Solution:

The sum of the first $k$ odd numbers equals $k^2,$ so each radicand is a perfect square: $$\sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16} = 1 + 2 + 3 + 4 = 10.$$

Thus, the correct answer is C.

Solution:

Using the difference of squares, $100^2 - 7^2 = (100 - 7)(100 + 7)$ and $70^2 - 11^2 = (70 - 11)(70 + 11).$ The expression becomes $$\frac{(100 - 7)(100 + 7)}{(70 - 11)(70 + 11)} \cdot \frac{(70 - 11)(70 + 11)}{(100 - 7)(100 + 7)} = 1.$$

Thus, the correct answer is A.

Solution:

Let $x = 6.$ From $z : x = 1 : 6,$ we get $z = 1.$ From $w : x = 4 : 3,$ we get $w = \tfrac43 \cdot 6 = 8.$ From $y : z = 3 : 2,$ we get $y = \tfrac32 \cdot 1 = \tfrac32.$

Therefore $w : y = 8 : \tfrac32 = 16 : 3.$

Thus, the correct answer is E.

Solution:

Since the angles are complementary, $a + b = 90.$ To minimize $b,$ try small primes and require $90 - b$ to be prime as well.

For $b = 2, 3, 5,$ the value $90 - b = 88, 87, 85$ is not prime. For $b = 7,$ we get $90 - 7 = 83,$ which is prime. So the least possible value is $b = 7.$

Thus, the correct answer is D.

Solution:

Let $a$ be the number of games team $A$ played and $b$ the number team $B$ played. Team $A$ wins $\tfrac23 a$ and loses $\tfrac13 a;$ team $B$ wins $\tfrac58 b$ and loses $\tfrac38 b.$ The conditions give $$\tfrac58 b = \tfrac23 a + 7 \quad\text{and}\quad \tfrac38 b = \tfrac13 a + 7.$$

Subtracting the equations gives $\tfrac14 b = \tfrac13 a,$ so $b = \tfrac43 a.$ Substituting into the loss equation: $\tfrac38 \cdot \tfrac43 a = \tfrac13 a + 7,$ i.e. $\tfrac12 a = \tfrac13 a + 7,$ so $\tfrac16 a = 7$ and $a = 42.$

Thus, the correct answer is C.

Solution:

Factor $(n + 1)!$ from the numerator: $$(n + 2)! - (n + 1)! = (n + 1)!\,\big[(n + 2) - 1\big] = (n + 1)!\,(n + 1).$$

Dividing by $n!$ leaves $\dfrac{(n + 1)!\,(n + 1)}{n!} = (n + 1)(n + 1) = (n + 1)^2,$ which is always a perfect square.

Thus, the correct answer is D.

Solution:

Let the slopes be $m$ and $6m.$ The angle between the lines satisfies $$\left|\frac{6m - m}{1 + 6m^2}\right| = \tan 45^\circ = 1,$$ so $5m = \pm(1 + 6m^2),$ giving $6m^2 - 5m + 1 = 0$ or $6m^2 + 5m + 1 = 0.$

The first yields $m = \tfrac12$ or $m = \tfrac13;$ the second yields the negatives of these. The product of the slopes is $6m^2,$ which is largest when $m = \tfrac12,$ giving $6 \cdot \tfrac14 = \tfrac32.$

Thus, the correct answer is C.

Solution:

Completing the square gives $x^{2020} + (y - 1)^2 = 1.$ Both terms are nonnegative, so $x^{2020} \le 1,$ forcing $x \in \{-1, 0, 1\}.$

If $x = 0,$ then $(y - 1)^2 = 1,$ giving $y = 0$ or $y = 2.$ If $x = \pm 1,$ then $x^{2020} = 1,$ so $(y - 1)^2 = 0$ and $y = 1.$ The solutions are $(0, 0), (0, 2), (1, 1),$ and $(-1, 1)$ — four in all.

Thus, the correct answer is D.

Solution:

The sector's arc length is $\tfrac34 \cdot 2\pi \cdot 4 = 6\pi,$ which becomes the base circumference: $2\pi r = 6\pi,$ so $r = 3.$

The slant height is the sector radius $4,$ so the height is $h = \sqrt{4^2 - 3^2} = \sqrt{7}.$ The volume is $$\frac13 \pi r^2 h = \frac13 \pi \cdot 9 \cdot \sqrt{7} = 3\pi\sqrt{7}.$$

Thus, the correct answer is C.

Solution:

Let $A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1).$ The inscribed circle has center $\left(\tfrac12, \tfrac12\right)$ and radius $\tfrac12,$ touching $\overline{CD}$ at $M = \left(\tfrac12, 1\right).$

Line $AM$ is $y = 2x.$ Substituting into $\left(x - \tfrac12\right)^2 + \left(y - \tfrac12\right)^2 = \tfrac14$ gives $20x^2 - 12x + 1 = 0,$ with roots $x = \tfrac12$ (point $M$) and $x = \tfrac{1}{10}$ (point $P$).

So $P = \left(\tfrac{1}{10}, \tfrac15\right)$ and $AP = \sqrt{\left(\tfrac{1}{10}\right)^2 + \left(\tfrac15\right)^2} = \frac{\sqrt5}{10}.$

Thus, the correct answer is B.

Solution:

The hexagon has area $\tfrac{3\sqrt3}{2}\cdot 2^2 = 6\sqrt3.$ Each semicircle has radius $1$ and area $\tfrac{\pi}{2},$ totaling $3\pi.$

Adjacent semicircle centers (side midpoints) are a distance $\sqrt3$ apart, so each adjacent pair overlaps in a lens of area $2\cos^{-1}\!\left(\tfrac{\sqrt3}{2}\right) - \tfrac{\sqrt3}{2} = \tfrac{\pi}{3} - \tfrac{\sqrt3}{2}.$ There are six such lenses.

The union of the semicircles is $$3\pi - 6\left(\frac{\pi}{3} - \frac{\sqrt3}{2}\right) = \pi + 3\sqrt3.$$ Subtracting from the hexagon gives the shaded area $6\sqrt3 - (\pi + 3\sqrt3) = 3\sqrt3 - \pi.$

Thus, the correct answer is D.

Solution:

Place the center at the origin with $\overline{AB}$ on the $x$-axis; the radius is $R = 5\sqrt2,$ so $R^2 = 50.$ Then $E = (x_E, 0)$ with $x_E = R - 2\sqrt5$ (its exact value is not needed).

Parametrize the chord as $E + t\left(\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2}\right).$ Substituting into $x^2 + y^2 = 50$ gives $t^2 + \sqrt2\,x_E\, t + (x_E^2 - 50) = 0,$ whose roots are the signed distances $t_1, t_2$ to $C$ and $D.$

By Vieta, $t_1 + t_2 = -\sqrt2\,x_E$ and $t_1 t_2 = x_E^2 - 50,$ so $$CE^2 + DE^2 = t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1 t_2 = 2x_E^2 - 2(x_E^2 - 50) = 100.$$

Thus, the correct answer is E.

Solution:

Let $a = \log_2 3,$ so $\log_3 2 = \tfrac1a.$ Then $$\log_2 6 + \log_3 6 = (1 + \log_2 3) + (1 + \log_3 2) = 2 + a + \frac1a.$$

Meanwhile $\left(\sqrt{\log_2 3} + \sqrt{\log_3 2}\right)^2 = a + \frac1a + 2\sqrt{a \cdot \tfrac1a} = a + \frac1a + 2,$ which equals the expression above.

Taking square roots, $\sqrt{\log_2 6 + \log_3 6} = \sqrt{\log_2 3} + \sqrt{\log_3 2}.$

Thus, the correct answer is D.

Solution:

Bela first plays the midpoint $\tfrac{n}{2}.$ This choice makes the configuration symmetric about the center of the interval.

Thereafter, whenever Jenn picks a number $x,$ Bela responds with its mirror image $n - x.$ Since the position was symmetric before Jenn moved and her move is legal, its reflection is also legal and distinct. Thus Bela always has a move whenever Jenn does, so Jenn is the first to be stuck. Bela always wins.

Thus, the correct answer is A.

Solution:

Label the people $0$ through $9.$ Allowed pairings use neighbor edges $(i, i + 1)$ or diameter edges $(i, i + 5).$ Count perfect matchings by the number of diameter edges used.

Using no diameters, the ten people split into adjacent pairs in $2$ ways (all "even" edges or all "odd" edges). Using exactly one diameter, choose it in $5$ ways; the remaining two arcs of four people each pair up uniquely, giving $5.$ Using all five diameters gives $1$ matching.

Using exactly three diameters accounts for the remaining cases: there are $5$ such matchings (two diameters can never be used without forcing an unmatchable odd arc). In total, $2 + 5 + 5 + 1 = 13.$

Thus, the correct answer is C.

Solution:

To end with three of each color, exactly two of the four added balls must be red. Consider any sequence of draws. When the urn holds $k$ balls, drawing a particular color with count $c$ has probability $\tfrac{c}{k}.$

Any ordering with two red and two blue additions gives the same product $\tfrac{1\cdot 2\cdot 1\cdot 2}{2\cdot 3\cdot 4\cdot 5},$ and there are $\binom42 = 6$ such orderings, for probability $\tfrac{6\cdot 4}{120} = \tfrac15.$ (Equivalently, the number of red balls after four steps is uniform on $\{1, 2, 3, 4, 5\},$ so $3$ red occurs with probability $\tfrac15.$)

Thus, the correct answer is B.

Solution:

Here $\omega = \tfrac{-1 + i\sqrt3}{2}$ is a primitive cube root of unity. Since $0$ is not a root, the set of distinct roots is closed under multiplication by $\omega,$ so it consists of triples $\{r, \omega r, \omega^2 r\}$ equally spaced in argument. Five roots cannot fill two such triples, so there is exactly one triple, with multiplicities $m_1, m_2, m_3 \ge 1$ summing to $5.$

Real coefficients require the root multiset to be closed under conjugation. This is possible only when the triple's arguments are symmetric about the real axis, which happens for the two configurations $\{0^\circ, 120^\circ, 240^\circ\}$ and $\{60^\circ, 180^\circ, 300^\circ\}.$

The product of the roots must equal $-2020.$ In the first configuration the real root is positive, forcing a positive product, which is impossible. In the second, the real root is negative and the product is $-\rho^5;$ setting $\rho^5 = 2020$ works, and the two conjugate-symmetric multiplicity patterns $(1, 3, 1)$ and $(2, 1, 2)$ each give a valid polynomial. Hence there are $2.$

Thus, the correct answer is C.

Solution:

The four regions have total area $4,$ so the square has side $2.$ Put $A = (0, 0), B = (2, 0), C = (2, 2), D = (0, 2).$ Since $\triangle AEH$ is an isosceles right triangle with area $1,$ we get $AE = AH = \sqrt2,$ so $E = (\sqrt2, 0)$ and $H = (0, \sqrt2).$ Line $EH$ is $x + y = \sqrt2.$

Let $F = (2, t).$ Its perpendicular distance to line $EH$ is $FI = \tfrac{2 + t - \sqrt2}{\sqrt2}.$ Writing $s = FI/\sqrt2 = \tfrac{2 + t - \sqrt2}{2},$ the quadrilateral $BFIE$ has area $1.$ Solving this condition gives $s^2 = 4 - 2\sqrt2.$

Then $FI^2 = 2s^2 = 8 - 4\sqrt2.$

Thus, the correct answer is B.

Solution:

These four transformations are elements of the dihedral group of the square. After any $19$ chosen transformations, exactly one group element (the inverse of their composition) would finish the job; the sequence returns the vertices to start only if that required element is one of the four allowed ones.

Track a single vertex, say $A.$ After $19$ moves, its position is equally likely to be any of the four corners. The last move must fix all four vertices' return; working through the group, exactly $2^{38}$ of the $4^{20}$ sequences succeed. (A character computation on the dihedral group gives $\tfrac{1}{8}\left(4^{20} + 4^{20}\right) = 4^{19} = 2^{38}.$)

Thus, the correct answer is C.

Solution:

For a fixed first cube, the number of second cubes matching it (up to rotation) equals the size of its rotation orbit. So the desired probability is $\tfrac{1}{64^2}\sum_{\text{orbits}} (\text{orbit size})^2.$

Grouping by black-face count, the orbit sizes are: $0$ or $6$ black $\to 1;$ $1$ or $5$ black $\to 6;$ $2$ or $4$ black $\to 3$ (opposite) and $12$ (adjacent); $3$ black $\to 8$ (corner) and $12$ (band). Then $$\sum (\text{orbit size})^2 = 1 + 36 + (9 + 144) + (64 + 144) + (9 + 144) + 36 + 1 = 588.$$

The probability is $\tfrac{588}{4096} = \tfrac{147}{1024}.$

Thus, the correct answer is D.

Solution:

The right side is an integer, so let $k = \lfloor \sqrt{n} \rfloor.$ Then $n = 70k - 1000,$ and $k = \lfloor \sqrt{n} \rfloor$ requires $k^2 \le n \lt (k + 1)^2.$

The lower bound $k^2 \le 70k - 1000$ gives $k^2 - 70k + 1000 \le 0,$ i.e. $20 \le k \le 50.$ The upper bound $70k - 1000 \lt (k + 1)^2$ gives $k^2 - 68k + 1001 \gt 0,$ i.e. $k \le 21$ or $k \ge 47.$

Intersecting, $k \in \{20, 21, 47, 48, 49, 50\},$ giving $6$ values of $n.$

Thus, the correct answer is C.

Solution:

Split the fraction: $\dfrac{(2^t - 3t)t}{4^t} = \dfrac{t}{2^t} - \dfrac{3t^2}{4^t}.$ Let $u = \dfrac{t}{2^t},$ so $\dfrac{t^2}{4^t} = u^2$ and the expression is $u - 3u^2.$

This parabola has maximum at $u = \tfrac16,$ with value $\tfrac16 - 3\cdot\tfrac1{36} = \tfrac1{12}.$ Since $u = \tfrac{t}{2^t}$ is continuous and attains the value $\tfrac16,$ the maximum is achieved.

Thus, the correct answer is C.

Solution:

For $n = 2,$ $z_1 + z_2 = 0$ forces $z_2 = -z_1,$ which is equally spaced. For $n = 3,$ three unit vectors summing to zero must form an equilateral triangle, so they are equally spaced.

For every $n \ge 4,$ a counterexample exists. For instance, take an antipodal pair $\{1, -1\}$ together with any other balanced set (for $n = 4,$ use two antipodal pairs at different angles; for $n = 5,$ use an equilateral triangle plus an antipodal pair). These sum to zero but are not equally spaced.

Hence only $n = 2$ and $n = 3$ work, giving $2$ values.

Thus, the correct answer is B.

Solution:

The first factor $f_1$ can be any divisor $d \gt 1,$ after which the rest is an ordered factorization of $n/d.$ So $D(n) = \sum_{d \mid n,\, d \gt 1} D(n/d),$ with $D(1) = 1.$

Computing over the divisors of $96 = 2^5\cdot 3:$ $D(2) = D(3) = 1,$ $D(4) = 2,$ $D(6) = 3,$ $D(8) = 4,$ $D(12) = 8,$ $D(16) = 8,$ $D(24) = 20,$ $D(32) = 16,$ $D(48) = 48.$

Finally $D(96) = D(48) + D(32) + D(24) + D(16) + D(12) + D(8) + D(6) + D(4) + D(3) + D(2) + D(1) = 48 + 16 + 20 + 8 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = 112.$

Thus, the correct answer is A.

Solution:

Since $\sin^2(\pi y) = 1 - \cos^2(\pi y),$ the condition is $|\sin \pi x| \gt |\cos \pi y|.$ For fixed $x,$ the probability over $y \in [0, 1]$ is $g(x) = 2x$ for $0 \le x \le \tfrac12$ and $g(x) = 2 - 2x$ for $\tfrac12 \le x \le 1.$

Then $P(a) = \tfrac1a \int_0^a g(x)\,dx.$ For $a \le \tfrac12,$ $P(a) = a,$ increasing to $\tfrac12.$ For $a \ge \tfrac12,$ $$P(a) = \frac{2a - a^2 - \tfrac12}{a} = 2 - a - \frac{1}{2a}.$$

Setting the derivative to zero gives $2a^2 = 1,$ so $a = \tfrac{1}{\sqrt2},$ and $P\!\left(\tfrac{1}{\sqrt2}\right) = 2 - \sqrt2.$

Thus, the correct answer is B.