Answer: 204

Small Hint:

Convert both trips into equations: $\frac{9}{s} + \frac{t}{60} = 4$ and $\frac{9}{s+2} + \frac{t}{60} = \frac{12}{5}$

Big Hint:

Subtracting the two equations eliminates $t$ and gives $s(s+2) = \frac{45}{4}$

Solution:

Measuring time in hours, the two scenarios say $$\frac{9}{s} + \frac{t}{60} = 4$$ and $$\frac{9}{s+2} + \frac{t}{60} = \frac{12}{5}.$$ Subtracting, $\frac{9}{s} - \frac{9}{s+2} = \frac{8}{5},$ so $\frac{18}{s(s+2)} = \frac{8}{5},$ giving $s(s+2) = \frac{45}{4}.$ The positive root of $s^2 + 2s - \frac{45}{4} = 0$ is $s = \frac{5}{2}.$

Then $\frac{t}{60} = 4 - \frac{9}{\frac{5}{2}} = \frac{2}{5},$ so $t = 24$ minutes. Walking at $s + \frac{1}{2} = 3$ kilometers per hour takes $\frac{9}{3} = 3$ hours, so the total is $180 + 24 = 204$ minutes.

Answer: 25

Small Hint:

Bring the exponents down: $x \log_x y = 10$ and $4y \log_y x = 10$

Big Hint:

Multiply the two equations: since $\log_x y \cdot \log_y x = 1,$ the logarithms disappear

Solution:

Pulling the exponents out of the logarithms, the conditions become $$x \log_x y = 10$$ and $$4y \log_y x = 10.$$ Multiplying these equations and using $\log_x y \cdot \log_y x = 1$ gives $4xy = 100,$ so $xy = 25.$

Such $x$ and $y$ do exist. Set $y = \frac{25}{x}.$ The continuous function $x\log_x(\frac{25}{x})$ tends to infinity as $x \to 1^+,$ while at $x = 2$ it is less than $10.$ Hence it equals $10$ for some $1 \lt x \lt 2,$ where $y \gt 1.$

Answer: 809

Small Hint:

Work out which small $n$ are losses for the player about to move: $n$ is a loss exactly when both $n - 1$ and $n - 4$ are wins

Big Hint:

The losing positions repeat with period $5:$ they are $n \equiv 0$ or $2 \pmod 5.$ Bob wins exactly for those $n.$

Solution:

Call $n$ a losing position if the player about to move loses with best play. We claim the losing positions are exactly $n \equiv 0$ or $2 \pmod 5.$ From such an $n,$ removing $1$ or $4$ tokens leaves $n \equiv 4, 1, 3 \pmod 5$ — never again $0$ or $2$ — while from any $n \equiv 1, 3, 4 \pmod 5$ one move reaches a position $\equiv 0$ or $2 \pmod 5$ (remove $1,$ $1,$ $4$ tokens respectively). Since $n = 0$ is a loss for the player to move, induction confirms the pattern.

Bob wins exactly when $n$ is a losing position for Alice. Among $1 \le n \le 2024$ there are $404$ multiples of $5$ and $405$ values $n \equiv 2 \pmod 5$ (from $2$ to $2022$), for a total of $404 + 405 = 809.$

Answer: 116

Small Hint:

Winning the grand prize forces winning a prize, so the conditional probability is $\frac{P(\text{grand})}{P(\text{prize})}$

Big Hint:

Count draws that share exactly $2,$ $3,$ or $4$ numbers with Jen’s ticket — every draw of $4$ numbers is equally likely

Solution:

All $\binom{10}{4} = 210$ draws are equally likely. The number of draws sharing exactly $k$ numbers with Jen’s ticket is $\binom{4}{k}\binom{6}{4-k},$ so the number winning a prize is $$\begin{aligned} &\binom{4}{2}\binom{6}{2} + \binom{4}{3}\binom{6}{1} \\ &\quad {}+ \binom{4}{4}\binom{6}{0} \\ &= 90 + 24 + 1 = 115, \end{aligned}$$ and exactly $1$ of these wins the grand prize.

Since the grand prize implies a prize, the conditional probability is $\frac{\frac{1}{210}}{\frac{115}{210}} = \frac{1}{115},$ so $m + n = 1 + 115 = 116.$

Answer: 104

Small Hint:

Put line $DF$ on the $x$-axis with $D$ at the origin, and write the coordinates of $A,$ $D,$ $H,$ and $G$ in terms of $DE$

Big Hint:

The circle’s center lies on $y = -8$ (bisector of $\overline{AD}$) and on the vertical bisector of $\overline{HG};$ equate its distances to $D$ and $H$

Solution:

Put line $DF$ on the $x$-axis with $D = (0, 0)$ and $C = (107, 0),$ so $A = (0, -16).$ Let $DE = e.$ Then $E = (e, 0),$ $F = (e + 184, 0),$ and the second rectangle sits above the line: $H = (e, 17)$ and $G = (e + 184, 17).$

The center of the circle through $A,$ $D,$ $H,$ $G$ lies on the perpendicular bisector of the vertical segment $\overline{AD},$ the line $y = -8,$ and on the perpendicular bisector of the horizontal segment $\overline{HG},$ the line $x = e + 92.$ Equating the center’s squared distances to $D$ and to $H,$ $$\begin{aligned} &(e + 92)^2 + 8^2 \\ &= 92^2 + 25^2 = 9089, \end{aligned}$$ so $(e + 92)^2 = 9025$ and $e + 92 = 95,$ giving $e = 3.$

Therefore $CE = DC - DE$ $= 107 - 3 = 104.$

Answer: 294

Small Hint:

A path with exactly four turns consists of five straight runs that alternate between rightward and upward

Big Hint:

If the path starts rightward, the three rightward runs and two upward runs are positive integers summing to $8$ each — count the compositions, then double

Solution:

A path that changes direction exactly four times consists of five maximal straight runs, alternating between rightward and upward moves. If the first run is rightward, the pattern is $R, U, R, U, R:$ three rightward runs with positive lengths summing to $8,$ and two upward runs with positive lengths summing to $8.$ The counts of such compositions are $\binom{7}{2} = 21$ and $\binom{7}{1} = 7,$ giving $21 \cdot 7 = 147$ paths.

Paths starting upward are counted symmetrically, another $147.$ The total is $147 + 147 = 294.$

Answer: 540

Small Hint:

Write $z = 4(\cos\theta + \mathrm{i}\sin\theta)$ and take the real part of each term

Big Hint:

You get an expression of the form $a\cos\theta + b\sin\theta,$ whose maximum is $\sqrt{a^2 + b^2}$

Solution:

Write $z = 4(\cos\theta + \mathrm{i}\sin\theta),$ so $\frac{1}{z} = \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta).$ The real part of $(75 + 117\mathrm{i})z$ is $4(75\cos\theta - 117\sin\theta)$ $= 300\cos\theta - 468\sin\theta,$ and the real part of $(96 + 144\mathrm{i}) \cdot \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta)$ is $24\cos\theta + 36\sin\theta.$

The total real part is $324\cos\theta - 432\sin\theta,$ whose maximum over $\theta$ is $$\begin{aligned} &\sqrt{324^2 + 432^2} \\ &= 108\sqrt{3^2 + 4^2} \\ &= 108 \cdot 5 = 540. \end{aligned}$$

Answer: 197

Small Hint:

The centers sit at height $\rho$ above $BC,$ spaced $2\rho$ apart, and the end centers lie at distances $\rho\cot\frac{B}{2}$ and $\rho\cot\frac{C}{2}$ from $B$ and $C$

Big Hint:

Both chains measure the same $BC,$ which determines $\cot\frac{B}{2} + \cot\frac{C}{2};$ the incircle satisfies $BC = r\left(\cot\frac{B}{2} + \cot\frac{C}{2}\right)$

Solution:

For a chain of $n$ circles of radius $\rho$ tangent to $\overline{BC},$ the centers lie at height $\rho$ with consecutive centers $2\rho$ apart. The first circle is tangent to $\overline{AB}$ and $\overline{BC},$ so its center lies on the bisector from $B,$ at horizontal distance $\rho\cot\frac{B}{2}$ from $B;$ similarly the last center is $\rho\cot\frac{C}{2}$ from $C.$ Hence with $k = \cot\frac{B}{2} + \cot\frac{C}{2},$ $$BC = \rho k + 2\rho(n - 1).$$

The two chains give $34k + 34 \cdot 14 = BC$ $= k + 2 \cdot 2023,$ so $33k = 3570$ and $k = \frac{1190}{11},$ whence $BC = k + 4046 = \frac{45696}{11}.$

The incircle is a chain of one circle of radius $r:$ $BC = rk.$ Therefore $$r = \frac{BC}{k} = \frac{45696}{1190} = \frac{192}{5},$$ and $m + n = 192 + 5 = 197.$

Answer: 480

Small Hint:

The diagonals are perpendicular lines through the origin; if $B = (x, y)$ with $y = mx,$ then $BD^2 = 4(x^2 + y^2)$

Big Hint:

The perpendicular diagonal must also meet the hyperbola, so both slopes squared lie strictly between $\frac{5}{6}$ and $\frac{6}{5};$ study $BD^2$ as the slope varies

Solution:

The diagonals of a rhombus are perpendicular bisectors of each other, so $C = -A,$ $D = -B,$ and $OA \perp OB.$ Let line $BD$ have slope $m,$ so $B = (x, mx)$ with $$x^2\left(\frac{1}{20} - \frac{m^2}{24}\right) = 1,$$ i.e. $$x^2 = \frac{120}{6 - 5m^2},$$ which requires $m^2 \lt \frac{6}{5}.$ Then $BD^2 = 4(x^2 + m^2x^2)$ $= \frac{480(1 + m^2)}{6 - 5m^2}.$ Line $AC$ has slope $-\frac{1}{m},$ so it meets the hyperbola only when $\frac{1}{m^2} \lt \frac{6}{5},$ that is $m^2 \gt \frac{5}{6}.$

On the interval $\frac{5}{6} \lt m^2 \lt \frac{6}{5},$ the quantity $\frac{480(1 + m^2)}{6 - 5m^2}$ is strictly increasing in $m^2:$ as $m^2 \to \frac{5}{6}$ it tends to $480 \cdot \frac{\frac{11}{6}}{\frac{11}{6}} = 480,$ and as $m^2 \to \frac{6}{5}$ it grows without bound. Hence $BD^2$ takes exactly the values in $(480, \infty)$ and never equals $480.$

The largest number that is less than $BD^2$ for every such rhombus is therefore $480.$

Answer: 113

Small Hint:

By the tangent-chord angle, $\angle DBC = \angle DCB = \angle A,$ so $DB = \frac{\frac{BC}{2}}{\cos A}$

Big Hint:

Power of the point gives $DP \cdot DA = DB^2,$ so $AP = DA - \frac{DB^2}{DA};$ compute $DA$ from the law of cosines in triangle $ABD$

Solution:

By the tangent-chord angle, $\angle DBC = \angle DCB = \angle A,$ so triangle $DBC$ is isosceles with $DB = \frac{\frac{BC}{2}}{\cos A}.$ The law of cosines gives $\cos A = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}$ and $\cos B = \frac{9^2 + 5^2 - 10^2}{2 \cdot 9 \cdot 5} = \frac{1}{15},$ so $DB = \frac{\frac{9}{2}}{\frac{11}{25}} = \frac{225}{22}.$

Since $D$ lies on the opposite side of $BC$ from $A,$ $\angle ABD = A + B,$ and $$\begin{aligned} &\cos(A + B) \\ &= \frac{11}{25} \cdot \frac{1}{15} \\ &\quad {}- \frac{6\sqrt{14}}{25} \cdot \frac{4\sqrt{14}}{15} \\ &= \frac{11 - 336}{375} \\ &= -\frac{13}{15}. \end{aligned}$$ The law of cosines in triangle $ABD$ then gives $$\begin{aligned} &DA^2 = 5^2 + \left(\tfrac{225}{22}\right)^2 \\ &\quad {}+ 2 \cdot 5 \cdot \tfrac{225}{22} \cdot \tfrac{13}{15} \\ &= \frac{105625}{484}, \end{aligned}$$ $$DA = \frac{325}{22}.$$

The power of $D$ gives $DP \cdot DA = DB^2,$ so $$\begin{aligned} &AP = DA - \frac{DB^2}{DA} \\ &= \frac{DA^2 - DB^2}{DA} \\ &= \frac{\frac{105625 - 50625}{484}}{\frac{325}{22}} \\ &= \frac{55000}{22 \cdot 325} \\ &= \frac{100}{13}, \end{aligned}$$ and $m + n = 100 + 13 = 113.$

Answer: 371

Small Hint:

With $b$ blue vertices, the overlaps $|B \cap (B + k)|$ summed over the seven nonzero rotations equal $b^2 - b,$ so every coloring with $b \le 3$ works

Big Hint:

For $b = 4$ a rotation must carry the blue set exactly onto its complement; count those sets separately for $k$ odd, $k \equiv 2 \pmod 4,$ and $k = 4$

Solution:

Label the vertices $0, \ldots, 7$ and let $B$ be the blue set, $b = |B|.$ Rotation by $k$ works exactly when $(B + k) \cap B = \varnothing.$ Since $B + k$ must fit inside the $8 - b$ red positions, $b \le 4.$ Summing $|B \cap (B + k)|$ over all eight rotations counts all pairs $(i, j) \in B \times B$ once (via $k = i - j$), a total of $b^2,$ and the $k = 0$ term contributes $b.$ So for $b \le 3$ the seven nonzero rotations share only $b^2 - b \le 6$ overlaps, and some rotation has none: all $1 + 8 + 28 + 56 = 93$ colorings with $b \le 3$ succeed.

For $b = 4,$ disjointness forces $B + k$ to be exactly the complement of $B.$ If $k$ is odd, the cycle $0, k, 2k, \ldots$ visits all vertices and must alternate between $B$ and its complement, so $B$ is the evens or the odds: $2$ sets. If $k \equiv 2 \pmod 4,$ then $B$ meets each of the $4$-cycles $\{0, 2, 4, 6\}$ and $\{1, 3, 5, 7\}$ in an antipodal pair: $2 \cdot 2 = 4$ sets, such as $\{0, 1, 4, 5\}.$ If $k = 4,$ then $B$ contains exactly one of each pair $\{i, i + 4\}:$ $2^4 = 16$ sets. The first two families contain both members of some antipodal pair while the third never does, and the evens/odds take both their antipodal pairs from one $4$-cycle, so the three families are disjoint: $2 + 4 + 16 = 22$ sets.

In total $93 + 22 = 115$ of the $2^8 = 256$ colorings work, so the probability is $\frac{115}{256}$ and $m + n = 115 + 256 = 371.$

Answer: 385

Small Hint:

As $u = |\sin 2\pi x|$ runs from $0$ to $1,$ the value $4||u - \frac{1}{2}| - \frac{1}{4}|$ zigzags between $1$ and $0$ four times; count the monotone arcs of each graph

Big Hint:

Each bottom-to-top arc of one graph crosses each left-to-right arc of the other exactly once — then examine the corner $(1, 1),$ which both graphs pass through

Solution:

Both right-hand sides take values in $[0, 1],$ so every intersection lies in the unit square, and there we may write both curves using $\varphi(u) = 4\left||u - \tfrac{1}{2}| - \tfrac{1}{4}\right|:$ the first is $y = \varphi(|\sin 2\pi x|)$ and the second is $x = \varphi(|\cos 3\pi y|).$ As $u$ increases from $0$ to $1,$ $\varphi(u)$ runs linearly $1 \to 0 \to 1 \to 0 \to 1$ with corners at $u = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}.$ For $x \in [0, 1],$ $|\sin 2\pi x|$ sweeps $[0, 1]$ monotonically $4$ times, so the first graph consists of $4 \cdot 4 = 16$ monotone arcs, each climbing or descending through the full range $0 \le y \le 1$ within a narrow vertical strip. Likewise $|\cos 3\pi y|$ sweeps $[0, 1]$ monotonically $6$ times for $y \in [0, 1],$ so the second graph consists of $24$ monotone arcs, each crossing the full range $0 \le x \le 1$ within a narrow horizontal strip.

Take one arc of each graph, living in the vertical strip $[a, b]$ and the horizontal strip $[c, d].$ Inside the rectangle $[a, b] \times [c, d],$ the first arc joins the bottom edge to the top edge and the second joins the left edge to the right edge, and each is monotone, so the two arcs cross exactly once. This yields $16 \cdot 24 = 384$ intersection points.

One further point hides at the corner $(1, 1),$ which lies on both graphs: $\varphi(|\sin 2\pi|) = \varphi(0) = 1$ and $\varphi(|\cos 3\pi|) = \varphi(1) = 1.$ Near it the first graph is $y \approx 1 - 8\pi(1 - x)$ while the second satisfies $x \approx 1 - 18\pi^2(1 - y)^2,$ so the two final arcs meet at their shared endpoint $(1, 1)$ in addition to the transversal crossing already counted. The total is $384 + 1 = 385.$

Answer: 110

Small Hint:

If $n^4 + 1$ is divisible by $p$ then $n$ has order $8$ modulo $p,$ so $p \equiv 1 \pmod 8;$ test the smallest such prime

Big Hint:

The fourth roots of $-1$ modulo $17$ are $\pm 2$ and $\pm 8;$ lift each to a solution modulo $289$ with a Hensel step and compare

Solution:

If $n^4 + 1$ is divisible by $p,$ then $n^8 \equiv 1$ and $n^4 \equiv -1 \pmod p,$ so $n$ has order $8$ modulo $p$ and $p - 1$ is divisible by $8$ (and $p = 2$ fails since $n^4 + 1 \equiv 2 \pmod 4$). The smallest prime $p \equiv 1 \pmod 8$ is $17,$ and indeed $2^4 = 16 \equiv -1 \pmod{17}.$ Because the derivative $4n^3$ is not divisible by $17$ at such an $n,$ each root lifts to a root modulo $17^2 = 289,$ so $p = 17.$