Solution:

The total juice is $3+\tfrac13=\tfrac{10}{3}$ glasses. Split evenly, each glass should hold $\tfrac{10}{3}\div 4=\tfrac{5}{6}$ of a glass. Each of the first three glasses must therefore give up $1-\tfrac{5}{6}=\tfrac{1}{6}.$

Thus, the correct answer is C.

Solution:

The final cost of a pair with original price $p$ is $0.8\times 1.075\times p=0.86p.$ Setting $0.86p\le 43$ gives $p\le 50,$ so the most expensive affordable pair originally cost $50$ dollars.

Thus, the correct answer is B.

Solution:

The hypotenuse of an inscribed right triangle is a diameter, so circle $A$ has diameter $5$ and circle $B$ has diameter $13.$ The ratio of areas is $\left(\tfrac{5}{13}\right)^2= \tfrac{25}{169}.$

Thus, the correct answer is D.

Solution:

Converting units, the strip is $0.65$ cm wide and $2500$ cm long, so its area is $0.65\times 2500=1625$ square centimeters.

Thus, the correct answer is C.

Solution:

A set of guessed squares is guaranteed to hit the domino if and only if the un-guessed squares contain no two adjacent squares, since otherwise the domino could hide on that adjacent pair. The largest set of pairwise non-adjacent squares in the $3\times 3$ grid is the $5$-square checkerboard (four corners plus the center). So at most $5$ squares can be left unguessed, and you must guess $9-5=4.$

Thus, the correct answer is C.

Solution:

The exponent of the factor $(x-k)$ is $k,$ so the sign of $P$ changes at $x=k$ only when $k$ is odd, i.e. at $1,3,5,7,9.$ For $x\gt 10$ every factor is positive, so $P\gt 0.$ Sweeping left and flipping at each odd root, the positive intervals are $(10,\infty),$ $(9,10),$ $(6,7),$ $(5,6),$ $(2,3),$ and $(1,2)$ — six intervals in all.

Thus, the correct answer is C.

Solution:

Write $L=\log n.$ Then $\log(n^2)-(\log n)^2=2L-L^2=L(2-L),$ and the fraction is $\dfrac{L(2-L)}{L-3}.$ A sign chart shows this is $\ge 0$ exactly when $L\le 0$ or $2\le L\lt 3.$ Since $n$ is a positive integer, $L\le 0$ forces $n=1,$ while $2\le L\lt 3$ gives $100\le n\le 999,$ which is $900$ values. In total $1+900=901.$

Thus, the correct answer is E.

Solution:

If the least element is $m\ (m\ge 1),$ then $|B|=m$ and the remaining $m-1$ elements come from $\{m+1,\ldots,12\},$ a set of size $12-m.$ The count is $$\sum_{m\ge 1}\binom{12-m}{m-1}=\binom{11}{0}+\binom{10}{1}+\binom{9}{2}+\binom{8}{3}+\binom{7}{4}+\binom{6}{5},$$ which equals $1+10+36+56+35+6=144.$

Thus, the correct answer is D.

Solution:

Replacing $x,y$ by $|x|,|y|,$ the condition $|x-1|+|y-1|\le 1$ describes a diamond centered at $(1,1)$ with diagonals of length $2,$ hence area $2.$ It lies entirely in the first quadrant (touching the axes only at single points), so reflecting across the two axes produces $4$ disjoint copies. The total area is $4\times 2=8.$

Thus, the correct answer is B.

Solution:

The circles are $(x-4)^2+y^2=16$ and $x^2+(y-10)^2=100,$ i.e. $x^2+y^2=8x$ and $x^2+y^2=20y.$ Subtracting gives $8x=20y,$ so the intersection points lie on $y=\tfrac{2}{5}x,$ which has slope $\tfrac{2}{5}.$

Thus, the correct answer is E.

Solution:

Let the bases be $a$ and $2a.$ Each leg has horizontal offset $\tfrac{a}{2},$ so the height is $\sqrt{1-\tfrac{a^2}{4}}$ and the area is $A=\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}}.$ Then $A^2=\tfrac94\left(a^2-\tfrac{a^4}{4}\right),$ maximized when $a^2=2.$ There the height is $\tfrac{1}{\sqrt2}$ and $A=\tfrac{3\sqrt2}{2}\cdot\tfrac{1}{\sqrt2}=\tfrac{3}{2}.$

Thus, the correct answer is D.

Solution:

With $z=a+bi,$ we have $z\otimes z=a^2+b^2 i$ and $z^2+40=(a^2-b^2+40)+2abi.$ The real parts give $a^2=a^2-b^2+40,$ so $b^2=40.$ The imaginary parts give $b^2=2ab,$ so $b=2a$ and $a^2=\tfrac{b^2}{4}=10.$ Then $|z|^2=a^2+b^2=50,$ so $|z|=5\sqrt{2}.$

Thus, the correct answer is E.

Solution:

From the edges, $4(a+b+c)=13,$ so $a+b+c=\tfrac{13}{4}.$ From the faces, $2(ab+bc+ca)=\tfrac{11}{2},$ so $ab+bc+ca=\tfrac{11}{4}.$ Then $$a^2+b^2+c^2=\left(\tfrac{13}{4}\right)^2-2\cdot\tfrac{11}{4}=\tfrac{169}{16}-\tfrac{88}{16}=\tfrac{81}{16},$$ so the diagonal is $\sqrt{\tfrac{81}{16}}=\tfrac{9}{4}.$

Thus, the correct answer is D.

Solution:

By Vieta, the three distinct integer roots multiply to $-6.$ The sets of three distinct integers with product $-6$ are $\{1,2,-3\},$ $\{1,-2,3\},$ $\{-1,2,3\},$ $\{-1,-2,-3\},$ and $\{1,-1,6\}.$ Each set determines $a=-(p+q+r)$ and $b=pq+qr+rp,$ and all five give different pairs, so there are $5$ ordered pairs $(a,b).$

Thus, the correct answer is A.

Solution:

Multiplying by $210$ gives $c=15a+14b.$ Since $15\equiv 1\pmod{14},$ we get $c\equiv a\pmod{14},$ so $\gcd(c,14)=1$ iff $\gcd(a,14)=1.$ Since $14\equiv -1\pmod{15},$ we get $c\equiv -b\pmod{15},$ so $\gcd(c,15)=1$ iff $\gcd(b,15)=1.$ As $210=14\cdot 15$ with $\gcd(14,15)=1,$ statement III follows: $\gcd(c,210)=1$ iff both hold. Statement II is the forward implication of III, hence true. Statement I is false: if $\gcd(a,14)=1$ but $\gcd(b,15)\ne 1,$ then $\gcd(c,15)\ne 1,$ so $\gcd(c,210)\ne 1.$ Only II and III are true.

Thus, the correct answer is E.

Solution:

The amounts $30,31,32,33,34,35$ are all attainable (for instance $30=6\cdot 5,$ $31=6+10+15,$ $32=6\cdot 2+10\cdot 2,$ $33=6\cdot 3+15,$ $34=6\cdot 4+10,$ $35=10+10+15$). Adding $6$'s then reaches every larger amount. Checking below, $29$ is impossible, since $29-6,\ 29-10,\ 29-15$ are all impossible. So the largest impossible amount is $29,$ whose digit sum is $2+9=11.$

Thus, the correct answer is D.

Solution:

Let the sides be $6,\ 6+d,\ 6+2d.$ The $120^\circ$ angle faces the longest side, so $(6+2d)^2=6^2+(6+d)^2-2\cdot 6\cdot(6+d)\cos 120^\circ.$ Using $\cos 120^\circ=-\tfrac12$ gives $3d^2+6d-72=0,$ so $d=4$ and the sides are $6,10,14.$ The area is $\tfrac12\cdot 6\cdot 10\cdot\sin 120^\circ=30\cdot\tfrac{\sqrt3}{2}=15\sqrt3.$

Thus, the correct answer is E.

Solution:

Set Zelda's first-semester average to $0.$ Then Yolanda's first semester is $3,$ her second semester is $3+18=21,$ and Zelda's second semester is $21-3=18.$ Each person's yearly average is a weighted average of their two semester averages, so Yolanda's year average lies between $3$ and $21,$ and Zelda's lies between $0$ and $18.$ The largest possible gap Yolanda $-$ Zelda is therefore at most $21-0=21,$ so it can never be $22.$ All the other statements are achievable for suitable quiz counts.

Thus, the correct answer is A.

Solution:

Counting assignments where all three bins are odd with the parity filter gives $$\frac{1}{8}\sum_{S\subseteq\{1,2,3\}}(-1)^{|S|}(3-2|S|)^n=\frac{3^n-3}{4}$$ for odd $n.$ Dividing by the $3^n$ total assignments, the probability is $\dfrac{3^n-3}{4\cdot 3^n},$ which for $n=2023$ is extremely close to $\tfrac14.$

Thus, the correct answer is E.

Solution:

Take the first jump as $(2,0)$ and the second as $(2\cos\theta,2\sin\theta)$ with $\theta$ uniform on $[0,2\pi).$ The landing distance satisfies $R^2=(2+2\cos\theta)^2+(2\sin\theta)^2= 8+8\cos\theta.$ We need $R\lt 1,$ i.e. $\cos\theta\lt-\tfrac78.$ The measure of such angles is $2\arccos\tfrac78,$ so the probability is $\dfrac{2\arccos\tfrac78}{2\pi}= \dfrac{\arccos\tfrac78}{\pi}.$ Using $\arccos(1-2x^2)=2\arcsin x$ with $x=\tfrac14$ gives $\arccos\tfrac78=2\arcsin\tfrac14,$ so the probability is $\dfrac{2\arcsin\tfrac14}{\pi}.$

Thus, the correct answer is E.

Solution:

Extend the frustum to a full cone. Since the radii are $3$ and $6$ with slant band $6,$ the apex is slant distance $6$ from the top rim and $12$ from the bottom rim. The bottom circumference $12\pi$ unrolls to a sector of radius $12$ and angle $\dfrac{12\pi}{12}=\pi.$ Place the bug at $(12,0)$ in this pattern; the honey, halfway around the base, is at radius $6$ and angle $\tfrac{\pi}{2}.$ The straight chord between them passes within radius $6$ (off the surface), so the geodesic goes tangent to the circle of radius $6:$ the tangent has length $\sqrt{12^2-6^2}=6\sqrt3$ and touches at angle $\tfrac{\pi}{3},$ after which the path follows the arc of angle $\tfrac{\pi}{6}$ on radius $6,$ of length $6\cdot\tfrac{\pi}{6}=\pi.$ The shortest path is $6\sqrt3+\pi.$

Thus, the correct answer is E.

Solution:

Setting $a=b=0$ gives $2f(0)=2f(0)^2,$ so $f(0)=0$ or $f(0)=1.$ If $f(0)=0,$ then setting $b=0$ forces $f\equiv 0,$ giving $f(1)=0.$ Otherwise $f(0)=1,$ and setting $a=b$ gives $f(2a)=2f(a)^2-1\ge -1$ for every $a.$ In particular, with $a=\tfrac12,$ $f(1)=2f\!\left(\tfrac12\right)^2-1\ge -1.$ So $f(1)\ge -1,$ and indeed every value in $[-1,\infty)$ is attainable (e.g. $f(x)=\cos(kx)$ or $\cosh(kx)$). Hence $-2$ is impossible.

Thus, the correct answer is E.

Solution:

Each die contributes an exponent vector in the primes $2,3,5$ (face $1\to(0,0,0),$ $2\to(1,0,0),$ $3\to(0,1,0),$ $4\to(2,0,0),$ $5\to(0,0,1),$ $6\to(1,1,0)$), and a product is determined by the sum of these vectors. Counting the distinct attainable sums for $n=1,2,3,\ldots$ gives $6,18,40,75,126,196,288,405,550,726,936,$ so $n=11.$

Thus, the correct answer is A.

Solution:

Handle each prime separately using the exponents of $a,b,c,d.$

Prime $2$ (total $6$): $\max(b,c)=1$ forces $a=3;$ then $b+c+d=3$ with $\max(b,d)=\max(c,d)=2$ gives $d=2$ and $\{b,c\}=\{0,1\},$ so the minimum exponent is $0.$

Prime $3$ (total $9$): $\max(a,b)=2$ with the other lcms equal to $3$ forces $c=d=3;$ then $a+b=3$ with $\max(a,b)=2$ gives $\{a,b\}=\{1,2\},$ so the minimum is $1.$

Prime $5$ (total $7$): $\max(a,b)=3$ with $\max(b,c),\max(b,d),\max(c,d)=2$ forces $a=3;$ then $b+c+d=4$ with each $\le 2$ and pairwise maxima $2$ gives two of them equal to $2$ and one equal to $0,$ so the minimum is $0.$

Therefore $\gcd(a,b,c,d)=2^0\cdot 3^1\cdot 5^0=3.$

Thus, the correct answer is C.

Solution:

Let the original pentagon have circumradius $R.$ Folding a vertex to the center creases along the perpendicular bisector of the segment from the center to that vertex, a line at distance $\tfrac{R}{2}$ from the center. The five creases bound a regular pentagon with apothem $\tfrac{R}{2},$ whereas the original has apothem $R\cos 36^\circ.$ Areas scale as the square of the apothem, so the ratio is $$\frac{(R/2)^2}{(R\cos 36^\circ)^2}=\frac{1}{4\cos^2 36^\circ}.$$ Since $\cos 36^\circ=\tfrac{1+\sqrt5}{4},$ this ratio is $\tfrac{4}{(1+\sqrt5)^2}= \tfrac{4}{6+2\sqrt5}=\tfrac{2}{3+\sqrt5}=\tfrac{3-\sqrt5}{2}.$ Multiplying by the original area $\sqrt5+1$ gives $\tfrac{(3-\sqrt5)(\sqrt5+1)}{2}=\tfrac{2\sqrt5-2}{2}=\sqrt5-1.$

Thus, the correct answer is B.