Answer: D

Solution:

Since $3\pi \approx 9.42,$ the inequality $|x| \lt 3\pi$ means $-9.42 \lt x \lt 9.42.$

The integers in this range run from $-9$ to $9,$ giving $19$ values.

Thus, the correct answer is D.

Answer: B

Solution:

The $23$ all-blue pairs account for $46$ blue students, leaving $57 - 46 = 11$ blue students.

Each of those $11$ blue students must be paired with a yellow student, so there are $11$ mixed pairs, using $11$ yellow students.

The remaining $75 - 11 = 64$ yellow students form $64 \div 2 = 32$ all-yellow pairs.

Thus, the correct answer is B.

Answer: A

Solution:

Working from the outside in, $\dfrac{144}{53}-2=\dfrac{38}{53},$ so the inner fraction equals $\dfrac{38}{53}.$

Its reciprocal gives $1+\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{53}{38},$ so $\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{15}{38}.$

Then $2+\dfrac{2}{3+x}=\dfrac{38}{15},$ so $\dfrac{2}{3+x}=\dfrac{8}{15},$ giving $3+x=\dfrac{15}{4}.$

Therefore $x=\dfrac{15}{4}-3=\dfrac{3}{4}.$

Thus, the correct answer is A.

Answer: C

Solution:

Suppose there are $3$ students in the morning class and $4$ in the afternoon class.

The total of all scores is $3\cdot 84+4\cdot 70=252+280=532.$

The overall mean is $\dfrac{532}{7}=76.$

Thus, the correct answer is C.

Answer: D

Solution:

A $90^\circ$ counterclockwise rotation about $(1,5)$ sends $(a,b)$ to $(1-(b-5),\,5+(a-1))=(6-b,\,4+a).$

Reflecting that about $y=-x$ (which maps $(x,y)$ to $(-y,-x)$) gives $(-(4+a),\,-(6-b))=(-4-a,\,b-6).$

Setting this equal to $(-6,3)$ gives $-4-a=-6$ and $b-6=3,$ so $a=2$ and $b=9.$

Therefore $b-a=9-2=7.$

Thus, the correct answer is D.

Answer: A

Solution:

The cone holds $\dfrac13\pi(12)^2(18)=864\pi$ cubic centimeters of water.

Poured into the cylinder, this fills to height $h$ where $\pi(24)^2 h=864\pi.$

Then $576h=864,$ so $h=1.5.$

Thus, the correct answer is A.

Answer: C

Solution:

Factoring, $34=2\cdot 17,$ $63=3^2\cdot 7,$ and $270=2\cdot 3^3\cdot 5,$ so $N=2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2.$

Let $M$ be the odd part $3^5\cdot 5\cdot 7\cdot 17^2.$ The sum of all divisors is $(1+2+4+8)\,\sigma(M)=15\,\sigma(M).$

The odd divisors sum to $\sigma(M),$ so the even divisors sum to $15\,\sigma(M)-\sigma(M)=14\,\sigma(M).$

The ratio is $\sigma(M):14\,\sigma(M)=1:14.$

Thus, the correct answer is C.

Answer: B

Solution:

Place the center at height $0.$ Two equal chords lie at equal distances from the center, so the three equally spaced lines are at heights $-\tfrac{d}{2},\tfrac{d}{2},\tfrac{3d}{2},$ with the two $38$-chords at $\pm\tfrac{d}{2}$ and the $34$-chord at $\tfrac{3d}{2}.$

Half-chord relations give $r^2-\left(\tfrac{d}{2}\right)^2=19^2$ and $r^2-\left(\tfrac{3d}{2}\right)^2=17^2.$

Subtracting, $2d^2=19^2-17^2=72,$ so $d^2=36$ and $d=6.$

Thus, the correct answer is B.

Answer: D

Solution:

Using $\dfrac{1}{\log_{40}2}=\log_2 40$ and $\dfrac{1}{\log_{20}2}=\log_2 20,$ the expression becomes $(\log_2 80)(\log_2 40)-(\log_2 160)(\log_2 20).$

Let $t=\log_2 5.$ Then $\log_2 80=4+t,$ $\log_2 40=3+t,$ $\log_2 160=5+t,$ $\log_2 20=2+t.$

The value is $(4+t)(3+t)-(5+t)(2+t)=(12+7t+t^2)-(10+7t+t^2)=2.$

Thus, the correct answer is D.

Answer: E

Solution:

The sum $1+2+\cdots+37=703.$ If the chosen numbers are $a$ and $b,$ then $703-a-b=ab.$

So $ab+a+b=703,$ and adding $1$ gives $(a+1)(b+1)=704=2^6\cdot 11.$

We need factors $a+1,b+1$ between $2$ and $38.$ The pair $22\cdot 32=704$ works, giving $a=21,$ $b=31.$

Their difference is $31-21=10.$

Thus, the correct answer is E.

Answer: D

Solution:

Place $A=(0,0)$ and $C=(15,0).$ Then $B=\left(\tfrac{33}{5},\tfrac{56}{5}\right),$ and since $PC=10,$ $P=(5,0).$ Line $BP$ has slope $7,$ so it is $y=7(x-5).$

For $ABCD$ to be a trapezoid with $D$ on line $BP,$ take $CD\parallel AB.$ The line through $C$ parallel to $AB$ meets line $BP$ at $(1.8,-22.4).$

For $ABCE$ with $E$ on line $BP,$ take $AE\parallel BC.$ The line through $A$ parallel to $BC$ meets line $BP$ at $(4.2,-5.6).$

The distance is $\sqrt{(4.2-1.8)^2+(-5.6+22.4)^2}=\sqrt{2.4^2+16.8^2}=\sqrt{288}=12\sqrt2.$

Thus, the correct answer is D.

Answer: D

Solution:

Let $n=|S|,$ let $T$ be the total, $M$ the greatest, and $L$ the least. Then $\dfrac{T-M}{n-1}=32,$ $\dfrac{T-M-L}{n-2}=35,$ and $\dfrac{T-L}{n-1}=40.$

Subtracting the first from the third: $\dfrac{M-L}{n-1}=8.$ Since $M-L=72,$ we get $n-1=9,$ so $n=10.$

Then $T-M=288$ and $T-L=360.$ The middle equation gives $T-M-L=35\cdot 8=280,$ so $L=288-280=8$ and $M=80.$

Thus $T=288+80=368,$ and the average is $\dfrac{368}{10}=36.8.$

Thus, the correct answer is D.

Answer: D

Solution:

Let $f(\theta)=1-3\sin\theta+5\cos 3\theta.$ The fast term $5\cos 3\theta$ completes three oscillations while $1-3\sin\theta$ stays between $-2$ and $4.$

Sampling $f$ at $0,30^\circ,60^\circ,\ldots,360^\circ,$ the values are $+,-,-,-,+,-,-,+,+,+,-,+,+,$ which shows six sign changes, hence six roots.

Each sign change corresponds to exactly one solution, so there are $6$ values of $\theta.$

Thus, the correct answer is D.

Answer: A

Solution:

Place $D$ at the origin with $A,$ $C$ along the rectangle's edges and $M$ directly above $D.$ Then $MA^2=AD^2+DM^2,$ $MC^2=CD^2+DM^2,$ and $MB^2=AD^2+CD^2+DM^2.$

Thus $MB^2=MA^2+MC^2-DM^2.$ Writing $MA,MC,MB=k,k+2,k+4,$ we get $DM^2=k^2+(k+2)^2-(k+4)^2=k^2-4k-12.$

This is a positive perfect square only for $k=7,$ giving $DM^2=9,$ so $DM=3,$ $MA=7,$ $MC=9.$ Then $AD^2=49-9=40$ and $CD^2=81-9=72.$

The base area is $AD\cdot CD=\sqrt{40}\cdot\sqrt{72}=\sqrt{2880}=24\sqrt5,$ and the volume is $\tfrac13\cdot 24\sqrt5\cdot 3=24\sqrt5.$

Thus, the correct answer is A.

Answer: D

Solution:

The eleven equal segments form two rhombi (each two equilateral triangles of side $2$) sharing the apex $A,$ with $C$ and $D$ joined by a final segment. The figure is symmetric about the vertical line through $A.$

Placing $A=(0,0)$ at the top, the two bottom vertices come out to $C=(-1,-\sqrt{11})$ and $D=(1,-\sqrt{11}),$ with $B$ and $E$ the outer corners at height $-\tfrac{\sqrt{11}}{2}+\tfrac{1}{2\sqrt3}.$

Applying the shoelace formula to pentagon $ABCDE$ gives area $\sqrt{11}+2\sqrt3=\sqrt{11}+\sqrt{12}.$

So $m+n=11+12=23.$

Thus, the correct answer is D.

Answer: A

Solution:

Let $f$ have roots $r,s,t.$ Since $g$ is monic with roots $\tfrac1r,\tfrac1s,\tfrac1t,$ $g(1)=\left(1-\tfrac1r\right)\left(1-\tfrac1s\right)\left(1-\tfrac1t\right)=\dfrac{(r-1)(s-1)(t-1)}{rst}.$

Now $f(1)=(1-r)(1-s)(1-t)=1+a+b+c,$ so $(r-1)(s-1)(t-1)=-(1+a+b+c).$ Also $rst=-c.$

Therefore $g(1)=\dfrac{-(1+a+b+c)}{-c}=\dfrac{1+a+b+c}{c}.$

Thus, the correct answer is A.

Answer: B

Solution:

Let $AB=a,$ $CD=b,$ and let the interior point be at heights $h_a$ from $AB$ and $h_b$ from $CD.$ The base triangles give $\tfrac12 a h_a=4$ and $\tfrac12 b h_b=2,$ so $a h_a=8$ and $b h_b=4.$

The total area is $2+3+4+5=14=\tfrac12(a+b)(h_a+h_b),$ so $(a+b)(h_a+h_b)=28.$ Expanding, $a h_a+b h_b+a h_b+b h_a=28,$ giving $a h_b+b h_a=16.$

Let $u=a h_b$ and $v=b h_a.$ Then $u+v=16$ and $uv=(a h_a)(b h_b)=32,$ so $u,v=8\pm 4\sqrt2.$

Finally $\dfrac{AB}{CD}=\dfrac{a}{b}=\dfrac{a h_b}{b h_b}=\dfrac{u}{4}=\dfrac{8+4\sqrt2}{4}=2+\sqrt2.$

Thus, the correct answer is B.

Answer: A

Solution:

Let $p=|z|^2=z\bar z$ and $s=z+\bar z.$ Then $|z+2|^2=p+2s+4,$ and $|z^2+1|^2=p^2+(z^2+\bar z^2)+1=p^2+(s^2-2p)+1.$

Substituting, $12p=2(p+2s+4)+p^2+s^2-2p+1+31,$ which simplifies to $p^2-12p+s^2+4s+40=0.$

Completing the square gives $(p-6)^2+(s+2)^2=0,$ so $p=6$ and $s=-2.$

Then $z+\dfrac{6}{z}=z+\dfrac{6\bar z}{|z|^2}=z+\bar z=-2.$

Thus, the correct answer is A.

Answer: B

Solution:

Let the dice have $a\le b$ faces. Since both have at least $6$ faces, a sum of $7$ occurs in exactly $6$ ways, so a sum of $10$ occurs in $6\div\tfrac34=8$ ways.

The number of ways to roll $10$ is $\min(a,9)-\max(1,10-b)+1=8.$ A sum of $12$ has probability $\tfrac{1}{12},$ so it occurs in $\tfrac{ab}{12}$ ways.

Trying $a=8,b=9$: sum $10$ has $8-1+1=8$ ways, and sum $12$ has $6=\tfrac{72}{12}$ ways. Both conditions hold, giving $a+b=17.$

Checking all smaller totals $a+b=16$ fails, so $17$ is minimal.

Thus, the correct answer is B.

Answer: A

Solution:

Since $z^3\equiv 1\pmod{z^2+z+1}$ and $2021=3\cdot 673+2,$ we have $z^{2021}\equiv z^2.$

So $z^{2021}+1\equiv z^2+1.$ Reducing further with $z^2\equiv -z-1,$ this is $-z-1+1=-z.$

Therefore $R(z)=-z.$

Thus, the correct answer is A.

Answer: D

Solution:

Taking $\log_2,$ the equation becomes $2^{\sqrt2}\log_2 x=2^{x-1}.$ Substituting $x=\sqrt2$ gives $2^{\sqrt2}\cdot\tfrac12=2^{\sqrt2-1},$ which holds, so $x=\sqrt2$ is a solution.

Let $f(x)=2^{x-1}-2^{\sqrt2}\log_2 x.$ Then $f(1)\gt 0,$ $f(\sqrt2)=0,$ $f(2)\lt 0,$ and $f(4)\gt 0,$ so there is a second root $x_0$ between $2$ and $4.$

Since $f$ has no other sign changes, there are exactly two solutions, and $S=\sqrt2+x_0\approx 1.41+3.1\approx 4.5,$ which lies in $[2,6).$

Thus, the correct answer is D.

Answer: B

Solution:

Treat each wall as a Nim-like heap with a Grundy value. A move removes $1$ or $2$ adjacent bricks, possibly splitting a wall into two, so $g(n)=\operatorname{mex}$ over all resulting XOR values.

Computing, $g(1)=1,$ $g(2)=2,$ $g(3)=3,$ $g(4)=1,$ $g(5)=4,$ $g(6)=3.$

The second player Beth wins exactly when the XOR of the walls' Grundy values is $0.$ Checking each option, only $(6,2,1)$ gives $g(6)\oplus g(2)\oplus g(1)=3\oplus 2\oplus 1=0.$

Thus, the correct answer is B.

Answer: A

Solution:

Evenly spaced distinct bins form an arithmetic progression $n,n+d,n+2d$ with $n,d\ge 1.$ The three labels sum to $3(n+d),$ so a fixed assignment of balls to these bins has probability $2^{-3(n+d)}.$

The three balls can be ordered in $3!=6$ ways, so the total probability is $6\sum_{n\ge 1}\sum_{d\ge 1}2^{-3(n+d)}=6\left(\sum_{n\ge 1}\tfrac{1}{8^n}\right)^2=6\cdot\tfrac17\cdot\tfrac17=\tfrac{6}{49}.$

Since $\gcd(6,49)=1,$ we get $p+q=6+49=55.$

Thus, the correct answer is A.

Answer: A

Solution:

Let the diagonals meet at $O$ at angle $\theta.$ The feet of the perpendiculars from $A$ and $C$ to $BD$ are symmetric about $O,$ so $PQ=AC\cos\theta=6;$ likewise $RS=BD\cos\theta=8.$