Solution:

The sums are $2+4+6=12$ and $1+3+5=9,$ so the expression equals $\dfrac{12}{9}-\dfrac{9}{12}=\dfrac{4}{3}-\dfrac{3}{4}.$

Over a common denominator this is $$\dfrac{16}{12}-\dfrac{9}{12}=\dfrac{7}{12}.$$

Thus, the correct answer is C.

Solution:

The five scores sum to $90+80+70+60+85=385,$ giving an average of $77.$ The goal is a new average of at least $80.$

Six tests averaging $80$ must total $6\cdot80=480,$ so the sixth score must be at least $480-385=95.$

Thus, the correct answer is E.

Solution:

The total cost is $A+B,$ so each person's fair share is $\dfrac{A+B}{2}.$

LeRoy paid $A,$ which is less than his share, so he must give Bernardo $$\dfrac{A+B}{2}-A=\dfrac{B-A}{2}.$$

Thus, the correct answer is C.

Solution:

Since $161=7\cdot23,$ the only two-digit factor is $23.$ This must be the reversed value of $a,$ so the true value of $a$ is $32,$ and $b=7.$

The correct product is $$32\cdot7=224.$$

Thus, the correct answer is E.

Solution:

A number divisible by every integer from $1$ to $6$ must be a multiple of $\operatorname{lcm}(1,2,3,4,5,6)=60.$

The second smallest positive multiple of $60$ is $120,$ whose digit sum is $1+2+0=3.$

Thus, the correct answer is A.

Solution:

Let $O$ be the center. The arcs measure $2x$ and $3x$ with $2x+3x=360^\circ,$ so $x=72^\circ$ and the minor arc $BC$ gives central angle $\angle BOC=144^\circ.$

The radii to $B$ and $C$ are perpendicular to the tangents, so $\angle ABO=\angle ACO=90^\circ.$ In quadrilateral $ABOC,$ $$\angle BAC=360^\circ-144^\circ-90^\circ-90^\circ=36^\circ.$$

Thus, the correct answer is C.

Solution:

Since $\dfrac{x+y}{2}=60,$ we have $x+y=120.$ To maximize $\dfrac{x}{y}$ we make $y$ small.

Because $x\le99,$ it follows that $y=120-x\ge21.$ Taking $x=99$ and $y=21$ gives the maximum $$\dfrac{99}{21}=\dfrac{33}{7}.$$

Thus, the correct answer is B.

Solution:

The straight sides are the same length for both paths, so the difference in length comes only from the two semicircular ends. If the inner radius is $r,$ those ends combine into a full circle, and the extra length is $$2\pi(r+6)-2\pi r=12\pi.$$

If her speed is $x$ meters per second, then the extra time gives $36x=12\pi,$ so $x=\dfrac{\pi}{3}.$

Thus, the correct answer is A.

Solution:

The interval has length $30,$ with $20$ of it negative and $10$ of it positive. So each number is positive with probability $\dfrac13$ and negative with probability $\dfrac23.$

The product is positive when both are positive or both are negative: $$\left(\dfrac13\right)^2+\left(\dfrac23\right)^2=\dfrac19+\dfrac49=\dfrac59.$$

Thus, the correct answer is D.

Solution:

Because $AB\parallel CD,$ we have $\angle CDM=\angle AMD.$ Combined with $\angle AMD=\angle CMD,$ this gives $\angle CDM=\angle CMD,$ so $\triangle CMD$ is isosceles with $CM=CD=6.$

Then $\triangle MBC$ is right-angled at $B$ with hypotenuse $CM=6$ and leg $BC=3,$ so it is a $30$-$60$-$90^\circ$ triangle with $\angle BMC=30^\circ.$

Finally, $\angle AMD+\angle CMD+\angle BMC=180^\circ,$ so $2\angle AMD+30^\circ=180^\circ,$ giving $\angle AMD=75^\circ.$

Thus, the correct answer is E.

Solution:

One jump cannot work, since $(0,0)$ and $(1,0)$ are only $1$ apart. Two jumps also fail: the intermediate point would be at distance $5$ from both, forcing it onto the perpendicular bisector $x=\dfrac12,$ which contains no lattice points.

Three jumps suffice, for example $$(0,0)\to(3,4)\to(6,0)\to(1,0),$$ where each step has length $5.$

Thus, the correct answer is B.

Solution:

Assume the octagon has edge length $1.$ The four corner triangles are right isosceles with legs $\dfrac{\sqrt2}{2}$ and area $\dfrac14$ each. The four rectangles are $1$ by $\dfrac{\sqrt2}{2}$ with area $\dfrac{\sqrt2}{2}$ each, and the center square has area $1.$

The total area is $$4\cdot\dfrac14+4\cdot\dfrac{\sqrt2}{2}+1=2+2\sqrt2.$$ The probability of hitting the center square is $$\dfrac{1}{2+2\sqrt2}=\dfrac{\sqrt2-1}{2}.$$

Thus, the correct answer is A.

Solution:

The largest difference is $w-z=9.$ Writing $9=(w-x)+(x-z)$ style splits, the interior differences pair as $3+6$ and $4+5,$ which forces the smallest difference $x-y=1.$

The second largest difference $6$ is either $w-y$ or $x-z.$ If $w-y=6,$ the numbers are $\{w,w-5,w-6,w-9\},$ so $4w-20=44$ and $w=16.$ If $x-z=6,$ the numbers are $\{w,w-3,w-4,w-9\},$ so $4w-16=44$ and $w=15.$

The possible values are $16$ and $15,$ which sum to $31.$

Thus, the correct answer is B.

Solution:

Let $p=FV$ and let the directrix be $\ell.$ Projecting $F$ and $B$ onto $\ell,$ the focus-directrix property gives $FB=2p$ (the distance from $B$ to $\ell$), and by the Pythagorean Theorem $$VB=\sqrt{FV^2+FB^2}=\sqrt{p^2+4p^2}=\sqrt5\,p.$$

Then $\cos(\angle FVB)=\dfrac{FV}{VB}=\dfrac{p}{\sqrt5\,p}=\dfrac{1}{\sqrt5}.$ Since $\angle AVB=2\angle FVB,$ $$\cos(\angle AVB)=2\cos^2(\angle FVB)-1=2\cdot\dfrac15-1=-\dfrac35.$$

Thus, the correct answer is D.

Solution:

Factoring, $$2^{24}-1=(2^{12}-1)(2^{12}+1)=(2^6-1)(2^6+1)(2^4+1)(2^8-2^4+1),$$ which equals $63\cdot65\cdot17\cdot241=3^2\cdot5\cdot7\cdot13\cdot17\cdot241.$

Since $241$ is a three-digit prime, the two-digit factors come from $3^2\cdot5\cdot7\cdot13\cdot17.$ They are $$13,15,17,21,35,39,45,51,63,65,85,91,$$ for a total of $12.$

Thus, the correct answer is D.

Solution:

Let $E$ and $H$ be the midpoints of $AB$ and $BC.$ The perpendicular bisector of $AB$ through $E$ meets diagonal $AC$ at $F,$ and the perpendicular bisector of $BC$ through $H$ meets $AC$ at $G.$ The region $R$ is the pentagon $BEFGH.$

Triangle $AFE$ is a $30$-$60$-$90^\circ$ triangle with $AE=1,$ so its area is $\dfrac12\cdot1\cdot\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{6}.$ Triangles $BFE$ and $BGH$ are congruent to it, and $\triangle FBG$ is equilateral, splitting into two more copies.

Hence $R$ consists of four congruent triangles, giving area $4\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}.$

Thus, the correct answer is C.

Solution:

First, $$h_1(x)=\log_{10}\!\left(\dfrac{10^{10x}}{10}\right)=\log_{10}\!\left(10^{10x-1}\right)=10x-1.$$

Iterating, $h_n(x)=10^n x-(1+10+\cdots+10^{n-1}).$ Therefore $h_n(1)$ is an $n$-digit integer whose units digit is $9$ and all of whose other digits are $8.$

For $n=2011,$ the digit sum is $$8\cdot2010+9=16{,}089.$$

Thus, the correct answer is B.

Solution:

Let the apex be $A$ and the base be square $BCDE.$ Then $AB=AD=1$ and $BD=\sqrt2,$ so $\triangle BAD$ is an isosceles right triangle.

Let the cube have edge length $x.$ Its intersection with the plane of $\triangle BAD$ is a rectangle of height $x$ and width $\sqrt2\,x,$ whose top corners lie on $AB$ and $AD.$ Because the legs $AB$ and $AD$ meet the base at $45^\circ,$ each portion of $BD$ outside the rectangle has length $x,$ so $$\sqrt2=BD=\sqrt2\,x+2x,$$ which reduces to $x=\dfrac{\sqrt2}{2+\sqrt2}=\sqrt2-1.$

The volume is $$(\sqrt2-1)^3=5\sqrt2-7.$$

Thus, the correct answer is A.

Solution:

For $0\lt x\le100,$ the nearest lattice point above the line $y=\tfrac12x+2$ is $\left(x,\tfrac12x+3\right)$ if $x$ is even and $\left(x,\tfrac12x+\tfrac52\right)$ if $x$ is odd.

The slope from $(0,2)$ to that point is $\dfrac12+\dfrac1x$ for even $x$ and $\dfrac12+\dfrac{1}{2x}$ for odd $x.$ The minimum such slope is $\dfrac{51}{100}$ for even $x$ and $\dfrac{50}{99}$ for odd $x.$

Since $\dfrac{50}{99}\lt\dfrac{51}{100},$ the line avoids all these lattice points exactly when $\dfrac12\lt m\lt\dfrac{50}{99},$ so the maximum is $a=\dfrac{50}{99}.$

Thus, the correct answer is B.

Solution:

Since $DE\parallel AC$ and $EF\parallel AB,$ we get $\angle BDE=\angle BAC=\angle EFC.$ By the Inscribed Angle Theorem, $\angle BXE=\angle BDE$ and $\angle EXC=\angle EFC,$ so $\angle BXE=\angle EXC.$ With $BE=EC,$ this forces $XB=XC.$

Also $\angle BXC=2\angle BAC,$ so by the Inscribed Angle Theorem $X$ is the circumcenter of $\triangle ABC.$ Hence $XA=XB=XC=R.$

The area of the $13$-$14$-$15$ triangle is $84$ by Heron's formula, so $$R=\dfrac{13\cdot14\cdot15}{4\cdot84}=\dfrac{65}{8},$$ and $XA+XB+XC=3R=\dfrac{195}{8}.$

Thus, the correct answer is C.

Solution:

Let the arithmetic mean be $10a+b$ and the geometric mean be $10b+a.$ Then $x+y=2(10a+b)$ and $xy=(10b+a)^2.$

Therefore $$(x-y)^2=(x+y)^2-4xy=396(a^2-b^2)=11\cdot6^2\cdot(a+b)(a-b).$$ This is a perfect square exactly when $a+b=11$ and $a-b$ is a perfect square. Among digit solutions, only $a-b=1$ works, giving $(a,b)=(6,5).$

Then $(x-y)^2=11\cdot6^2\cdot11=66^2,$ so $|x-y|=66.$ (Indeed $\{x,y\}=\{32,98\}.$)

Thus, the correct answer is D.

Solution:

For a triangle with sides $a,b,c,$ the tangent lengths are $AD=\tfrac12(b+c-a),$ $BE=\tfrac12(a+c-b),$ and $CF=\tfrac12(a+b-c).$ If $T_n$ has sides $(y-1,y,y+1),$ then $T_{n+1}$ has sides $\left(\tfrac{y}{2}-1,\tfrac{y}{2},\tfrac{y}{2}+1\right).$

Starting from $T_1$ with middle side $2012,$ the middle side halves each step and the perimeter of $T_{n+1}$ is $\tfrac12$ the perimeter of $T_n.$ A triangle of this form exists only while its middle side exceeds $2.$

The middle side of $T_n$ is $\dfrac{2012}{2^{n-1}}.$ This first drops to $2$ or below at $n=11,$ so the last valid triangle is $T_{10},$ whose middle side is $\dfrac{2012}{2^9}$ and whose perimeter is $$3\cdot\dfrac{2012}{2^9}=\dfrac{6036}{512}=\dfrac{1509}{128}.$$

Thus, the correct answer is D.

Solution:

A lattice point $X=(x,y)$ lies on some path exactly when $$d=|x-3|+|x+3|+|y-2|+|y+2|\le20.$$ This expression is unchanged when $x\to-x$ or $y\to-y,$ so we count points with $x\ge0,$ $y\ge0,$ multiply by $4,$ and correct for the axes.

Splitting into the four regions determined by whether $x\le3$ and $y\le2$ gives $12+20+15+10=57$ points in the first quadrant (including axis points). By symmetry the total is $$4\cdot57-2\cdot15-3=195.$$

Thus, the correct answer is C.

Solution:

Factoring in $z^4,$ $$P(z)=(z^4-1)\big(z^4+(4\sqrt3+7)\big).$$ The first factor gives the roots $1,-1,i,-i.$ Since $4\sqrt3+7=(\sqrt3+2)^2$ and $2(\sqrt3+2)=(\sqrt3+1)^2,$ writing $w=\tfrac12(\sqrt3+1)$ the other four roots are $w(\pm1\pm i).$

The eight roots are symmetric about the origin with $4$-fold symmetry, and every segment joining two of them has length at least $\sqrt2.$ Thus any such polygon has perimeter at least $8\sqrt2,$ and the polygon with vertices $1, w(1+i), i, w(-1+i), -1, w(-1-i), -i, w(1-i)$ achieves it.

Thus, the correct answer is B.

Solution:

Because $\left[\dfrac{n+mk}{k}\right]=\left[\dfrac{n}{k}\right]+m,$ whether $n$ satisfies the identity depends only on $n\bmod k.$ Since $k\mid99!$ for $1\le k\le99,$ every residue class is equally likely.

Write $100=qk+r$ with $|r|\le\dfrac{k-1}{2}.$ Analyzing the carry in $[\,(100-n)/k\,]$ shows the identity holds precisely for the residues in an interval of the appropriate length, giving $$P(k)=1-\dfrac{|r|}{k}.$$

To minimize $P(k)$ we maximize $\dfrac{|r|}{k}.$ Taking $r=\dfrac{k-1}{2}$ gives $201=k(2q+1),$ and the largest odd $k\le99$ dividing $201=3\cdot67$ is $k=67.$ Then $$P(67)=\dfrac12+\dfrac{1}{2\cdot67}=\dfrac{34}{67},$$ which is smaller than the values from all other cases.

Thus, the correct answer is D.