Solution:

The bug crawls $|-6-(-2)| = 4$ units on the first leg and $|5-(-6)| = 11$ units on the second leg.

The total distance is $4 + 11 = 15.$

Thus, the correct answer is E.

Solution:

In $5$ minutes there are $300$ seconds. Cagney frosts $\dfrac{300}{20} = 15$ cupcakes and Lacey frosts $\dfrac{300}{30} = 10$ cupcakes.

Together they frost $15 + 10 = 25$ cupcakes.

Thus, the correct answer is D.

Solution:

The second box has $2 \times 3 \times 1 = 6$ times the volume of the first, so it holds $6$ times as much clay.

Therefore $n = 6 \cdot 40 = 240.$

Thus, the correct answer is D.

Solution:

Suppose there are $5$ marbles: $3$ blue and $2$ red. Doubling the red gives $4$ red while the blue stays at $3.$

The total is now $3 + 4 = 7,$ so the fraction that is red is $\dfrac{4}{7}.$

Thus, the correct answer is C.

Solution:

Let $b$ be the number of blueberries. Then there are $2b$ raspberries, $4 \cdot 2b = 8b$ cherries, and $3 \cdot 8b = 24b$ grapes.

The total is $b + 2b + 8b + 24b = 35b = 280,$ so $b = 8$ and there are $8b = 64$ cherries.

Thus, the correct answer is D.

Solution:

Let the numbers be $a \lt b \lt c.$ Adding the three pairwise sums gives $2(a+b+c) = 12+17+19 = 48,$ so $a+b+c = 24.$

Then $a = 24-19 = 5,$ $b = 24-17 = 7,$ and $c = 24-12 = 12.$ The middle number is $7.$

Thus, the correct answer is D.

Solution:

Let $a$ be the smallest angle and $d \ge 0$ the common difference. The sum of the angles is $12a + 66d = 360,$ so $2a + 11d = 60.$

To make $a$ small, take $d$ large. Since $11d$ must be even, $d$ is even, and $d = 4$ gives $2a = 60 - 44 = 16,$ so $a = 8.$ A larger even $d$ makes $a$ non-positive.

Thus, the correct answer is C.

Solution:

For the order $a, b, c, d, e,$ the iterative average is $$\frac{a + b + 2c + 4d + 8e}{16}.$$ The later positions carry the most weight.

The largest value uses $(a,b,c,d,e) = (1,2,3,4,5),$ giving $\dfrac{65}{16},$ and the smallest uses $(5,4,3,2,1),$ giving $\dfrac{31}{16}.$

The difference is $\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}.$

Thus, the correct answer is C.

Solution:

From February $7, 1812$ to February $7, 2012$ there are $200 \cdot 365 = 73000$ ordinary days plus one for each leap day.

One quarter of the $200$ years contain a leap day, except $1900,$ giving $\tfrac14 \cdot 200 - 1 = 49$ leap days. So the span is $73049$ days.

Since $73049 = 7 \cdot 10435 + 4,$ the birth day was $4$ days before a Tuesday, which is a Friday.

Thus, the correct answer is A.

Solution:

The median divides the triangle into two triangles of equal area $15.$ One of them has the two sides of length $5$ (half the base) and $9$ (the median) meeting at angle $\theta.$

Its area is $\tfrac12 \cdot 5 \cdot 9 \sin\theta = 15,$ so $\sin\theta = \dfrac{2 \cdot 15}{5 \cdot 9} = \dfrac{2}{3}.$

Thus, the correct answer is D.

Solution:

Since Alex wins with probability $\tfrac12,$ the others share the remaining $\tfrac12.$ With Mel twice as likely as Chelsea, $P(\text{Mel}) = \tfrac13$ and $P(\text{Chelsea}) = \tfrac16.$

The number of orderings of the wins $AAAMMC$ is $\dfrac{6!}{3!\,2!\,1!} = 60.$ The probability is $$60 \cdot \left(\tfrac12\right)^3 \left(\tfrac13\right)^2 \left(\tfrac16\right) = \frac{60}{432} = \frac{5}{36}.$$

Thus, the correct answer is B.

Solution:

By symmetry let $A = (a, b)$ with $a \gt 0$ and $B = (-a, b).$ The square sits on the tangent point $(0,1),$ so its horizontal width is $2a$ and its height is $b - 1.$

Since these are equal, $2a = b - 1,$ giving $b = 2a + 1.$

Substituting into $a^2 + b^2 = 4$ yields $5a^2 + 4a - 3 = 0.$ The positive root is $a = \dfrac{\sqrt{19} - 2}{5},$ so the side length is $2a = \dfrac{2\sqrt{19} - 4}{5}.$

Thus, the correct answer is D.

Solution:

Let $m$ be the lunch length in minutes. The three worked $480 - m$ minutes Monday, the helpers $372 - m$ minutes Tuesday, and Paula $672 - m$ minutes Wednesday.

If Paula paints $p\%$ per minute and the helpers together paint $h\%$ per minute, then $$(p+h)(480-m) = 50,\quad h(372-m) = 24,\quad p(672-m) = 26.$$

Adding the last two equations and subtracting from the first gives $108h - 192p = 0,$ so $h = \tfrac{16}{9}p.$ Solving the system gives $p = \tfrac{1}{24}$ and $m = 48.$

Thus, the correct answer is D.

Solution:

Each arc has length $\dfrac{2\pi}{3}$ on a unit circle, so it is a $120^\circ$ sector. The nine equal sectors can be reassembled so that the enclosed region equals the regular hexagon of side $2$ plus one full circle of radius $1.$

A regular hexagon of side $2$ splits into $6$ equilateral triangles of side $2,$ so its area is $6 \cdot \dfrac{\sqrt3}{4} \cdot 2^2 = 6\sqrt3.$

Adding the unit circle's area $\pi$ gives $\pi + 6\sqrt3.$

Thus, the correct answer is E.

Solution:

The four corners form one cycle under the rotation, the four edge squares form another, and the center is fixed. These three groups are independent.

For the four corners, checking the $2^4 = 16$ colorings shows that $7$ of them end all black, so the corners are all black with probability $\dfrac{7}{16}.$ The same holds for the four edges.

The center is black at the end only if it started black, with probability $\dfrac12.$ Multiplying, the whole grid is black with probability $$\frac12 \cdot \left(\frac{7}{16}\right)^2 = \frac{49}{512}.$$

Thus, the correct answer is A.

Solution:

Let $r$ be the radius of $C_1,$ so $OX = OY = r.$ These are equal chords of $C_2,$ so they subtend equal angles at $Z:$ $\angle XZO = \angle OZY.$

Applying the Law of Cosines to triangles $XZO$ and $YZO,$ $$\frac{13^2 + 11^2 - r^2}{2 \cdot 13 \cdot 11} = \frac{7^2 + 11^2 - r^2}{2 \cdot 7 \cdot 11}.$$

Clearing denominators and solving gives $r^2 = 30,$ so $r = \sqrt{30}.$

Thus, the correct answer is E.

Solution:

Group $\{1, \ldots, 30\}$ by residue modulo $5;$ each class has $6$ numbers. A sum is divisible by $5$ when the residues are $0{+}0,$ $1{+}4,$ or $2{+}3.$

So $S$ can use at most one number $\equiv 0,$ and only one of the classes $\{1\}, \{4\}$ and only one of $\{2\}, \{3\}.$ That allows at most $1 + 6 + 6 = 13$ numbers.

The set $\{1, 2, 6, 7, 11, 12, 16, 17, 21, 22, 26, 27, 30\}$ achieves $13,$ so the maximum is $13.$

Thus, the correct answer is B.

Solution:

Let $D$ be the foot of the perpendicular from the incenter $I$ to $BC.$ The tangent length $BD = s - AC,$ where $s = \tfrac12(25 + 26 + 27) = 39,$ so $BD = 39 - 26 = 13.$

By Heron's formula the area is $\sqrt{39 \cdot 14 \cdot 13 \cdot 12},$ and the inradius satisfies $r^2 = \dfrac{(s-a)(s-b)(s-c)}{s} = \dfrac{14 \cdot 13 \cdot 12}{39} = 56.$

In right triangle $BDI,$ $BI^2 = r^2 + BD^2 = 56 + 169 = 225,$ so $BI = 15.$

Thus, the correct answer is A.

Solution:

Model people as vertices of a graph, with edges for friendships. Everyone has the same degree $n$ with $1 \le n \le 4.$ The cases $n$ and $6 - 1 - n$ are complementary graphs, so $n = 1$ pairs with $n = 4$ and $n = 2$ with $n = 3.$

For $n = 1$ the graph is a perfect matching: $5 \cdot 3 = 15$ ways. Thus $n = 4$ also gives $15.$

For $n = 2$ the graph is a union of cycles: either two triangles $\left(\binom{5}{2} = 10\right)$ or one hexagon $\left(\dfrac{6!}{12} = 60\right),$ totaling $70.$ Thus $n = 3$ also gives $70.$

The total is $15 + 15 + 70 + 70 = 170.$

Thus, the correct answer is B.

Solution:

Expanding the product, a term of degree $2012$ comes from choosing $x^{2^k}$ from some factors so that the exponents sum to $2012.$ Since powers of two are distinct, this corresponds to the binary representation $2012 = 11111011100_2.$

That representation is unique, so exactly one term gives $x^{2012},$ and its coefficient is the product of the constants $2^k$ from the remaining factors: those with $k \in \{0, 1, 5\}.$

The coefficient is $2^0 \cdot 2^1 \cdot 2^5 = 2^6,$ so $a = 6.$

Thus, the correct answer is B.

Solution:

Adding the two equations gives $2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 14,$ that is, $$(a-b)^2 + (b-c)^2 + (c-a)^2 = 14.$$

The only way to write $14$ as a sum of three squares is $9 + 4 + 1.$ Since $a \ge b \ge c,$ we get $a - c = 3,$ with either $(a-b, b-c) = (2,1)$ or $(1,2).$

Substituting $(a, b, c) = (c+3, c+1, c)$ into the first equation gives $3(2c+3) + 2(c+1) = 2011,$ so $c = 250$ and $(a, b, c) = (253, 251, 250).$ The other case has no integer solution.

Thus, the correct answer is E.

Solution:

On every face, the required segments join midpoints of edges. A plane cutting the cube meets the faces in one of four symmetric shapes: a square through midpoints ($3$ such planes), a rectangle per edge ($12$ planes), a triangle per vertex ($8$ planes), or a regular hexagon per pair of opposite vertices ($4$ planes).

Using all of them gives the maximum $k = 3 + 12 + 8 + 4 = 27.$

The full figure consists of $24$ short segments and $12$ long segments. The $4$ hexagon planes together contain all $24$ short segments, and the $3$ square planes contain all $12$ long segments, so the minimum is $k = 4 + 3 = 7.$

The difference is $27 - 7 = 20.$

Thus, the correct answer is C.

Solution:

The diagonal from $(0.1, 0.7)$ to $(-0.1, -0.7)$ has length $\sqrt{0.2^2 + 1.4^2} = \sqrt2,$ so $S$ is a square of area $1.$ The translate $T(v)$ contains a lattice point exactly when $v$ lies inside the copy of $S$ centered at that point.

Containing exactly two interior lattice points requires $v$ to lie in the overlap of two copies centered at adjacent lattice points. By periodicity the answer is the total such overlap area within one unit cell.

The overlap of two unit-area copies whose centers are one unit apart has area $0.08.$ Summing over the horizontal and vertical adjacencies gives probability $2 \cdot 0.08 = 0.16.$

Thus, the correct answer is C.

Solution:

Because each base lies strictly between $0$ and $1,$ the function $t \mapsto (\text{base})^t$ is decreasing, while $t \mapsto t^b$ is increasing for $b \gt 0.$ Comparing terms shows the sequence orders as $$1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010} \gt a_{2011} \gt a_{2009} \gt \cdots \gt a_1 \gt 0.$$

So in the decreasing arrangement, the even-indexed terms come first, then the odd-indexed terms in reverse. A term satisfies $a_k = b_k$ exactly when its position equals its index, which for the descending odd tail requires $2(k - 1006) = 2011 - k.$

Solving gives $3k = 4023,$ so $k = 1341,$ the unique fixed index, and the sum is $1341.$

Thus, the correct answer is C.

Solution:

Since $0 \le f(x) \le 1,$ every solution lies in $[0, n].$ The function $f$ is a triangular wave of period $1,$ and $g(x) = xf(x)$ is monotonic on each half-integer interval, mapping it onto an interval on which $f(g(x))$ oscillates.

Counting the oscillations, on the intervals $[a, a + \tfrac12)$ and $[a + \tfrac12, a+1)$ the curve $y = f(g(x))$ meets the line $y = \tfrac{x}{n}$ a total of $2a$ and $2(a+1)$ times. Summing over $a = 0, \ldots, n-1$ gives $$\sum_{a=0}^{n-1}\bigl(2a + 2(a+1)\bigr) = 2n^2$$ real solutions.

The smallest $n$ with $2n^2 \ge 2012$ is $n = 32,$ since $2 \cdot 31^2 = 1922$ and $2 \cdot 32^2 = 2048.$

Thus, the correct answer is C.