Answer: 150

Small Hint:

Tom runs $5$ times as fast as he swims and bicycles $10$ times as fast, so write each leg’s time using only his swimming speed

Big Hint:

The total time is $\frac{0.5}{s} + \frac{30}{10s} + \frac{8}{5s} = 4.25$ hours; solve for the swimming speed $s$

Solution:

Let Tom’s swimming speed be $s$ miles per hour. Then he runs at $5s$ and bicycles at $10s.$ The total time in hours is $$\begin{aligned} &\frac{0.5}{s} + \frac{30}{10s} + \frac{8}{5s} \\ &= \frac{0.5 + 3 + 1.6}{s} \\ &= \frac{5.1}{s} = 4.25, \end{aligned}$$ so $s = \frac{5.1}{4.25} = 1.2$ miles per hour.

He bicycles at $12$ miles per hour, so the ride takes $\frac{30}{12} = 2.5$ hours, which is $150$ minutes.

Answer: 200

Small Hint:

Divisibility by $5$ makes the last digit $0$ or $5,$ and the first digit cannot be $0$

Big Hint:

With both outer digits $5,$ pick the second and third digits freely; exactly $2$ values of the fourth digit make the digit sum a multiple of $5$

Solution:

Since $n$ is divisible by $5,$ its last digit is $0$ or $5;$ since the first digit equals the last digit and cannot be $0,$ both are $5.$ The outer digits contribute $10$ to the digit sum, so the three middle digits must also sum to a multiple of $5.$

Choose the second and third digits freely, in $10 \cdot 10 = 100$ ways. Whatever their sum is, the fourth digit must land in a prescribed residue class modulo $5,$ and exactly $2$ of the digits $0$ through $9$ lie in each class. The count is $10 \cdot 10 \cdot 2 = 200.$

Answer: 18

Small Hint:

Let $AE = x$ and $EB = y;$ the two squares then have sides $x$ and $y,$ and $ABCD$ has side $x + y$

Big Hint:

Expand $x^2 + y^2 = \frac{9}{10}(x+y)^2$ and notice that the requested quantity equals $\frac{x^2 + y^2}{xy}$

Solution:

Let $AE = x$ and $EB = y,$ so the square has side $x + y$ and the two smaller squares have sides $x$ and $y.$ The condition says $$x^2 + y^2 = \frac{9}{10}(x + y)^2.$$ Multiplying by $10$ and expanding, $10x^2 + 10y^2$ $= 9x^2 + 18xy + 9y^2,$ so $x^2 + y^2 = 18xy.$

Dividing by $xy$ gives $$\begin{aligned} \frac{AE}{EB} + \frac{EB}{AE} &= \frac{x}{y} + \frac{y}{x} \\ &= \frac{x^2 + y^2}{xy} = 18. \end{aligned}$$

Answer: 429

Small Hint:

A coloring unchanged by the rotation is determined by one L-shaped arm: the four arms must be colored identically

Big Hint:

To get $8$ red and $5$ blue, the center must be blue and each arm needs $2$ red and $1$ blue; compare with $\binom{13}{5}$ total colorings

Solution:

The rotation cycles the four L-shaped arms, so a symmetric coloring colors all four arms identically, and the $12$ outer squares contain $4$ copies of whatever the arm shows. The number of red squares among the outer twelve is therefore a multiple of $4.$ Since there are $8$ red squares in all, the center must be blue and each arm must contain exactly $2$ red squares and $1$ blue square.

The blue square within the arm can be chosen in $3$ ways, so exactly $3$ of the $\binom{13}{5} = 1287$ equally likely colorings are symmetric. The probability is $\frac{3}{1287} = \frac{1}{429},$ so $n = 429.$

Answer: 98

Small Hint:

Move $x^3 + 3x^2 + 3x + 1$ to one side: the equation becomes $(x+1)^3 = 9x^3$

Big Hint:

Take cube roots to get $x = \frac{1}{\sqrt[3]{9} - 1},$ then rationalize using $u^3 - 1 = (u - 1)(u^2 + u + 1)$

Solution:

Rewrite the equation as $9x^3 = x^3 + 3x^2 + 3x + 1$ $= (x + 1)^3.$ Taking real cube roots, $\sqrt[3]{9}\,x = x + 1,$ so $$x = \frac{1}{\sqrt[3]{9} - 1}.$$

Multiply numerator and denominator by $\sqrt[3]{81} + \sqrt[3]{9} + 1;$ the denominator becomes $(\sqrt[3]{9})^3 - 1 = 8,$ so $$x = \frac{\sqrt[3]{81} + \sqrt[3]{9} + 1}{8}.$$ Thus $a + b + c = 81 + 9 + 8 = 98.$

Answer: 47

Small Hint:

Compute separately the probability that all three math books land in the $3$-box, the $4$-box, and the $5$-box

Big Hint:

For the box holding $k$ books, that probability is $\frac{\binom{9}{k-3}}{\binom{12}{k}};$ add the three fractions

Solution:

Focus on one box at a time. The box of $k$ books receives a uniformly random $k$-subset of the $12$ books, so the probability that it contains all three math books is $\frac{\binom{9}{k-3}}{\binom{12}{k}}.$ For $k = 3,$ $4,$ and $5$ this gives $\frac{1}{220},$ $\frac{9}{495} = \frac{1}{55},$ and $\frac{36}{792} = \frac{1}{22}.$

The events are disjoint, so the total probability is $$\begin{aligned} \frac{1}{220} + \frac{1}{55} + \frac{1}{22} &= \frac{1 + 4 + 10}{220} \\ &= \frac{15}{220} = \frac{3}{44}, \end{aligned}$$ and $m + n = 3 + 44 = 47.$

Answer: 41

Small Hint:

Place the corner at the origin so the box is $[0,12] \times [0,16] \times [0,h],$ and write down the three face centers

Big Hint:

The centers are $(6,8,0),$ $(0,8,\frac{h}{2}),$ $(6,0,\frac{h}{2});$ a cross product gives the triangle’s area as $\frac{1}{2}\sqrt{25h^2 + 2304}$

Solution:

Let the height be $h$ and place the corner at the origin, so the box is $[0,12] \times [0,16] \times [0,h].$ The three faces meeting at the origin have centers $P = (6, 8, 0),$ $Q = \left(0, 8, \tfrac{h}{2}\right),$ and $R = \left(6, 0, \tfrac{h}{2}\right).$

Then $\overrightarrow{PQ} = \left(-6, 0, \tfrac{h}{2}\right)$ and $\overrightarrow{PR} = \left(0, -8, \tfrac{h}{2}\right),$ whose cross product is $(4h, 3h, 48).$ The area is $$\begin{aligned} &\frac{1}{2}\sqrt{16h^2 + 9h^2 + 48^2} \\ &= \frac{1}{2}\sqrt{25h^2 + 2304} \\ &= 30, \end{aligned}$$ so $25h^2 = 3600 - 2304 = 1296$ and $h = \frac{36}{5}.$

Therefore $m + n = 36 + 5 = 41.$

Answer: 371

Small Hint:

The domain requires $-1 \le \log_m(nx) \le 1,$ so $x$ runs over $\left[\frac{1}{mn}, \frac{m}{n}\right],$ which has length $\frac{m^2 - 1}{mn}$

Big Hint:

Then $mn = 2013(m^2 - 1).$ Since $\gcd(m, m^2 - 1) = 1,$ $m$ must divide $2013,$ and the smallest $m \gt 1$ minimizes the sum.

Solution:

The function is defined when $-1 \le \log_m(nx) \le 1,$ that is $\frac{1}{m} \le nx \le m,$ so the domain is $\left[\frac{1}{mn}, \frac{m}{n}\right],$ with length $$\frac{m}{n} - \frac{1}{mn} = \frac{m^2 - 1}{mn} = \frac{1}{2013}.$$ Hence $n = \frac{2013(m^2 - 1)}{m}.$ Since $m$ is relatively prime to $m^2 - 1,$ $m$ must divide $2013 = 3 \cdot 11 \cdot 61.$

Because $n \approx 2013m,$ the sum $m + n$ grows with $m,$ so take the smallest factor $m = 3:$ then $n = \frac{2013 \cdot 8}{3} = 5368$ and $m + n = 5371.$

The remainder upon division by $1000$ is $371.$

Answer: 113

Small Hint:

Folding preserves distances: if the crease meets $\overline{AB}$ at $P,$ then $P$ is equidistant from $A$ and from the landing point on $\overline{BC}$

Big Hint:

The law of cosines in the $60^\circ$ corner triangles at $B$ and $C$ locates both crease endpoints; then apply the law of cosines once more at $A$

Solution:

Let $A'$ be the landing point, with $BA' = 9$ and $CA' = 3,$ and let the crease meet $\overline{AB}$ at $P$ and $\overline{AC}$ at $Q.$ Folding preserves distances, so $PA' = PA = x$ and $QA' = QA = y.$ In triangle $PBA',$ with $PB = 12 - x$ and $\angle B = 60^\circ,$ the law of cosines gives $$\begin{aligned} x^2 &= (12 - x)^2 + 81 \\ &\quad {}- 9(12 - x), \end{aligned}$$ which simplifies to $15x = 117,$ so $x = \frac{39}{5}.$ Similarly, in triangle $QCA',$ $y^2 = (12 - y)^2 + 9 - 3(12 - y)$ gives $21y = 117,$ so $y = \frac{39}{7}.$

Finally, in triangle $APQ$ with $\angle A = 60^\circ,$ $$\begin{aligned} PQ^2 &= x^2 + y^2 - xy \\ &= 39^2\left(\frac{1}{25} + \frac{1}{49} - \frac{1}{35}\right) \\ &= 39^2 \cdot \frac{49 + 25 - 35}{1225} \\ &= \frac{39^3}{1225}, \end{aligned}$$ so $PQ = \frac{39\sqrt{39}}{35}.$ Thus $m + n + p = 39 + 35 + 39$ $= 113.$

Answer: 80

Small Hint:

Real coefficients force the zeros to be $r \pm si$ and a real $q$ with $q(r^2 + s^2) = 65$

Big Hint:

List the ways a factor of $65$ is a sum of two nonzero squares; for each, the choices $\pm r$ cancel in the sum, leaving only the $q$’s

Solution:

Since $P$ has real coefficients, $r - si$ is also a zero, and the third zero $q$ is real. The sum of the zeros is $q + 2r = a,$ so $q = a - 2r$ is a nonzero integer. Their product is $q(r^2 + s^2) = 65,$ so $r^2 + s^2$ is a factor of $65.$ With $r$ and $s$ nonzero, the possibilities are $r^2 + s^2 = 5 = 1^2 + 2^2$ (with $q = 13$), $13 = 2^2 + 3^2$ (with $q = 5$), and $65 = 1^2 + 8^2 = 4^2 + 7^2$ (with $q = 1$).

For each representation $\{u, v\},$ the zero $r + si$ can have $r = \pm u$ or $\pm v,$ giving $4$ distinct polynomials (the sign of $s$ changes nothing). The sum of the zeros is $p_{a,b} = q + 2r,$ and over the four choices the $2r$ terms cancel, leaving $4q$ from each representation.

The total is $4 \cdot 13 + 4 \cdot 5 + 4 \cdot 1 + 4 \cdot 1 = 80.$

Answer: 148

Small Hint:

$N$ is a multiple of $\operatorname{lcm}(14, 15, 16) = 1680,$ and every positive integer below $14$ except $9,$ $11,$ and $13$ divides $1680$

Big Hint:

So $x,$ $y,$ and $z$ must be $9,$ $11,$ and $13:$ solve $1680m \equiv 3$ modulo $9,$ $11,$ and $13$ by the Chinese remainder theorem

Solution:

Divisibility by $16,$ $15,$ and $14$ means $N = 1680m$ where $1680 = \operatorname{lcm}(14, 15, 16)$ $= 2^4 \cdot 3 \cdot 5 \cdot 7.$ Every positive integer less than $14$ divides $1680$ except $9,$ $11,$ and $13,$ and a divisor of $N$ leaves remainder $0,$ not $3.$ So necessarily $\{x, y, z\} = \{9, 11, 13\},$ and we need $1680m \equiv 3$ modulo each of $9,$ $11,$ $13.$

Since $1680 \equiv 6 \pmod 9,$ the first congruence is $6m \equiv 3 \pmod 9,$ i.e. $m \equiv 2 \pmod 3.$ Since $1680 \equiv 8 \pmod{11},$ we need $8m \equiv 3 \pmod{11},$ i.e. $m \equiv 10 \pmod{11}.$ Since $1680 \equiv 3 \pmod{13},$ we need $m \equiv 1 \pmod{13}.$ By the Chinese remainder theorem these combine to $m \equiv 131 \pmod{429},$ so the least $m$ is $131.$

Then $N = 1680 \cdot 131$ $= 2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 131,$ and since $131$ is prime, the sum of the distinct prime divisors is $2 + 3 + 5 + 7 + 131 = 148.$

Answer: 21

Small Hint:

The hexagon’s $120^\circ$ angles make the corner triangle at $Q$ equilateral, so $QB = QC = 1;$ set up coordinates with $Q$ at the origin and $QR$ along the $x$-axis

Big Hint:

Vertex $F$ sits directly above $D$ at height $\sqrt{3},$ and the $45^\circ$ line $RP$ through $F$ gives $QR = 2 + \sqrt{3}$

Solution:

Note $\angle R = 45^\circ.$ Because the hexagon’s interior angles are $120^\circ,$ segment $\overline{BC}$ cuts off a corner triangle at $Q$ with two $60^\circ$ base angles, so triangle $BQC$ is equilateral and $QB = QC = 1.$ Put $Q$ at the origin with $QR$ along the positive $x$-axis. Then $C = (1, 0),$ $D = (2, 0),$ and the hexagon’s vertices are $B = \left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right),$ $A = (1, \sqrt{3}),$ $F = (2, \sqrt{3}),$ $E = \left(\tfrac{5}{2}, \tfrac{\sqrt{3}}{2}\right).$

Since $\angle R = 45^\circ,$ line $RP$ has slope $-1.$ If it passed through $E,$ it would be $x + y = \tfrac{5 + \sqrt{3}}{2},$ which puts $F$ (with $x + y = 2 + \sqrt{3}$) outside the triangle; so the vertex on $\overline{RP}$ is $F,$ and $RP$ is the line $x + y = 2 + \sqrt{3}.$ It meets the $x$-axis at $R = (2 + \sqrt{3},\ 0)$ and the line $y = \sqrt{3}\,x$ (line $QP$) where $x(1 + \sqrt{3}) = 2 + \sqrt{3},$ giving $P$ height $$y = \frac{\sqrt{3}(2 + \sqrt{3})}{1 + \sqrt{3}} = \frac{3 + \sqrt{3}}{2}.$$

The area is $\frac{1}{2} \cdot QR \cdot y$ $= \frac{1}{2}(2 + \sqrt{3}) \cdot \frac{3 + \sqrt{3}}{2}$ $= \frac{9 + 5\sqrt{3}}{4},$ so $a + b + c + d = 9 + 5 + 3 + 4$ $= 21.$

Answer: 961

Small Hint:

Match up corresponding sides: $\triangle B_0C_1C_0 \sim \triangle AB_0C_0$ has ratio $\frac{B_0C_0}{AC_0} = \frac{17}{25}$

Big Hint:

With $r = \frac{17}{25},$ triangle $AB_1C_1$ has ratio $1 - r^2,$ so the disjoint triangles’ areas form a geometric series with first term $90r^2(1 - r^2)$ and ratio $(1 - r^2)^2$

Solution:

By Heron’s formula with $s = 27,$ the area of $\triangle AB_0C_0$ is $\sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90.$ In the similarity $\triangle B_0C_1C_0 \sim \triangle AB_0C_0,$ side $B_0C_0$ corresponds to $AC_0,$ so the ratio is $r = \frac{17}{25},$ and $C_1C_0$ (corresponding to $B_0C_0$) equals $17r.$ Hence $$\frac{AC_1}{AC_0} = \frac{25 - 17r}{25} = 1 - r^2,$$ which is the similarity ratio of $\triangle AB_1C_1$ to $\triangle AB_0C_0.$