Small Hint:

$a^{-2} = \dfrac{1}{a^2}$

Big Hint:

$(-2)^2 = 4,$ so subtract $\dfrac14$ from $2$

Solution:

Since $(-2)^{-2} = \dfrac{1}{(-2)^2} = \dfrac14,$ we get $2 - \dfrac14 = \dfrac74.$

Thus, the correct answer is C.

Small Hint:

The first two tasks span $1{:}00$ to $2{:}40,$ which is $100$ minutes

Big Hint:

Divide by $2$ to get one task’s length, then add it after $2{:}40$

Solution:

The first two tasks together take $100$ minutes, so each task takes $50$ minutes.

The third task finishes $50$ minutes after $2{:}40$ PM, at $3{:}30$ PM.

Thus, the correct answer is B.

Small Hint:

Let $x$ appear twice and $y$ appear three times, so $2x + 3y = 100$

Big Hint:

Test whether $28$ is the number written two times or three times, and keep the integer case

Solution:

Write $2x + 3y = 100.$ If $28$ were written twice, then $3y = 100 - 56 = 44,$ which is not a multiple of $3.$

So $28$ is written three times: $2x = 100 - 84 = 16,$ giving $x = 8.$

Thus, the correct answer is A.

Small Hint:

Start from Marta in $6$th place and use each stated gap

Big Hint:

Marta is $1$ behind Jack, so Jack is $5$th; chain to Todd, then Rand, then Hikmet

Solution:

Marta is $6$th, so Jack is $5$th. Jack is $2$ behind Todd, so Todd is $3$rd. Todd is $1$ behind Rand, so Rand is $2$nd.

Rand is $6$ ahead of Hikmet, so Hikmet is $8$th. (David is $10$th.)

Thus, the correct answer is B.

Small Hint:

To minimize $N,$ let the Sharks win all $N$ additional games

Big Hint:

Solve $\dfrac{1 + N}{3 + N} \ge \dfrac{95}{100}$

Solution:

The Sharks won $1$ of the first $3$ games. To reach $95\%$ with the fewest extra games, they should win all $N$ additional games, giving a win fraction $\dfrac{1 + N}{3 + N}.$

Requiring $\dfrac{1 + N}{3 + N} \ge \dfrac{19}{20}$ gives $20 + 20N \ge 57 + 19N,$ so $N \ge 37.$

Thus, the correct answer is B.

Small Hint:

A product is odd only when both factors are odd

Big Hint:

Count the odd numbers from $0$ to $12,$ then compare with the $13 \times 13$ entries

Solution:

The body has $13 \times 13 = 169$ entries. A product is odd exactly when both factors are odd.

There are $6$ odd numbers among $0, 1, \ldots, 12,$ giving $6 \times 6 = 36$ odd entries. The fraction is $\dfrac{36}{169} = 0.213\ldots \approx 0.21.$

Thus, the correct answer is A.

Small Hint:

A regular $n$-gon has $n$ lines of symmetry

Big Hint:

Its smallest rotational symmetry angle is $\dfrac{360^\circ}{n}$

Solution:

A regular $15$-gon has $L = 15$ lines of symmetry, and its smallest angle of rotational symmetry is $R = \dfrac{360}{15} = 24$ degrees.

Then $L + R = 15 + 24 = 39.$

Thus, the correct answer is D.

Small Hint:

$625 = 5^4$

Big Hint:

Use $5^{\log_5 x} = x$ after collecting the exponents

Solution:

Since $625 = 5^4,$ we have $$\begin{gathered} 625^{\log_5 2015} = 5^{4\log_5 2015} \\ = \left(5^{\log_5 2015}\right)^4 = 2015^4. \end{gathered}$$

Taking the fourth root gives $\left(2015^4\right)^{\frac{1}{4}} = 2015.$

Thus, the correct answer is D.

Small Hint:

Larry wins immediately with probability $\dfrac12$

Big Hint:

If both miss (probability $\dfrac14$), the game resets; solve $x = \dfrac12 + \dfrac14 x$

Solution:

Let $x$ be the probability Larry wins. He wins right away with probability $\dfrac12,$ or both players miss (probability $\dfrac14$) and the game restarts.

So $x = \dfrac12 + \dfrac14 x,$ giving $\dfrac34 x = \dfrac12$ and $x = \dfrac23.$

Thus, the correct answer is C.

Small Hint:

With sides $a \lt b \lt c,$ the triangle inequality gives perimeter $\gt 2c$

Big Hint:

This forces $c \le 6;$ list scalene triples and drop the right triangle $3, 4, 5$

Solution:

Let the distinct sides be $a \lt b \lt c.$ Since $a + b \gt c,$ the perimeter exceeds $2c,$ so $2c \lt 15$ and $c \le 6.$

The scalene triples with perimeter less than $15$ are $(6,5,3),$ $(6,5,2),$ $(6,4,3),$ $(5,4,3),$ $(5,4,2),$ and $(4,3,2).$ Of these, only $(5,4,3)$ is a right triangle, leaving $5.$

Thus, the correct answer is C.

Small Hint:

The intercepts are $(5, 0)$ and $(0, 12),$ so the legs are $5$ and $12$

Big Hint:

Two altitudes are the legs; the third is $\dfrac{2 \cdot \text{area}}{\text{hypotenuse}}$

Solution:

The line meets the axes at $(5, 0)$ and $(0, 12),$ so the triangle is right with legs $5$ and $12$ and hypotenuse $13.$ Its area is $30.$

Two altitudes are the legs $5$ and $12;$ the altitude to the hypotenuse is $\dfrac{2 \cdot 30}{13} = \dfrac{60}{13}.$ The sum is $17 + \dfrac{60}{13} = \dfrac{281}{13}.$

Thus, the correct answer is E.

Small Hint:

Factor out the common $(x - b)$

Big Hint:

The roots are $b$ and $\dfrac{a + c}{2};$ maximize $b + \dfrac{a + c}{2}$ over distinct digits

Solution:

Factoring gives $(x - b)\bigl(2x - (a + c)\bigr) = 0,$ so the roots are $b$ and $\dfrac{a + c}{2}.$ Their sum is $b + \dfrac{a + c}{2}.$

Using distinct digits, take $b = 9$ and $a + c = 8 + 7 = 15,$ giving $9 + 7.5 = 16.5.$

Thus, the correct answer is D.

Small Hint:

$\angle BAC$ and $\angle BDC$ subtend the same arc $BC$

Big Hint:

Find $\angle ABC,$ then compare it with $\angle BAC$ to show $\triangle ABC$ is isosceles

Solution:

Angles $BAC$ and $BDC$ subtend arc $BC,$ so $\angle BDC = 70^\circ.$ Then $\angle ADC = \angle ADB$ $+ \angle BDC = 110^\circ.$

Since $ABCD$ is cyclic, $\angle ABC = 180^\circ - 110^\circ = 70^\circ$ $= \angle BAC.$ Thus $\triangle ABC$ is isosceles with $AC = BC = 6.$

Thus, the correct answer is B.

Small Hint:

Let $z$ be the overlap area; the requested difference is $(\text{circle} - z) - (\text{triangle} - z)$

Big Hint:

That equals circle area minus triangle area; use $\pi r^2$ and $\dfrac{\sqrt3}{4}s^2$

Solution:

Let $z$ be the area shared by the circle and triangle. The requested difference is $$\begin{aligned} &(\text{circle} - z) - (\text{triangle} - z) \\ &= \text{circle} - \text{triangle}. \end{aligned}$$

The circle has area $\pi \cdot 2^2 = 4\pi,$ and the equilateral triangle has area $\dfrac{\sqrt3}{4}\cdot 4^2 = 4\sqrt3.$ The difference is $4\pi - 4\sqrt3 = 4(\pi - \sqrt3).$

Thus, the correct answer is D.

Small Hint:

A $3.5$ GPA needs $14$ total points, so English and History must give at least $6$ points

Big Hint:

Find each class’s chance of a C, then add the probabilities of totals $6,$ $7,$ and $8$

Solution:

Math and Science give $8$ points, so Rachelle needs at least $6$ more from English and History. The chance of a C is $1 - \dfrac16 - \dfrac14 = \dfrac{7}{12}$ in English and $1 - \dfrac14 - \dfrac13 = \dfrac{5}{12}$ in History.

Working over a denominator of $144:$ $8$ points has probability $\dfrac16\cdot\dfrac14 = \dfrac{6}{144};$ $7$ points has $\dfrac16\cdot\dfrac13 + \dfrac14\cdot\dfrac14 = \dfrac{17}{144};$ and $6$ points has $\dfrac16\cdot\dfrac{5}{12} + \dfrac14\cdot\dfrac13$ $+ \dfrac{7}{12}\cdot\dfrac14 = \dfrac{43}{144}.$

The total is $\dfrac{6 + 17 + 43}{144} = \dfrac{66}{144} = \dfrac{11}{24}.$

Thus, the correct answer is D.

Small Hint:

In a regular hexagon of side $6,$ the distance from the center to a vertex is also $6$

Big Hint:

The apex sits above the center at height $\sqrt{8^2 - 6^2};$ use $V = \dfrac13 \cdot (\text{base area}) \cdot h$

Solution:

The distance from the hexagon’s center to a vertex is $6.$ A lateral edge has length $8,$ so the pyramid’s height is $\sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt7.$

The hexagon’s area is $\dfrac{3\sqrt3}{2}\cdot 6^2 = 54\sqrt3.$ Thus the volume is $\dfrac13 \cdot 54\sqrt3 \cdot 2\sqrt7 = 36\sqrt{21}.$

Thus, the correct answer is C.

Small Hint:

Set $\binom{n}{2}\left(\tfrac14\right)^2\left(\tfrac34\right)^{n-2}$ $= \binom{n}{3}\left(\tfrac14\right)^3\left(\tfrac34\right)^{n-3}$

Big Hint:

Cancel common factors to reach $\binom{n}{2}\cdot 3 = \binom{n}{3}$

Solution:

Setting the two probabilities equal and cancelling the common powers of $\tfrac14$ and $\tfrac34$ gives $\binom{n}{2}\cdot\dfrac34 = \binom{n}{3}\cdot\dfrac14.$

This becomes $\dfrac{n(n-1)}{2}\cdot 3$ $= \dfrac{n(n-1)(n-2)}{6},$ so $\dfrac32 = \dfrac{n-2}{6},$ giving $n - 2 = 9$ and $n = 11.$

Thus, the correct answer is D.

Small Hint:

The smallest value comes from the smallest composite, $4 = 2 \cdot 2$

Big Hint:

Show every integer greater than $3$ is attained, e.g. $r(2^k) = 2k$ and $r(2^k\cdot 3) = 2k + 3$

Solution:

A composite number has at least two prime factors (with multiplicity), and the smallest prime is $2,$ so the least possible value is $2 + 2 = 4.$

Every integer greater than $3$ is attained: $r(2^k) = 2k$ covers the even values $\ge 4,$ and $r(2^k\cdot 3) = 2k + 3$ covers the odd values $\ge 5.$ So the range is the integers greater than $3.$

Thus, the correct answer is D.

Small Hint:

The circle’s center $O$ is equidistant from $X, Y, Z, W,$ so it is the circumcenter of $\triangle ABC$

Big Hint:

Since $\angle C = 90^\circ,$ $O$ is the midpoint of $AB;$ set $a = \tfrac12 BC,$ $b = \tfrac12 CA$ and use $OX = OW$

Solution:

The center $O$ of the circle lies on the perpendicular bisectors of $XY$ and $ZW,$ which are the same as those of $AB$ and $AC.$ So $O$ is the circumcenter of $\triangle ABC,$ and since $\angle C = 90^\circ,$ $O$ is the midpoint of $AB.$

Let $a = \tfrac12 BC$ and $b = \tfrac12 CA.$ Then $a^2 + b^2 = 6^2,$ and computing $OX^2 = OW^2$ gives $12^2 + 6^2 = b^2 + (a + 2b)^2.$ Because the left side is $5(a^2+b^2),$ subtracting gives $4a(b-a)=0.$ Thus $a=b,$ and $2a^2=36$ gives $a=b=3\sqrt2.$ Therefore $BC=CA=6\sqrt2,$ and the perimeter is $12+12\sqrt2.$