Answer: 790

Solution:

The four eyelets on each $80$ mm side are equally spaced with one at each vertex, so consecutive eyelets on a side are $\frac{80}{3}$ mm apart. The lace consists of one segment across the $50$ mm width, six crisscross pieces (after the width crossing, each of the two strands makes three crossings to reach the top), and two free ends of at least $200$ mm each. The lace is shortest when every piece is a straight segment.

Each crisscross piece spans the full width and rises one gap, so its length is $$\sqrt{50^2 + \left(\tfrac{80}{3}\right)^2} = \sqrt{\tfrac{28900}{9}} = \frac{170}{3}.$$

The minimum length is $50 + 6 \cdot \frac{170}{3} + 2 \cdot 200 = 50 + 340 + 400 = 790.$

Answer: 144

Solution:

Both balls are green with probability $\frac{4}{10} \cdot \frac{16}{16+N},$ and both are blue with probability $\frac{6}{10} \cdot \frac{N}{16+N}.$ The condition is $$\frac{64 + 6N}{10(16 + N)} = \frac{29}{50}.$$

Clearing denominators, $5(64 + 6N) = 29(16 + N),$ so $320 + 30N = 464 + 29N,$ giving $N = 144.$

Answer: 200

Solution:

Write $r = \frac{a}{b}$ in lowest terms with $a + b = 1000;$ since $0 \lt r \lt 1,$ we need $1 \le a \le 499.$ Because $\gcd(a, b) = \gcd(a, 1000 - a) = \gcd(a, 1000),$ the fraction is in lowest terms exactly when $a$ is coprime to $1000.$

There are $\varphi(1000) = 1000 \cdot \frac{1}{2} \cdot \frac{4}{5} = 400$ integers in $[1, 999]$ coprime to $1000,$ and they pair up as $a \leftrightarrow 1000 - a$ (note $a = 500$ is not coprime to $1000$), so exactly $200$ of them are less than $500.$ The answer is $200.$

Answer: 49

Solution:

Let the eastbound and westbound trains have speeds $v_1$ and $v_2$ miles per hour and common length $L$ miles. A train passes a rider in time $L$ divided by their relative speed. Passing Jon (riding east at $20$) in $\frac{1}{60}$ hour gives $$\frac{L}{v_1 - 20} = \frac{L}{v_2 + 20} = \frac{1}{60},$$ so $v_1 = 60L + 20$ and $v_2 = 60L - 20.$

Relative to Steve (riding west at $20$), the speeds are $v_1 + 20$ and $v_2 - 20,$ and the westbound train takes $10$ times as long: $\frac{L}{v_2 - 20} = \frac{10L}{v_1 + 20},$ so $v_1 + 20 = 10(v_2 - 20).$ Substituting, $60L + 40 = 600L - 400,$ so $540L = 440$ and $L = \frac{22}{27}.$

Since $\gcd(22, 27) = 1,$ the answer is $22 + 27 = 49.$

Answer: 134

Solution:

A subset $Q$ is communal exactly when its vertices are consecutive around the $12$-gon. Indeed, a separating circle meets the circumcircle of the $12$-gon in at most two points, so the vertices inside it form a contiguous arc. Conversely, any run of consecutive vertices can be separated from the remaining vertices by a line, and a sufficiently large circle on the proper side of that line contains exactly that run.

For each size $k$ with $1 \le k \le 11$ there are $12$ runs of $k$ consecutive vertices (one starting at each vertex), giving $12 \cdot 11 = 132$ subsets, and the empty set and all of $S$ are also communal. The total is $132 + 2 = 134.$

Answer: 36

Solution:

Setting $x = 0$ gives $3h^2 + j = 2013$ and $2h^2 + k = 2014.$ Expanding, the first graph is $y = 3x^2 - 6hx + 2013,$ whose roots are positive integers with sum $2h$ and product $\frac{2013}{3} = 671 = 11 \cdot 61.$ Similarly the second is $y = 2x^2 - 4hx + 2014,$ with integer roots of sum $2h$ and product $\frac{2014}{2} = 1007 = 19 \cdot 53.$

The first pair of roots is $\{11, 61\}$ or $\{1, 671\},$ so $2h = 72$ or $672;$ the second pair is $\{19, 53\}$ or $\{1, 1007\},$ so $2h = 72$ or $1008.$ The only common value is $2h = 72,$ so $h = 36,$ which indeed gives $x$-intercepts $11, 61$ and $19, 53.$

Answer: 100

Solution:

Since $\frac{w-z}{z} = \frac{w}{z} - 1,$ and $\frac{w}{z}$ can be any complex number of modulus $\frac{1}{10},$ the point $\zeta = \frac{w-z}{z}$ ranges over the circle of radius $\frac{1}{10}$ centered at $-1.$

Because $\tan^2\theta$ is unchanged when $\theta$ shifts by $180^\circ,$ we want the largest angle $\alpha$ that a ray from the origin to this circle makes with the real axis. The extreme rays are tangent to the circle, where $\sin \alpha = \frac{1/10}{1} = \frac{1}{10}.$

Then $$\tan^2\theta = \frac{\sin^2\alpha}{1 - \sin^2\alpha} = \frac{1/100}{99/100} = \frac{1}{99},$$ so $p + q = 1 + 99 = 100.$

Answer: 937

Solution:

The condition is $N^2 \equiv N \pmod{10^4},$ that is, $N(N-1) \equiv 0 \pmod{2^4 \cdot 5^4}.$ Since consecutive integers are coprime, $16$ divides one of $N,$ $N - 1$ and $625$ divides one of them. This gives four cases modulo $10000:$ $N \equiv 0,$ $N \equiv 1,$ $N \equiv 625$ (which is $0$ mod $625$ and $1$ mod $16$), and $N \equiv 9376$ (which is $0$ mod $16$ and $1$ mod $625$).

The last four digits $abcd$ must have $a \ne 0,$ which rules out $0000,$ $0001,$ and $0625.$ So $abcd = 9376$ — for instance $9376^2 = 87909376$ — and $abc = 937.$

Answer: 2

Solution:

Write $a = \sqrt{2014},$ so the equation is $ax^3 - (2a^2 + 1)x^2 + 2 = 0.$ It factors as $$\left(ax - 1\right)\left(x^2 - 2ax - 2\right) = 0,$$ as expanding confirms. So one root is $\frac{1}{a},$ and the other two are $a \pm \sqrt{a^2 + 2},$ with product $-2$ and sum $2a.$

Since $a - \sqrt{a^2+2} \lt 0 \lt \frac{1}{a} \lt a + \sqrt{a^2+2},$ the middle root is $x_2 = \frac{1}{a},$ and $x_1 + x_3 = 2a.$ Therefore $x_2(x_1 + x_3) = \frac{1}{a} \cdot 2a = 2.$

Answer: 58

Solution:

Place $E$ at the origin with $C = (6, 0),$ so $A = (5, 0).$ When a circle of radius $1$ rolls without slipping outside a fixed circle of radius $5$ and its center sweeps an angle $\varphi$ about $E,$ the rolling contact turns the disk through $5\varphi$ relative to the line of centers, and the revolution of that line adds $\varphi$ more, so the disk turns $6\varphi$ in the ground frame. Turning through $360^\circ$ therefore means $\varphi = 60^\circ,$ so $D = 6(\cos 60^\circ, \sin 60^\circ) = (3, 3\sqrt{3}).$

Having turned through a full $360^\circ,$ the disk is back in its original orientation, so the vector from its center to the marked point is unchanged: $B = D + (A - C) = (3 - 1, 3\sqrt{3}) = (2, 3\sqrt{3}).$ (In particular $\overline{BD}$ is parallel to $\overline{AC},$ as the problem states.)

The ray $EA$ is the positive $x$-axis, so $$\sin^2(\angle BEA) = \frac{(3\sqrt{3})^2}{2^2 + (3\sqrt{3})^2} = \frac{27}{31},$$ and $m + n = 27 + 31 = 58.$

Answer: 391

Solution:

Work in the diagonal coordinates $u = x + y$ and $v = x - y.$ Each of the four moves changes $u$ by $\pm 1$ and $v$ by $\pm 1,$ and the four moves realize all four sign combinations equally often — so $u$ and $v$ perform independent six-step $\pm 1$ walks. The token ends on $|y| = |x|$ exactly when $y = \pm x,$ that is, when $u = 0$ or $v = 0.$

Each of $u = 0$ and $v = 0$ requires three $+1$s and three $-1$s, with probability $\binom{6}{3}/2^6 = \frac{20}{64} = \frac{5}{16}.$ By independence and inclusion-exclusion, the probability is $$\frac{5}{16} + \frac{5}{16} - \left(\frac{5}{16}\right)^2 = \frac{160 - 25}{256} = \frac{135}{256}.$$

Thus $m + n = 135 + 256 = 391.$

Answer: 453

Solution:

Condition on the range of $f.$ If it has $k$ elements, then the range of $g$ is disjoint from it exactly when $g$ maps $A$ into the remaining $4 - k$ elements, which happens for $(4-k)^4$ of the $4^4 = 256$ functions $g.$

Count functions $f$ by range size: $4$ constant functions; $\binom{4}{2}(2^4 - 2) = 84$ with range size $2;$ $\binom{4}{3} \cdot 36 = 144$ with range size $3$ (there are $36$ surjections from four elements onto three); and $4! = 24$ bijections. The number of favorable pairs is $$4 \cdot 3^4 + 84 \cdot 2^4 + 144 \cdot 1^4 + 24 \cdot 0^4 = 324 + 1344 + 144 = 1812.$$

The probability is $\frac{1812}{4^8} = \frac{1812}{65536} = \frac{453}{16384},$ and since $16384$ is a power of $2$ while $453 = 3 \cdot 151$ is odd, this is in lowest terms. Thus $m = 453.$

Answer: 850

Solution:

Place $B = (0,0),$ $C = (s, 0),$ $D = (s, s),$ $A = (0, s),$ with $E = (0, e),$ $F = (f, 0),$ $G = (s, g),$ $H = (h, s).$ The regions $AEPH$ and $DHPG$ together form the trapezoid $AEGD,$ of area $\frac{s\,((s - e) + (s - g))}{2};$ since $\frac{269 + 411}{1360} = \frac{1}{2},$ this is half of $s^2,$ forcing $e + g = s,$ i.e. $EG$ passes through the center. Likewise $AEPH$ and $BFPE$ form the trapezoid $ABFH$ of area $\frac{s(f + h)}{2} = \frac{269 + 275}{1360}\,s^2 = \frac{2}{5}s^2,$ so $f + h = \frac{4s}{5}.$ Perpendicularity of the directions $(s, g - e)$ and $(h - f, s)$ gives $h - f = e - g.$ Writing $\delta = g - e,$ we get $E = \left(0, \frac{s - \delta}{2}\right),$ $G = \left(s, \frac{s + \delta}{2}\right),$ $F = \left(\frac{2s}{5} + \frac{\delta}{2}, 0\right),$ $H = \left(\frac{2s}{5} - \frac{\delta}{2}, s\right),$ and $EG = 34$ gives $s^2 + \delta^2 = 1156.$

Intersecting lines $EG$ and $FH$ (and simplifying with $s^2 + \delta^2 = 1156$) yields $$P = \left(\frac{s}{2} - \frac{s^3}{11560},\; \frac{s}{2} - \frac{s^2\delta}{11560}\right),$$ and the shoelace formula on $A, E, P, H$ then gives $[AEPH] = \frac{s^2}{5} - \frac{s^3\delta}{231200}.$ Setting this equal to $\frac{269}{1360}s^2$ leaves $\frac{3s^2}{1360} = \frac{s^3\delta}{231200},$ so $s\delta = 510.$

Now $s^2 + \delta^2 = 1156$ and $s^2\delta^2 = 260100,$ so $s^2$ and $\delta^2$ are the roots of $t^2 - 1156t + 260100 = 0,$ which are $\frac{1156 \pm 544}{2} = 850$ and $306.$ Since $|\delta| = |g - e| \lt s,$ the area is $s^2 = 850.$

Answer: 263

Solution:

Add $4$ to both sides, giving one unit to each fraction: since $\frac{k}{x-k} + 1 = \frac{x}{x-k},$ the equation becomes $$x\left(\frac{1}{x-3} + \frac{1}{x-5} + \frac{1}{x-17} + \frac{1}{x-19}\right) = x^2 - 11x = x(x - 11).$$ Besides $x = 0,$ we can divide by $x$ and substitute $t = x - 11,$ which pairs the fractions symmetrically: $$\frac{2t}{t^2 - 64} + \frac{2t}{t^2 - 36} = t.$$

Besides $t = 0,$ dividing by $t$ gives $\frac{2}{t^2 - 64} + \frac{2}{t^2 - 36} = 1.$ With $u = t^2,$ clearing denominators gives $2(u - 36) + 2(u - 64) = (u - 36)(u - 64),$ i.e. $u^2 - 104u + 2504 = 0,$ so $u = 52 \pm \sqrt{200}.$

The largest solution is $m = 11 + \sqrt{52 + \sqrt{200}} \approx 19.1,$ which exceeds the other candidates $0,$ $11,$ and $11 \pm \sqrt{52 \pm \sqrt{200}}.$ Therefore $a + b + c = 11 + 52 + 200 = 263.$

Answer: 41

Solution:

Since $3^2 + 4^2 = 5^2,$ angle $B$ is right, and as $\angle EBD = 90^\circ$ is inscribed in $\omega,$ the chord $ED$ is a diameter. Hence $\angle EFD = \angle EGD = 90^\circ.$ From $EF = DF,$ triangle $EFD$ is an isosceles right triangle, so $EF = DF = \frac{DE}{\sqrt{2}}$ and $\angle FED = 45^\circ;$ from $DG : EG = 3 : 4$ and $DG^2 + EG^2 = DE^2$ we get $DG = \frac{3}{5}DE$ and $EG = \frac{4}{5}DE.$ On $\omega$ the order is $E, F, G, D$ (both $F$ and $G$ lie on the arc opposite $B,$ and $\angle FED = 45^\circ$ exceeds $\angle GED = \arcsin\frac{3}{5},$ so $F$ is farther from $D$).

Line $AC$ is the line $FG,$ so the distances from $E$ and $D$ to it follow from the angles of cyclic quadrilateral $EFGD.$ At $F:$ $\angle EFG = 180^\circ - \angle GDE$ and $\sin\angle GDE = \frac{EG}{DE} = \frac{4}{5},$ so the distance from $E$ is $EF \sin\angle EFG = \frac{DE}{\sqrt{2}} \cdot \frac{4}{5} = \frac{2\sqrt{2}}{5}DE.$ At $G:$ $\angle FGD = 180^\circ - \angle FED = 135^\circ,$ so the distance from $D$ is $DG \sin\angle FGD = \frac{3}{5}DE \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{10}DE.$

Now place $B = (0,0),$ $C = (4,0),$ $A = (0,3),$ so $E = (0, e),$ $D = (d, 0),$ line $AC:$ $3x + 4y = 12,$ and $DE^2 = d^2 + e^2.$ Setting $k = \frac{\sqrt{2}}{2}DE,$ the two distance formulas read $\frac{12 - 4e}{5} = \frac{2\sqrt{2}}{5}DE$ and $\frac{12 - 3d}{5} = \frac{3\sqrt{2}}{10}DE,$ which give $e = 3 - k$ and $d = 4 - k.$ Then $DE^2 = 2k^2$ becomes $2k^2 = (3-k)^2 + (4-k)^2 = 2k^2 - 14k + 25,$ so $k = \frac{25}{14}$ and $DE = \sqrt{2}\,k = \frac{25\sqrt{2}}{14}.$ Therefore $a + b + c = 25 + 2 + 14 = 41.$