With $a=\tfrac12,$ we have $a^{-1}=2.$ The numerator is $2\cdot2+\dfrac{2}{2}=4+1=5,$ and dividing by $a=\tfrac12$ gives $\dfrac{5}{1/2}=10.$

Thus, the correct answer is D.

The harmonic mean is $\dfrac{2\cdot1\cdot2016}{1+2016}=\dfrac{4032}{2017}.$ Since $\dfrac{2016}{2017}$ is very close to $1,$ this is just under $2,$ so the closest integer is $2.$

Thus, the correct answer is A.

Since $x=-2016,$ $|x|=2016.$ The innermost expression is $|x|-x=2016+2016=4032.$ Then $\big|4032\big|-|x|=4032-2016$ $=2016,$ and the outer absolute value leaves $2016.$ Finally subtracting $x$ gives $2016-(-2016)=4032.$

Thus, the correct answer is D.

Let the angles be $\alpha\lt\beta$ with $\beta=\tfrac54\alpha.$ The larger complement belongs to the smaller angle, so $90-\alpha=2(90-\beta)=180-\tfrac52\alpha.$ This gives $\tfrac32\alpha=90,$ so $\alpha=60^\circ$ and $\beta=75^\circ.$ The sum is $135^\circ.$

Thus, the correct answer is C.

Because $919=7\cdot131+2,$ the treaty was signed $131$ full weeks plus $2$ days after Thursday. Two days beyond Thursday is Saturday.

Thus, the correct answer is B.

Let the vertex in the first quadrant be $(x,x^2).$ By symmetry the base is $BC=2x$ and the height is $x^2,$ so $\tfrac12\cdot2x\cdot x^2=x^3=64.$ Thus $x=4$ and $BC=2x=8.$

Thus, the correct answer is C.

The first pass removes the odd numbers, leaving the multiples of $2.$ The second pass removes $2,6,10,\ldots,$ leaving the multiples of $4.$ In general, after the $n$th pass only the multiples of $2^n$ remain. The surviving number is the highest power of $2$ not exceeding $100,$ which is $2^6=64.$

Thus, the correct answer is D.

Weight is proportional to area, and area scales with the square of the side length. The second side is $\tfrac53$ times the first, so its weight is $12\cdot\left(\dfrac53\right)^2=\dfrac{100}{3}\approx33.3$ ounces.

Thus, the correct answer is D.

Let the shorter side have $x$ posts, so the longer side has $2x.$ Counting all posts and subtracting the four corners counted twice, $2x+2(2x)-4=20,$ giving $x=4.$ The shorter side has $4$ posts, or $(4-1)\cdot4=12$ yards, and the longer side has $8$ posts, or $(8-1)\cdot4=28$ yards. The area is $12\cdot28=336.$

Thus, the correct answer is B.

The sides $\overline{PQ}$ and $\overline{RS}$ have slope $-1,$ and $\overline{QR}$ and $\overline{PS}$ have slope $1,$ so $PQRS$ is a rectangle with sides $(a-b)\sqrt2$ and $(a+b)\sqrt2.$ Its area is $2(a-b)(a+b)=2(a^2-b^2)$ $=16,$ so $a^2-b^2=8.$ The only perfect squares differing by $8$ are $9$ and $1,$ giving $a=3,$ $b=1,$ and $a+b=4.$

Thus, the correct answer is A.

A unit square in the strip $k\le x\le k+1$ fits below $y=\pi x$ up to height $\lfloor\pi k\rfloor.$ Counting $1\times1$ squares in the strips $1\le x\le5$ gives $3+6+9+12=30.$ The $2\times2$ squares give $2+5+8=15,$ and the $3\times3$ squares give $1+4=5.$ There are no larger squares, so the total is $30+15+5=50.$

Thus, the correct answer is D.

Color the grid like a checkerboard so the four corners and the center share one color. Since consecutive numbers occupy adjacent (opposite colored) squares, the numbers alternate parity along the chain, so the five same-colored cells contain the five odd numbers $1,3,5,7,9,$ which sum to $25.$ The four corners add to $18,$ so the center is $25-18=7.$

Thus, the correct answer is C.

Let the airplane be at $C,$ directly above point $D$ on the ground at altitude $h.$ Triangles $ACD$ and $BCD$ are $30$-$60$-$90$ right triangles, so $AD=\sqrt3\,h$ and $BD=\dfrac{h}{\sqrt3}.$ Since Alice looks north and Bob looks west, $\angle ADB=90^\circ,$ so $AD^2+BD^2=AB^2=100.$ Then $3h^2+\dfrac{h^2}{3}=\dfrac{10h^2}{3}=100,$ giving $h=\sqrt{30}\approx5.48,$ closest to $5.5.$

Thus, the correct answer is E.

Let $r$ be the common ratio. Since the second term is $1,$ the first term is $\dfrac1r,$ so $S=\dfrac{1/r}{1-r}=\dfrac{1}{r-r^2}.$ Because $S\gt0,$ it is smallest when $r-r^2$ is largest. The parabola $r-r^2$ peaks at $r=\tfrac12,$ where it equals $\tfrac14,$ so the smallest value of $S$ is $\dfrac{1}{1/4}=4.$

Thus, the correct answer is E.

Pair the opposite faces as $(a,b),(c,d),(e,f).$ Each vertex product uses one face from each pair, so the sum of all eight products factors as $(a+b)(c+d)(e+f).$ The three factors have fixed total $2+3+4+5+6+7=27,$ and a product with fixed sum is largest when the factors are equal, at $9$ each. This balance is achievable with $(2,7),(3,6),(4,5),$ giving $9\cdot9\cdot9=729.$

Thus, the correct answer is D.

A sum of consecutive integers equals the count times the median. For an odd number of terms, the median is an integer divisor of $345,$ giving runs of $3$ (median $115$), $5$ (median $69$), $15$ (median $23$), and $23$ (median $15$) terms. For an even number of terms $2k,$ the median is a half-integer, giving runs of $2,$ $6,$ and $10$ terms. Longer runs would force negative terms. This gives $4+3=7$ ways.

Thus, the correct answer is E.

Let $x=BH.$ Then $CH=8-x,$ and from the two right triangles $AH^2=7^2-x^2=9^2-(8-x)^2.$ This gives $x=2$ and $AH=\sqrt{45}.$ By the angle bisector theorem in $\triangle ACH,$ $\dfrac{AP}{PH}=\dfrac{CA}{CH}=\dfrac96,$ so $AP=\dfrac35 AH.$ Similarly in $\triangle ABH,$ $\dfrac{AQ}{QH}=\dfrac{BA}{BH}=\dfrac72,$ so $AQ=\dfrac79 AH.$ Then $$\begin{aligned} PQ &= AQ-AP \\ &= \left(\dfrac79-\dfrac35\right)AH \\ &= \dfrac{8}{45}\sqrt{45} \\ &= \dfrac{8}{15}\sqrt5. \end{aligned}$$

Thus, the correct answer is D.

By symmetry, consider the first quadrant, where the equation is $x^2+y^2=x+y,$ or $\left(x-\tfrac12\right)^2+\left(y-\tfrac12\right)^2=\tfrac12.$ This is a circle centered at $\left(\tfrac12,\tfrac12\right)$ passing through $(1,0)$ and $(0,1);$ since the center is the midpoint of that chord, the enclosed first-quadrant region is the right triangle with legs to $(1,0)$ and $(0,1)$ (area $\tfrac12$) plus a semicircle of radius $\dfrac{\sqrt2}{2}$ (area $\dfrac\pi4$). Multiplying by $4$ for all quadrants gives $4\left(\tfrac12+\tfrac\pi4\right)=\pi+2.$

Thus, the correct answer is B.

A player's first head comes on flip $n$ with probability $\left(\tfrac12\right)^n.$ All three stopping on the same flip $n$ has probability $\left(\left(\tfrac12\right)^n\right)^3=\left(\tfrac18\right)^n.$ Summing over $n\ge1,$ $\displaystyle\sum_{n=1}^\infty\left(\tfrac18\right)^n =\frac{1/8}{1-1/8}=\frac17.$

Thus, the correct answer is B.

Since each team won $10$ and lost $10,$ there are $21$ teams and $\binom{21}{3}=1330$ triples. A triple is not cyclic exactly when one team beats both others. Choosing that team ($21$ ways) and $2$ of the $10$ teams it beat gives $21\cdot\binom{10}{2}=21\cdot45=945$ non-cyclic triples. Thus the cyclic triples number $1330-945=385.$

Thus, the correct answer is A.