Answer: A

Solution:

The pan has area $20\cdot18=360$ square inches, and each piece has area $2\cdot2=4$ square inches.

The number of pieces is $$\dfrac{360}{4}=90.$$

Thus, the correct answer is A.

Answer: D

Solution:

In the first $30$ minutes Sam covered $60\cdot\tfrac12=30$ miles, and in the second he covered $65\cdot\tfrac12=32.5$ miles.

The last $30$ minutes covered $96-30-32.5=33.5$ miles, so the speed was $$\dfrac{33.5}{1/2}=67\text{ mph}.$$

Thus, the correct answer is D.

Answer: B

Solution:

The line of slope $2$ is $y-30=2(x-40);$ setting $y=0$ gives $x=25.$ The line of slope $6$ is $y-30=6(x-40);$ setting $y=0$ gives $x=35.$

The distance between the intercepts is $|35-25|=10.$

Thus, the correct answer is B.

Answer: B

Solution:

Dropping a perpendicular from the center to the chord bisects it, forming a right triangle with legs $5$ (half the chord) and $5$ (the distance), and hypotenuse $r.$

Then $r^2=5^2+5^2=50,$ so the area is $\pi r^2=50\pi.$

Thus, the correct answer is B.

Answer: D

Solution:

The set has $8$ elements, giving $2^8=256$ subsets. The subsets with no prime use only the four non-primes $\{4,6,8,9\},$ and there are $2^4=16$ of these.

So the number containing at least one prime is $256-16=240.$

Thus, the correct answer is D.

Answer: B

Solution:

One can costs $\tfrac{Q}{S}$ quarters, which is $\tfrac{Q}{4S}$ dollars. The number of cans that $D$ dollars can buy is

$$\dfrac{D}{\tfrac{Q}{4S}}=\dfrac{4DS}{Q}.$$

Thus, the correct answer is B.

Answer: C

Solution:

The factors split into two telescoping chains. The odd-position factors form $$\log_3 7\cdot\log_7 11\cdot\log_{11} 15\cdots\log_{23} 27=\log_3 27=3,$$ and the even-position factors form $$\log_5 9\cdot\log_9 13\cdots\log_{21} 25=\log_5 25=2.$$

The product is $3\cdot2=6.$

Thus, the correct answer is C.

Answer: C

Solution:

Let $O$ be the center of the circle. The centroid of $\triangle ABC$ is the average of $A,$ $B,$ and $C;$ since $O$ is the midpoint of $\overline{AB},$ the centroid lies one-third of the way from $O$ to $C.$

As $C$ traces the circle of radius $12,$ the centroid traces a circle of radius $\tfrac13\cdot12=4.$ Its area is $16\pi\approx50.$

Thus, the correct answer is C.

Answer: E

Solution:

Splitting the sum, $$\sum_{i=1}^{100}\sum_{j=1}^{100}(i+j)=\sum_{i=1}^{100}\sum_{j=1}^{100}i+\sum_{i=1}^{100}\sum_{j=1}^{100}j=100\sum_{i=1}^{100}i+100\sum_{j=1}^{100}j.$$

Since $\sum_{k=1}^{100}k=5050,$ this equals $100\cdot5050+100\cdot5050=1{,}010{,}000.$

Thus, the correct answer is E.

Answer: D

Solution:

The mode uses $10$ of the entries, leaving $2008.$ Because the mode is unique, every other value appears at most $9$ times, so at least $\left\lceil\tfrac{2008}{9}\right\rceil=224$ distinct non-mode values are needed.

Adding the mode gives $224+1=225.$ This is achievable: use $9$ copies each of $1$ through $223,$ ten copies of $224,$ and one copy of $225.$

Thus, the correct answer is D.

Answer: A

Solution:

Following a fold from a corner of the paper to the center of the box top, the distance from a corner of the sheet to its center is $$\dfrac{w}{2}+h+\dfrac{w}{2}=w+h.$$

That segment is a leg of a $45$-$45$-$90$ triangle whose hypotenuse is a full side of the square sheet, so the side length is $\sqrt2\,(w+h).$

The area of the sheet is $\left(\sqrt2\,(w+h)\right)^2=2(w+h)^2.$

Thus, the correct answer is A.

Answer: C

Solution:

Let $q=AC$ and $r=BD.$ The angle bisector theorem gives $\tfrac{q}{3}=\tfrac{10}{r},$ so $r=\tfrac{30}{q}.$

Applying the triangle inequalities to sides $q,$ $10,$ and $3+r$ and substituting $r=\tfrac{30}{q}$ yields $(q-15)(q+2)\lt0$ and $(q-3)(q+10)\gt0$ (the third inequality holds automatically). Together these force $3\lt q\lt15.$

So $(m,n)=(3,15)$ and $m+n=18.$

Thus, the correct answer is C.

Answer: C

Solution:

Place $A=(0,30),$ $B=(0,0),$ $C=(30,0),$ $D=(30,30),$ and $P=(3x,3y).$ Averaging the vertices, the four centroids are $$(x,\,y+10),\ (x+10,\,y),\ (x+20,\,y+10),\ (x+10,\,y+20).$$

These form a square whose diagonals, one horizontal and one vertical, each have length $20.$ Its area is $\tfrac12\cdot20\cdot20=200,$ independent of where $P$ lies.

Thus, the correct answer is C.

Answer: E

Solution:

Let Chloe be $n$ today, so she is $n-1$ years older than Zoe. In $y$ years Chloe's age $n+y$ is a multiple of Zoe's age $1+y$ exactly when $1+y$ divides $n-1.$ Having $9$ such birthdays means $n-1$ has exactly $9$ divisors.

A number with exactly $9$ divisors has the form $p^2q^2$ or $p^8;$ the only two-digit case is $2^2\cdot3^2=36.$ So Chloe is $37$ and Joey is $38.$

Joey's age $38+y$ is a multiple of $1+y$ exactly when $1+y$ divides $37.$ The next time is $y=36,$ making Joey $74,$ with digit sum $7+4=11.$

Thus, the correct answer is E.

Answer: A

Solution:

Write the number as $\overline{abc}.$ The hundreds digit $a$ has $8$ choices ($1,2,4,5,6,7,8,9$), and the units digit $c$ has $4$ choices ($1,5,7,9$).

The tens digit $b$ may be any of $\{0,1,2,4,5,6,7,8,9\}.$ These split into three residue classes mod $3$ of equal size $\{0,6,9\},\{1,4,7\},\{2,5,8\},$ so exactly $3$ choices of $b$ make $a+b+c$ divisible by $3.$

The count is $8\cdot4\cdot3=96.$

Thus, the correct answer is A.

Answer: B

Solution:

Translating by $6,$ the solutions of $z^8=81$ are eight points on a circle of radius $81^{1/8}=\sqrt3,$ forming a regular octagon. The minimum-area triangle uses three consecutive vertices.

Take $A=\left(\tfrac12\sqrt6,\tfrac12\sqrt6\right),$ $B=(\sqrt3,0),$ and $C=\left(\tfrac12\sqrt6,-\tfrac12\sqrt6\right).$ Then $AC=\sqrt6$ and the height is $\sqrt3-\tfrac12\sqrt6,$ so the area is $$\tfrac12\cdot\sqrt6\left(\sqrt3-\tfrac12\sqrt6\right)=\tfrac{3}{2}\sqrt2-\tfrac{3}{2}.$$

Thus, the correct answer is B.

Answer: A

Solution:

From $\tfrac59\lt\tfrac pq$ we get $9p-5q\ge1,$ and from $\tfrac pq\lt\tfrac47$ we get $4q-7p\ge1.$ Now $$\dfrac{1}{63}=\dfrac47-\dfrac59=\dfrac{4q-7p}{7q}+\dfrac{9p-5q}{9q}\ge\dfrac{1}{7q}+\dfrac{1}{9q}=\dfrac{16}{63q}.$$

Hence $q\ge16.$ With $q=16,$ the fraction $\tfrac{9}{16}$ lies strictly between $\tfrac59$ and $\tfrac47,$ so $p=9$ and $q-p=16-9=7.$

Thus, the correct answer is A.

Answer: B

Solution:

Repeatedly substituting the recursion into itself gives $$f(n)=f(n-6)+6.$$ So $f$ increases by $6$ every time $n$ increases by $6.$

Since $2018=2+6\cdot336,$ we have $f(2018)=f(2)+6\cdot336=1+2016=2017.$

Thus, the correct answer is B.

Answer: C

Solution:

Let $d$ be the next divisor after $323.$ If $\gcd(d,323)=1,$ then $323d$ divides $n,$ forcing $n\ge323d\gt323^2\gt9999,$ impossible for a $4$-digit number. So $d$ shares a prime factor with $323=17\cdot19.$

Then $d-323\ge\gcd(d,323)\ge17,$ so $d\ge340.$ Indeed $d=340=17\cdot20$ occurs for $n=17\cdot19\cdot20=6460,$ which is even and $4$-digit.

Thus, the correct answer is C.

Answer: C

Solution:

Both $\triangle ACE$ and $\triangle XYZ$ are equilateral, and $\triangle ACE$ has half the area of the hexagon. The vertices where the two triangles cut each other let the shaded hexagon be measured against the midpoint triangle $U$ of $\triangle ACE.$

That midpoint triangle has $\tfrac14$ the area of $\triangle ACE,$ hence $\tfrac18$ of the hexagon. The shaded region equals $\tfrac52$ of the midpoint triangle, so it is $\tfrac52\cdot\tfrac18=\tfrac{5}{16}$ of the hexagon.

The hexagon has area $6\cdot\tfrac{\sqrt3}{4}\cdot1^2=\tfrac{3\sqrt3}{2},$ so the shaded area is $\tfrac{5}{16}\cdot\tfrac{3\sqrt3}{2}=\tfrac{15}{32}\sqrt3.$

Thus, the correct answer is C.

Answer: E

Solution:

Since $5^2+12^2=13^2,$ the triangle is right-angled at $C.$ Set $C=(0,0),$ $A=(12,0),$ and $B=(0,5).$ Then $O$ is the midpoint of $\overline{AB},$ namely $O=\left(6,\tfrac52\right),$ with circumradius $\tfrac{13}{2}.$ The inradius is $\tfrac{\text{area}}{s}=\tfrac{30}{15}=2,$ so $I=(2,2).$

Because $M$'s circle is tangent to both legs, $M=(\rho,\rho).$ Internal tangency to the circumcircle gives $MO=\tfrac{13}{2}-\rho.$ Setting this equal to $\sqrt{(\rho-6)^2+\left(\rho-\tfrac52\right)^2}$ and solving gives $\rho=4,$ so $M=(4,4).$

The shoelace formula on $M=(4,4),$ $O=\left(6,\tfrac52\right),$ $I=(2,2)$ gives area $\tfrac72.$

Thus, the correct answer is E.

Answer: D

Solution:

Write $P(x)=ax^3+bx^2+cx+d$ with each of $a,b,c,d$ in $\{0,\ldots,9\}.$ The condition is $-a+b-c+d=-9.$

Let $a'=9-a$ and $c'=9-c,$ both in $[0,9].$ Then $a'+b+c'+d=9.$ By stars and bars the number of nonnegative solutions is $\binom{9+3}{3}=\binom{12}{3}=220,$ and each automatically satisfies the upper bounds since the sum is $9.$

Thus, the correct answer is D.

Answer: C

Solution:

The longitudes differ by $360^\circ-(110^\circ+115^\circ)=135^\circ,$ and $B$ is at latitude $45^\circ$ N. Place $A=(1,0,0)$ on the unit sphere.

Then $B=\left(\cos45^\circ\cos135^\circ,\ \cos45^\circ\sin135^\circ,\ \sin45^\circ\right)=\left(-\tfrac12,\tfrac12,\tfrac{\sqrt2}{2}\right).$ The dot product is $A\cdot B=-\tfrac12,$ so $\cos\angle ACB=-\tfrac12$ and $\angle ACB=120^\circ.$

Thus, the correct answer is C.

Answer: C

Solution:

Let $\{x\}=x-\lfloor x\rfloor.$ The equation becomes $x^2=10{,}000\{x\},$ so $\tfrac{x^2}{10{,}000}=\{x\}.$ Since $0\le\{x\}\lt1,$ we need $0\le x^2\lt10{,}000,$ i.e. $-100\lt x\lt100.$

On each interval $[k,k+1)$ the increasing parabola $\tfrac{x^2}{10{,}000}$ meets the segment $\{x\}$ exactly once. These intervals run for $k=-100,-99,\ldots,98,$ giving $199$ solutions.

Thus, the correct answer is C.

Answer: D

Solution:

Let $O_i$ be the center of $\omega_i,$ and let $K$ be the intersection of lines $O_1P_1$ and $O_2P_2.$ Because $\angle P_1P_2P_3=60^\circ,$ triangle $P_2KP_1$ is a $30$-$60$-$90^\circ$ triangle. With $d=P_1K,$ we get $P_2K=2d$ and $P_1P_2=\sqrt3\,d.$

The Law of Cosines in $\triangle O_1KO_2$ (with $O_1O_2=8$) gives $$8^2=(d+4)^2+(2d-4)^2-2(d+4)(2d-4)\cos60^\circ,$$ which simplifies to $3d^2-12d-16=0,$ so $d=2+\tfrac23\sqrt{21}.$

Then $P_1P_2=\sqrt3\,d=2\sqrt3+2\sqrt7,$ and the area is $$\dfrac{\sqrt3}{4}\left(2\sqrt3+2\sqrt7\right)^2=10\sqrt3+6\sqrt7=\sqrt{300}+\sqrt{252}.$$

So $a+b=300+252=552.$

Thus, the correct answer is D.