Solution:

The areas are proportional to the squares of the radii, so the ratio of the larger area to the smaller is $\dfrac{16}{9}.$

The percent increase is $$\left(\dfrac{16}{9} - 1\right) \times 100 = \dfrac{700}{9} \approx 77.8.$$ The closest integer is $78.$

Thus, the correct answer is E.

Solution:

Since $a = 1.5b,$ we have $$\dfrac{3b}{a} = \dfrac{3b}{1.5b} = 2.$$

As a percentage, $3b$ is $200\%$ of $a.$

Thus, the correct answer is D.

Solution:

In the worst case, we draw $14$ each of red, green, and yellow, plus all of the blue $(13),$ white $(11),$ and black $(9),$ without reaching $15$ of any color.

That is $$14 + 14 + 14 + 13 + 11 + 9 = 75$$ balls.

The next ball must complete a set of $15,$ so $76$ balls are needed.

Thus, the correct answer is B.

Solution:

Negative integers are allowed. The integers from $-44$ to $44$ sum to $0,$ so the integers from $-44$ to $45$ sum to $45.$

This run has $45 - (-44) + 1 = 90$ integers, and no longer run can work.

Thus, the correct answer is D.

Solution:

The two lines are $y = \tfrac{1}{2}x + 1$ and $y = 2x - 2.$ Intersecting each with $x + y = 10$ gives the points $(6, 4)$ and $(4, 6).$

The triangle has vertices $(2, 2),$ $(6, 4),$ and $(4, 6).$ By the shoelace formula,

$$\tfrac{1}{2}\left| 2(4 - 6) + 6(6 - 2) + 4(2 - 4) \right| = \tfrac{1}{2}\left| -4 + 24 - 8 \right| = 6.$$

Thus, the correct answer is C.

Solution:

A translation by one full period maps the figure to itself, so translation works.

A $180^\circ$ rotation about a suitable point on $\ell$ sends each square above the line to the square below it, with the diagonal segments matching, so this rotation works.

Reflection across $\ell$ sends the top-right diagonals to top-right diagonals below the line, but the actual below-line diagonals point to the bottom-left, so it fails. A reflection across a perpendicular line fails for the same reason. Only $2$ of the four transformations work.

Thus, the correct answer is C.

Solution:

The values $1$ through $28$ each appear $12$ times and are the modes, so $d = \dfrac{14 + 15}{2} = 14.5.$

The $183$rd of the $365$ ordered values is the median. Values $1$ through $15$ fill the first $180$ positions, so position $183$ is $16;$ thus $M = 16.$

The total of all values is $12(1 + \cdots + 28) + 11(29 + 30) + 7 \cdot 31 = 5738,$ so $\mu = \dfrac{5738}{365} \approx 15.7.$

Therefore $d \lt \mu \lt M.$

Thus, the correct answer is E.

Solution:

With four lines, the number of intersection points ranges over the achievable configurations. All parallel gives $0;$ all concurrent gives $1.$

Working through the cases (parallel classes and points of concurrency), the achievable values are $0, 1, 3, 4, 5, 6;$ the value $2$ is impossible.

The sum is $0 + 1 + 3 + 4 + 5 + 6 = 19.$

Thus, the correct answer is D.

Solution:

Taking reciprocals, $$\dfrac{1}{a_n} = \dfrac{2a_{n-2} - a_{n-1}}{a_{n-2}a_{n-1}} = \dfrac{2}{a_{n-1}} - \dfrac{1}{a_{n-2}}.$$

Let $b_n = \dfrac{1}{a_n}.$ Then $b_n = 2b_{n-1} - b_{n-2},$ so $b_n$ is arithmetic with $b_1 = 1,$ $b_2 = \dfrac{7}{3},$ and common difference $\dfrac{4}{3}.$

Thus $b_{2019} = 1 + 2018 \cdot \dfrac{4}{3} = \dfrac{8075}{3},$ so $a_{2019} = \dfrac{3}{8075}.$ Since these are relatively prime, $p + q = 8078.$

Thus, the correct answer is E.

Solution:

Place a unit circle at the center, six around it with centers at distance $2$ (a hexagon), and six more with centers at distance $2\sqrt{3}$ in the outer gaps. That is $1 + 6 + 6 = 13$ circles.

The outermost circles are tangent to the big circle, whose radius is therefore $2\sqrt{3} + 1.$ Its area is $$\pi(2\sqrt{3} + 1)^2 = \pi(13 + 4\sqrt{3}).$$

Subtracting the $13$ unit circles leaves $\pi(13 + 4\sqrt{3}) - 13\pi = 4\pi\sqrt{3}.$

Thus, the correct answer is A.

Solution:

The repeating block gives $$0.\overline{23}_k = \dfrac{2k + 3}{k^2 - 1} = \dfrac{7}{51}.$$

Cross-multiplying, $51(2k + 3) = 7(k^2 - 1),$ so $7k^2 - 102k - 160 = 0.$

The quadratic formula gives $k = \dfrac{102 + \sqrt{14884}}{14} = \dfrac{102 + 122}{14} = 16.$

Thus, the correct answer is D.

Solution:

Let $a = \log_2 x$ and $b = \log_2 y.$ Then $\log_y 16 = \dfrac{4}{b},$ so $a = \dfrac{4}{b},$ giving $ab = 4.$

Since $xy = 64,$ we have $a + b = 6.$

Therefore $$\left(\log_2 \tfrac{x}{y}\right)^2 = (a - b)^2 = (a + b)^2 - 4ab = 36 - 16 = 20.$$

Thus, the correct answer is B.

Solution:

The primes $5$ and $7$ have no proper divisors here, giving $3$ choices each.

Along the chain $2 \to 4 \to 8,$ there are $3 \cdot 2 \cdot 1 = 6$ colorings. Number $9$ must differ from $3,$ giving $2$ choices once $3$ is set.

Number $6$ must differ from both $2$ and $3.$ Summing over the colors of $2$ and $3$ (equal in $3$ pairs, unequal in $6$ pairs), the combined factor for $4, 8, 9, 6$ totals $2 \cdot 2 \cdot (3 \cdot 2 + 6 \cdot 1) = 48.$

Multiplying by the $9$ ways for $5$ and $7$ gives $48 \cdot 9 = 432.$

Thus, the correct answer is E.

Solution:

The factors $x^2 - 2x + 2$ and $x^2 - 4x + 8$ have roots $1 \pm i$ and $2 \pm 2i,$ which are $4$ distinct values.

For $P$ to have exactly $4$ distinct roots, the roots of $x^2 - cx + 4$ must lie among these. Their product must equal $4,$ and the only such pair is one root from each factor, for example $(1 + i)(2 - 2i) = 4.$

Then $c = (1 + i) + (2 - 2i) = 3 - i,$ so $|c| = \sqrt{3^2 + 1^2} = \sqrt{10}.$

Thus, the correct answer is E.

Solution:

Let $\sqrt{\log a} = p$ and $\sqrt{\log b} = q,$ so $\log a = p^2$ and $\log\sqrt{a} = \dfrac{p^2}{2}.$ For this to be an integer, $p$ is even; likewise $q.$

Writing $p = 2m,$ $q = 2n,$ the equation $p + q + \dfrac{p^2}{2} + \dfrac{q^2}{2} = 100$ becomes $m(m+1) + n(n+1) = 50.$

The only solution is $\{m, n\} = \{4, 5\},$ giving $\log(ab) = p^2 + q^2 = 4(16 + 25) = 164.$

Therefore $ab = 10^{164}.$

Thus, the correct answer is D.

Solution:

There are $5$ odd and $4$ even numbers. Each row and column must contain an odd number of odd entries.

The only way to place $5$ odd entries with every row and column odd is to fill one complete row and one complete column (a plus shape of $3 + 3 - 1 = 5$ cells). There are $3 \cdot 3 = 9$ such patterns.

Each pattern admits $5!$ placements of the odd numbers and $4!$ of the even numbers, so the probability is $$\dfrac{9 \cdot 5! \cdot 4!}{9!} = \dfrac{1}{14}.$$

Thus, the correct answer is B.

Solution:

Every root $r$ satisfies $r^3 = 5r^2 - 8r + 13,$ so $r^{k+1} = 5r^k - 8r^{k-1} + 13r^{k-2}.$

Summing over the three roots gives $s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2},$ so $a = 5,$ $b = -8,$ $c = 13.$

Therefore $a + b + c = 5 - 8 + 13 = 10.$

Thus, the correct answer is D.

Solution:

The sphere intersects the triangle's plane in a circle of radius $\sqrt{36 - d^2},$ where $d$ is the distance from $O$ to the plane. Since each side is tangent to the sphere, this circle is the triangle's incircle.

The triangle has area $\tfrac{1}{2} \cdot 24 \cdot 9 = 108$ and semiperimeter $27,$ so its inradius is $\dfrac{108}{27} = 4.$

Thus $\sqrt{36 - d^2} = 4,$ giving $d^2 = 20$ and $d = 2\sqrt{5}.$

Thus, the correct answer is D.

Solution:

Each sine is $\sqrt{1 - \cos^2}:$ $\sin A = \dfrac{3\sqrt{15}}{16},$ $\sin B = \dfrac{2\sqrt{15}}{16},$ $\sin C = \dfrac{4\sqrt{15}}{16}.$

By the Law of Sines the sides are in ratio $3 : 2 : 4.$ The smallest integer sides are $3, 2, 4,$ which satisfy the triangle inequality.

The least perimeter is $3 + 2 + 4 = 9.$

Thus, the correct answer is A.

Solution:

Each variable equals $0$ with probability $\tfrac14,$ equals $1$ with probability $\tfrac14,$ and is uniform on $[0, 1]$ with probability $\tfrac12.$

Considering the nine combinations of types: the pairs $(0, 1)$ and $(1, 0)$ each contribute $\tfrac{1}{16}.$ Each of the four point-versus-uniform cases contributes $\tfrac{1}{16}.$ The uniform-versus-uniform case contributes $\tfrac14 \cdot \tfrac14 = \tfrac{1}{16}.$

The total is $\dfrac{2 + 4 + 1}{16} = \dfrac{7}{16}.$

Thus, the correct answer is B.

Solution:

Since $z = e^{i\pi/4},$ we have $z^{k^2} = e^{i\pi k^2/4},$ depending only on $k^2 \bmod 8.$

For $k = 1$ to $12,$ the residue $k^2 \bmod 8$ is $1$ (giving $z$) six times, $4$ (giving $-1$) three times, and $0$ (giving $1$) three times. So the first sum is $6z - 3 + 3 = 6z.$

The second sum is likewise $\dfrac{6}{z} - 3 + 3 = \dfrac{6}{z}.$ Their product is $6z \cdot \dfrac{6}{z} = 36.$

Thus, the correct answer is C.

Solution:

Let $s = AB.$ Since $\dfrac{BP}{CP} = 3,$ we have $BP = \dfrac{3s}{4}$ and $CP = \dfrac{s}{4}.$ Put $P$ at the origin with $BC$ on the $x$-axis, $B = \left(-\tfrac{3s}{4}, 0\right),$ $C = \left(\tfrac{s}{4}, 0\right),$ and apex $A = \left(-\tfrac{s}{4}, \tfrac{s\sqrt{3}}{2}\right).$

Points $P, O, A$ are collinear, so $O = t \cdot A$ for some scalar $t.$ Two conditions pin it down: $O$ is at distance $17$ from line $BC,$ giving $|t| \cdot \dfrac{s\sqrt{3}}{2} = 17,$ and $A$ is on $\omega,$ giving $|t - 1| \cdot \dfrac{s\sqrt{13}}{4} = 20$ since $|A| = \dfrac{s\sqrt{13}}{4}.$

Solving, $|t| s = \dfrac{34}{\sqrt{3}}$ and $|t - 1| s = \dfrac{80}{\sqrt{13}}.$ The valid configuration gives $$AB = s = \dfrac{80}{\sqrt{13}} - \dfrac{34}{\sqrt{3}}.$$

Then $m + n + p + q = 80 + 13 + 34 + 3 = 130.$

Thus, the correct answer is E.

Solution:

Let $L(x) = \log_7 x.$ Then $L(a \diamondsuit b) = L(a)L(b)$ and $L(a \heartsuit b) = \dfrac{L(a)}{L(b)}.$

So $L(a_3) = \dfrac{L(3)}{L(2)},$ and $L(a_n) = \dfrac{L(n)}{L(n-1)} \cdot L(a_{n-1}).$ The product telescopes: $$L(a_N) = \dfrac{L(3)}{L(2)} \cdot \dfrac{L(N)}{L(3)} = \dfrac{L(N)}{L(2)}.$$

Hence $L(a_{2019}) = \dfrac{\log_7 2019}{\log_7 2} = \log_2 2019 \approx 10.98,$ which rounds to $11.$

Thus, the correct answer is D.

Solution:

For a prime $p,$ the condition to be an integer reduces (via Legendre's formula) to $n \cdot s_p(n) \ge s_p(n^2 - 1) + 1,$ where $s_p$ is the base-$p$ digit sum. This can fail only when $s_p(n) = 1,$ i.e. $n = p^a$ is a prime power.

For $n = p^a,$ the requirement becomes $p^a - 1 \ge 2a(p - 1).$ Checking prime powers up to $50,$ this fails exactly for every prime $n$ and for $n = 4.$

There are $15$ primes at most $50,$ plus $n = 4,$ giving $16$ failures. Hence $50 - 16 = 34$ values of $n$ work.

Thus, the correct answer is D.

Solution:

For an acute triangle, the orthic triangle (feet of the altitudes) has angles $180^\circ - 2\alpha$ for each original angle $\alpha.$

Writing an angle as $60^\circ + x,$ the new angle is $60^\circ - 2x,$ so each deviation from $60^\circ$ is multiplied by $-2.$ The initial deviations are $\pm 0.001^\circ.$

After $n$ steps a deviation has magnitude $0.001 \cdot 2^n$ degrees. The triangle first becomes obtuse when this exceeds $30^\circ,$ i.e. $2^n \gt 30000.$ Since $2^{14} = 16384$ and $2^{15} = 32768,$ the least such $n$ is $15.$

Thus, the correct answer is E.