Solution:

The target person has $1013 - 1 = 1012$ people to the left and $1010 - 1 = 1009$ people to the right. Including the person themself, the line has $1012 + 1009 + 1 = 2022$ people.

Thus, the correct answer is B.

Solution:

Note that $6! = 720 = 8 \cdot 9 \cdot 10.$ Therefore $$7! \cdot 6! = 7! \cdot (8 \cdot 9 \cdot 10) = 10!.$$ So $10! - 7! \cdot 6! = 10! - 10! = 0.$

Thus, the correct answer is B.

Solution:

The inequality $|2x| \le 7\pi$ is equivalent to $|x| \le \dfrac{7\pi}{2} \approx 10.99.$ The integers satisfying this run from $-10$ to $10,$ which is $10 + 10 + 1 = 21$ values.

Thus, the correct answer is E.

Solution:

Step $k$ deposits $k$ balls, so after step $k$ a total of $\dfrac{k(k+1)}{2}$ balls have been placed. Since $\dfrac{63 \cdot 64}{2} = 2016$ and $\dfrac{64 \cdot 65}{2} = 2080,$ ball $2024$ falls in step $64.$

The steps cycle through the bins $A, B, C, D, E,$ so step $k$ uses position $(k-1) \bmod 5.$ Here $(64 - 1) \bmod 5 = 63 \bmod 5 = 3,$ which is the fourth bin, $D.$

Thus, the correct answer is D.

Solution:

The original expression sums the first $50$ odd numbers, giving $50^2 = 2500.$ Changing a term of value $v$ from $+$ to $-$ decreases the total by $2v,$ so to make the result negative the flipped terms must total more than $\dfrac{2500}{2} = 1250.$

To use as few terms as possible, flip the largest odd numbers $99, 97, 95, \ldots$ The largest $k$ of them sum to $k(100 - k).$ With $k = 14$ this is $14 \cdot 86 = 1204 \le 1250,$ but with $k = 15$ it is $15 \cdot 85 = 1275 \gt 1250.$ So $15$ sign changes suffice and $14$ do not.

Thus, the correct answer is B.

Solution:

The number of digits of $N$ in base $5$ is $\lfloor \log_5 N \rfloor + 1.$ With $N = 5 \cdot 10^{13},$ $$\log_{10} N = 13 + \log_{10} 5 = 13.7.$$ Converting bases, $\log_5 N = \dfrac{13.7}{\log_{10} 5} = \dfrac{13.7}{0.7} = 19.57\ldots$

Thus the number of digits is $\lfloor 19.57 \rfloor + 1 = 19 + 1 = 20.$

Thus, the correct answer is B.

Solution:

Set $X = (0,0),$ $W = (0,4),$ $Y = (8,0),$ $Z = (8,4),$ with $M = (m, 0)$ on $\overline{XY}$ and $A = (8, a)$ on $\overline{YZ}.$

Since $\angle WMA = 90^\circ,$ $\overrightarrow{MW} \cdot \overrightarrow{MA} = (-m)(8-m) + 4a = 0,$ so $4a = m(8-m).$ The areas give $[\triangle WXM] = \tfrac12 \cdot 4 \cdot m = 2m$ and $[\triangle WAZ] = \tfrac12 \cdot 8 \cdot (4 - a) = 4(4 - a).$ Setting these equal yields $m = 8 - 2a.$

Substituting $a = \tfrac{8-m}{2}$ into $4a = m(8-m)$ gives $2(8-m) = m(8-m),$ so $m = 2$ and $a = 3.$ Then with $W = (0,4),$ $M = (2,0),$ $A = (8,3),$ $$[\triangle WMA] = \tfrac12\,\bigl|2(3 - 4) + 8(4 - 0)\bigr| = \tfrac12 (30) = 15.$$

Thus, the correct answer is C.

Solution:

Dividing top and bottom by $\log_2 x \cdot \log_3 x,$ the left side becomes $$\frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x}} = \frac{1}{\log_x 2 + \log_x 3} = \frac{1}{\log_x 6}.$$ So $\dfrac{1}{\log_x 6} = 2,$ meaning $\log_x 6 = \dfrac12,$ i.e. $x^{1/2} = 6.$

Therefore $x = 36.$

Thus, the correct answer is C.

Solution:

The dartboard $|x| + |y| \le 8$ is a square with diagonals $16,$ so its area is $\tfrac12 \cdot 16 \cdot 16 = 128.$ The target condition $(x^2 + y^2 - 25)^2 \le 49$ means $-7 \le x^2 + y^2 - 25 \le 7,$ i.e. $18 \le x^2 + y^2 \le 32,$ an annulus of area $\pi(32 - 18) = 14\pi.$

The distance from the origin to a side of the square (for instance $x + y = 8$) is $\dfrac{8}{\sqrt2} = \sqrt{32},$ exactly the annulus's outer radius. So the annulus is tangent to the square and lies entirely within $B.$ The probability is $\dfrac{14\pi}{128} = \dfrac{7}{64}\pi,$ giving $m + n = 7 + 64 = 71.$

Thus, the correct answer is B.

Solution:

The six fixed numbers sum to $24.8.$ The mean $\dfrac{24.8 + x + y + z}{9}$ is an integer exactly when $x + y + z$ has fractional part $0.2.$ The fixed numbers span $[1, 7],$ so to make the range $7$ the extremes must be pushed apart by one more unit.

Checking the possibilities gives exactly three valid triples:

$(x, y, z) = (0, 5, 6.2)$ with mean $4$ and median $5;$ $(x, y, z) = (0.1, 4, 7.1)$ with mean $4$ and median $4;$ and $(x, y, z) = (6, 6.2, 8)$ with mean $5$ and median $6.$ Each has range $7$ and integer mean and median.

Thus, the correct answer is C.

Solution:

Using $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2},$ $$\sum_{n=1}^{90} \sin^2(n^\circ) = \frac{90}{2} - \frac12 \sum_{n=1}^{90}\cos(2n^\circ).$$ In the cosine sum, the terms for $n$ and $90 - n$ satisfy $\cos(2n^\circ) + \cos(180^\circ - 2n^\circ) = 0,$ and $\cos 90^\circ = 0,$ so everything cancels except $\cos 180^\circ = -1.$

Hence the sum is $45 - \tfrac12(-1) = 45.5,$ and the mean is $\dfrac{45.5}{90} = \dfrac{91}{180}.$

Thus, the correct answer is E.

Solution:

For vertices $0, z, z^2, z^3$ the shoelace formula gives area $$\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\,z^3)\bigr| = \tfrac12\,\bigl|\operatorname{Im}\bigl((|z|^2 + |z|^4)z\bigr)\bigr| = \tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z).$$

With $|z| = 2,$ this is $\tfrac12(4 + 16)\operatorname{Im}(z) = 10\operatorname{Im}(z).$ Setting $10\operatorname{Im}(z) = 15$ gives $\operatorname{Im}(z) = \dfrac32.$ (Then $\operatorname{Re}(z) = \sqrt{4 - \tfrac94} = \tfrac{\sqrt7}{2} \gt 1,$ as required.)

Thus, the correct answer is D.

Solution:

Adding the equations, $$h + k = 2x^2 + 2y^2 - 16x - 4y = 2(x - 4)^2 + 2(y - 1)^2 - 34.$$ Both squared terms are nonnegative, so the minimum occurs at $x = 4,$ $y = 1,$ giving $h + k = -34.$

Thus, the correct answer is C.

Solution:

If $n$ is coprime to $5,$ then since $\varphi(125) = 100,$ Euler's theorem gives $n^{100} \equiv 1 \pmod{125}.$ If $n$ is a multiple of $5,$ then $n^{100}$ is divisible by $5^{100},$ hence by $125,$ leaving remainder $0.$

So the only possible remainders are $0$ and $1,$ which is $2$ distinct values.

Thus, the correct answer is B.

Solution:

The vertices are $A = (0, 1),$ $B = (\log_2 3, 2),$ $C = (\log_2 7, 3).$ By the shoelace formula, $$[\triangle ABC] = \tfrac12\,\bigl|0(2 - 3) + \log_2 3\,(3 - 1) + \log_2 7\,(1 - 2)\bigr| = \tfrac12\,\bigl|2\log_2 3 - \log_2 7\bigr|.$$

This equals $\tfrac12\log_2 \dfrac{9}{7} = \log_2 \sqrt{\tfrac{9}{7}} = \log_2 \dfrac{3}{\sqrt7}.$

Thus, the correct answer is B.

Solution:

The number of ways to split $16$ people into $4$ indistinguishable groups of $4$ is $\dfrac{16!}{(4!)^4\, 4!}.$ Each committee then chooses a chairperson and a secretary in $4 \cdot 3 = 12$ ways, contributing $12^4.$ So the total is $\dfrac{16!}{(4!)^4\,4!}\cdot 12^4.$

Counting factors of $3:$ $16!$ contributes $\lfloor 16/3\rfloor + \lfloor 16/9\rfloor = 6.$ The denominator $(4!)^4\,4!$ contributes $4 + 1 = 5.$ And $12^4 = (2^2\cdot 3)^4$ contributes $4.$ Thus $r = 6 - 5 + 4 = 5.$

Thus, the correct answer is A.

Solution:

The set has $21$ integers, so there are $21 \cdot 20 = 420$ ordered choices of $(a, b).$ If the polynomial has distinct integer roots $p, q, r,$ then $pqr = -6,$ $a = -(p + q + r),$ and $b = pq + qr + rp.$

The triples of distinct integers with product $-6$ are $\{1, 2, -3\},$ $\{1, -2, 3\},$ $\{-1, 2, 3\},$ $\{-1, -2, -3\},$ and $\{1, -1, 6\}.$ These give $(a, b) = (0, -7),$ $(-2, -5),$ $(-4, 1),$ $(6, 11),$ and $(-6, -1).$ The fourth has $b = 11 \gt 10,$ so it is invalid; the other four are valid and distinct.

The probability is $\dfrac{4}{420} = \dfrac{1}{105}.$

Thus, the correct answer is C.

Solution:

Since $F_{2k} = F_k L_k$ where $L_k$ is the $k$th Lucas number, each term $\dfrac{F_{2k}}{F_k} = L_k.$ The sum is $$L_1 + L_2 + \cdots + L_{10} = 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319.$$ (Equivalently, $L_1 + \cdots + L_{10} = L_{12} - 3 = 322 - 3 = 319.$)

Thus, the correct answer is B.

Solution:

The six vertices lie on the circumcircle of radius $R = \dfrac{14}{\sqrt3},$ so $R^2 = \dfrac{196}{3}.$ Going around, the central angles alternate between $\theta$ (three times) and $120^\circ - \theta$ (three times). The cyclic-hexagon area is $$\tfrac12 R^2\bigl(3\sin\theta + 3\sin(120^\circ - \theta)\bigr) = 98\bigl(\sin\theta + \sin(120^\circ - \theta)\bigr).$$

By sum-to-product, $\sin\theta + \sin(120^\circ - \theta) = 2\sin 60^\circ\cos(\theta - 60^\circ) = \sqrt3\cos(60^\circ - \theta).$ Setting the area to $91\sqrt3$ gives $\sqrt3\cos(60^\circ - \theta) = \dfrac{91\sqrt3}{98} = \dfrac{13\sqrt3}{14},$ so $\cos(60^\circ - \theta) = \dfrac{13}{14}$ and $\sin(60^\circ - \theta) = \dfrac{3\sqrt3}{14}.$

Then $\tan(60^\circ - \theta) = \dfrac{3\sqrt3}{13},$ and $$\tan\theta = \tan\bigl(60^\circ - (60^\circ - \theta)\bigr) = \frac{\sqrt3 - \frac{3\sqrt3}{13}}{1 + \sqrt3\cdot\frac{3\sqrt3}{13}} = \frac{\frac{10\sqrt3}{13}}{\frac{22}{13}} = \frac{5\sqrt3}{11}.$$

Thus, the correct answer is B.

Solution:

Let $a = BC.$ The median length gives $x^2 = \dfrac{2\cdot 1600 + 2\cdot 1764 - a^2}{4} = \dfrac{6728 - a^2}{4}.$ The triangle inequality requires $2 \lt a \lt 82,$ i.e. $4 \lt a^2 \lt 6724,$ which translates to $1 \lt x \lt 41.$ So $(p, q) = (1, 41).$

With $AB = 40$ and $AC = 42$ fixed, the area $\tfrac12\cdot 40\cdot 42\sin A$ is largest when $\angle A = 90^\circ,$ giving $r = 840.$ Then $a^2 = 40^2 + 42^2 = 3364,$ so $x^2 = \dfrac{6728 - 3364}{4} = 841,$ i.e. $s = 29.$

Thus $p + q + r + s = 1 + 41 + 840 + 29 = 911.$

Thus, the correct answer is C.

Solution:

The smallest angles $\alpha, \beta$ of the $3$-$4$-$5$ and $5$-$12$-$13$ triangles have $\tan\alpha = \tfrac34$ and $\tan\beta = \tfrac{5}{12}.$ By the tangent addition formula, $$\tan(\alpha + \beta) = \frac{\tfrac34 + \tfrac{5}{12}}{1 - \tfrac34\cdot\tfrac{5}{12}} = \frac{\tfrac{14}{12}}{\tfrac{33}{48}} = \frac{56}{33}.$$

The third smallest angle $\gamma$ satisfies $\gamma = 90^\circ - (\alpha + \beta),$ so $\tan\gamma = \dfrac{33}{56}.$ The right triangle with legs $33$ and $56$ has hypotenuse $\sqrt{33^2 + 56^2} = \sqrt{4225} = 65,$ a primitive triple. Its perimeter is $33 + 56 + 65 = 154.$

Thus, the correct answer is C.

Solution:

When $\angle B = 2\angle A,$ the side lengths satisfy $b^2 = a(a + c),$ where $a = BC,$ $b = CA,$ $c = AB.$ So $c = \dfrac{b^2 - a^2}{a}$ must be a positive integer, and the sides must form a valid triangle.

Trying small values, $a = 4,$ $b = 6$ gives $c = \dfrac{36 - 16}{4} = 5,$ and the sides $4, 5, 6$ form a valid triangle with $6^2 = 4(4 + 5).$ Its perimeter is $15,$ and a search shows no smaller perimeter works.

Thus, the correct answer is C.

Solution:

Let $R$ be the circumradius of the octagon and $L$ the length of each lateral edge, so $L^2 = h^2 + R^2.$ Since $\angle AVD = 90^\circ,$ $AD^2 = 2L^2.$

Vertices $A$ and $D$ are three steps apart, a central angle of $135^\circ,$ so $AD^2 = 2R^2(1 - \cos 135^\circ) = R^2(2 + \sqrt2).$ Setting $R^2(2 + \sqrt2) = 2(h^2 + R^2)$ gives $2h^2 = R^2\sqrt2.$

For a regular octagon of side $1,$ $R^2 = \dfrac{1}{2\sin^2(22.5^\circ)} = \dfrac{2 + \sqrt2}{2}.$ Therefore $h^2 = \dfrac{R^2\sqrt2}{2} = \dfrac{(2 + \sqrt2)\sqrt2}{4} = \dfrac{2\sqrt2 + 2}{4} = \dfrac{1 + \sqrt2}{2}.$

Thus, the correct answer is B.

Solution:

Writing each side as $\dfrac{2[\triangle]}{h},$ the semiperimeter is $[\triangle]\bigl(\tfrac1a + \tfrac1b + \tfrac1c\bigr),$ so the inradius $r = \dfrac{[\triangle]}{s}$ satisfies $\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c.$ We need this to be $\dfrac1r$ for a positive integer $r,$ with the sides (proportional to $\tfrac1a, \tfrac1b, \tfrac1c$) forming a non-degenerate triangle, requiring $\tfrac1a \lt \tfrac1b + \tfrac1c.$

Searching $a \le b \le c \le 9,$ the triples with $\tfrac1a + \tfrac1b + \tfrac1c = \tfrac1r$ that also satisfy the triangle inequality are exactly the equilateral ones: $(3,3,3)$ with $r = 1,$ $(6,6,6)$ with $r = 2,$ and $(9,9,9)$ with $r = 3.$ Other solutions such as $(2,3,6)$ or $(4,8,8)$ give degenerate triangles. So there are $3$ triples.