2}.$ By symmetry the outer semicircles each contribute the same overlap $R=\dfrac{7\pi}{12}-2+\dfrac{\sqrt3}{2}$ inside the circle.

The shaded area is $$4\pi-\dfrac{\pi}{2}-2R=\dfrac{7}{3}\pi-\sqrt3+4.$$ Hence $a=7,\ b=3,\ c=3,\ d=4,$ so $a+b+c+d=17.$

Thus, E is the correct answer.

Solution:

Let $p(n)$ be the probability of landing on pad $n$ without first landing on pad $3$ or $6.$ Each pad sends probability $\dfrac12$ to the next pad and $\dfrac12$ two pads ahead, and pads $3$ and $6$ pass nothing on.

Then $p(0)=1,\ p(1)=\dfrac12,\ p(2)=\dfrac34,$ and (skipping $3$) $p(4)=\dfrac38,\ p(5)=\dfrac{3}{16},$ then (skipping $6$) $p(7)=\dfrac{3}{32},\ p(8)=\dfrac{3}{64},\ p(9)=\dfrac{9}{128}.$

Finally $$p(10)=\dfrac12 p(8)+\dfrac12 p(9)=\dfrac{3}{128}+\dfrac{9}{256}=\dfrac{15}{256}.$$

Thus, A is the correct answer.

Solution:

The three points form an equilateral triangle iff $|z|=|z^3|=|z^3-z|.$ From $|z|=|z^3|=|z|^3$ we get $|z|=1.$

Then $|z^3-z|=|z|\,|z^2-1|=|z^2-1|,$ so we need $|z^2-1|=1.$ Writing $z=e^{i\theta},$ $|z^2-1|=2|\sin\theta|=1,$ so $|\sin\theta|=\dfrac12.$

This gives $\theta=30^\circ,150^\circ,210^\circ,330^\circ,$ four values of $z,$ all yielding distinct vertices.

Thus, D is the correct answer.

Solution:

Place $A=(0,0,0),\ B=(3,0,0),\ C=(3,3,0),\ D=(0,3,0),\ E=(0,0,6).$ Then $$P=(2,0,2),\quad Q=(0,2,2),\quad R=(1,1,4).$$

So $\vec{PQ}=(-2,2,0)$ and $\vec{PR}=(-1,1,2),$ giving $\vec{PQ}\times\vec{PR}=(4,4,0).$

The area is $\dfrac12\bigl|(4,4,0)\bigr|=\dfrac12\cdot4\sqrt2=2\sqrt2.$

Thus, C is the correct answer.

Solution:

From $(1,1,1),$ each of the three players gives to one of two others, so there are $8$ equally likely outcomes; only the $2$ cyclic gift patterns return to $(1,1,1),$ a probability of $\dfrac14.$

From a $(2,1,0)$ state the broke player gives nothing, and checking the $4$ equally likely choices of the other two shows exactly one yields $(1,1,1),$ again probability $\dfrac14.$

So after any ring the probability of $(1,1,1)$ is $\dfrac14,$ including after $2019$ rings.

Thus, B is the correct answer.

Solution:

Let $P=(x,0)$ be the intersection. Equal tangent lengths give $PA=PB,$ so $(x-6)^2+13^2=(x-12)^2+11^2,$ yielding $x=5$ and $P=(5,0).$

The center $O=(h,k)$ satisfies $OA\perp PA$ and $OB\perp PB.$ With $PA=(1,13)$ and $PB=(7,11),$ these give $h+13k=175$ and $7h+11k=205,$ so $O=\left(\dfrac{37}{4},\dfrac{51}{4}\right).$

Then $r^2=OA^2=\left(\dfrac{13}{4}\right)^2+\left(\dfrac14\right)^2=\dfrac{170}{16}=\dfrac{85}{8},$ so the area is $\dfrac{85\pi}{8}.$

Thus, C is the correct answer.

Solution:

The set $\{a,b,c\}$ must equal the two-element set $\{r,s\},$ so at least two coefficients coincide, and the roots are the two distinct coefficient values. By Vieta's formulas $r+s=-\dfrac{b}{a}$ and $rs=\dfrac{c}{a}.$

Working through the cases of which coefficients are equal yields the polynomials $x^2+x-2,$ $-x^2-x,$ $x^2-\dfrac12 x-\dfrac12,$ and $ux^2+\dfrac1u x+u$ where $u$ is the unique real root of $u^3+u+1=0.$

That is $4$ polynomials in all.

Thus, B is the correct answer.

Solution:

Let $a_n=x_n-4.$ A short computation gives $$a_{n+1}=x_{n+1}-4=\dfrac{(x_n+5)(x_n-4)}{x_n+6}=a_n\cdot\dfrac{x_n+5}{x_n+6}.$$

Starting from $a_0=1,$ the terms stay positive and decrease. Because $x_n$ decreases from $5$ toward $4,$ each ratio $\dfrac{x_n+5}{x_n+6}$ lies strictly between $\dfrac{9}{10}$ and $\dfrac{10}{11}.$

Hence $a_m$ is squeezed between $\left(\dfrac{9}{10}\right)^m$ and $\left(\dfrac{10}{11}\right)^m.$ Solving $a_m\le2^{-20}$ puts $m$ between about $132$ and $146,$ which lies in $[81,242].$

Thus, C is the correct answer.

Solution:

No two $0$s are adjacent, so the $0$s are separated by blocks of $1$s, each of size $1$ or $2$ (never $3$). If there are $k$ zeros, there are $k-1$ such blocks summing to $19-k$ ones.

The number of size-$2$ blocks is $(19-k)-(k-1)=20-2k,$ which must satisfy $0\le20-2k\le k-1,$ i.e. $7\le k\le10.$

Summing $\binom{k-1}{20-2k}$ over $k=7,8,9,10$ gives $\binom66+\binom74+\binom82+\binom90=1+35+28+1=65.$

Thus, C is the correct answer.

Solution:

As $a,b,c$ range over $[0,1],$ the set $S$ is the Minkowski sum of the three unit segments along $v_1=1=(1,0),\ v_2=\omega=\left(-\dfrac12,\dfrac{\sqrt3}{2}\right),\ v_3=\omega^2=\left(-\dfrac12,-\dfrac{\sqrt3}{2}\right).$

This is a zonogon whose area is the sum of the cross-product magnitudes over pairs. Each pair gives $|v_i\times v_j|=\dfrac{\sqrt3}{2}.$

Therefore the area is $3\cdot\dfrac{\sqrt3}{2}=\dfrac{3\sqrt3}{2}.$

Thus, C is the correct answer.

Solution:

The centroids are $\dfrac{A+B+C}{3},\ \dfrac{B+C+D}{3},\ \dfrac{A+C+D}{3}.$ Their pairwise differences are $\dfrac{A-D}{3},\ \dfrac{B-A}{3},\ \dfrac{B-D}{3},$ so an equilateral centroid triangle forces $AB=BD=DA;$ that is, $\triangle ABD$ is equilateral with side $s=BD.$

Splitting along $BD,$ $$[ABCD]=[ABD]+[BCD]=\dfrac{\sqrt3}{4}s^2+\dfrac12\cdot2\cdot6\sin C,$$ where $C=\angle BCD.$ By the Law of Cosines $s^2=40-24\cos C,$ so $$[ABCD]=10\sqrt3-6\sqrt3\cos C+6\sin C.$$