Solution:

Measuring time in hours, the two scenarios say $$\frac{9}{s} + \frac{t}{60} = 4 \qquad \text{and} \qquad \frac{9}{s+2} + \frac{t}{60} = \frac{12}{5}.$$ 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 $\frac{t}{60} = 4 - \frac{9}{5/2} = \frac{2}{5},$ 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.

Solution:

Pulling the exponents out of the logarithms, the conditions become $$x \log_x y = 10 \qquad \text{and} \qquad 4y \log_y x = 10.$$ 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: the system solves to $\log_x y = \frac{10}{x}$ with $y = \frac{25}{x},$ which has a solution with $x, y \gt 1,$ so the answer is $25.$

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 $404 + 405 = 809.$

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 $$\binom{4}{2}\binom{6}{2} + \binom{4}{3}\binom{6}{1} + \binom{4}{4}\binom{6}{0} = 90 + 24 + 1 = 115,$$ and exactly $1$ of these wins the grand prize.

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

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 $H,$ $$(e + 92)^2 + 8^2 = 92^2 + 25^2 = 9089,$$ so $(e + 92)^2 = 9025$ and $e + 92 = 95,$ giving $e = 3.$

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

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 $147 + 147 = 294.$

Solution:

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

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

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 $m + n = 192 + 5 = 197.$

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 $$x^2\left(\frac{1}{20} - \frac{m^2}{24}\right) = 1, \qquad \text{i.e.} \qquad x^2 = \frac{120}{6 - 5m^2},$$ 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 $480 \cdot \frac{11/6}{11/6} = 480,$ 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 $480.$

Solution:

By the tangent-chord angle, $\angle DBC = \angle DCB = \angle A,$ so triangle $DBC$ is isosceles with $DB = \frac{BC/2}{\cos A}.$ 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 $DB = \frac{9/2}{11/25} = \frac{225}{22}.$

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

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

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 $m + n = 115 + 256 = 371.$

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 $384 + 1 = 385.$

Solution:

If $p \mid n^4 + 1,$ then $n^8 \equiv 1$ and $n^4 \equiv -1 \pmod p,$ so $n$ has order $8$ modulo $p$ and $8 \mid p - 1$ (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 $p = 17.$

The fourth roots of $-1$ modulo $17$ are $\pm 2$ and $\pm 8.$ To lift $n = 8,$ set $n = 8 + 17t:$ modulo $289,$ $$n^4 + 1 \equiv 8^4 + 1 + 4 \cdot 8^3 \cdot 17t = 17(241 + 2048t),$$ so we need $241 + 2048t \equiv 3 + 8t \equiv 0 \pmod{17},$ giving $t \equiv 6$ and $n \equiv 8 + 102 = 110 \pmod{289}.$

The same computation lifts $2,$ $15,$ and $9$ to $155,$ $134,$ and $179$ respectively, so the least positive $m$ is $110.$ Indeed $110^4 + 1 = 146410001 = 289 \cdot 506609.$

Solution:

A tetrahedron with equal opposite edges embeds in a rectangular box with the six edges as face diagonals. If the box has dimensions $a \times b \times c,$ then $a^2 + b^2 = 41,$ $a^2 + c^2 = 80,$ and $b^2 + c^2 = 89.$ Adding gives $a^2 + b^2 + c^2 = 105,$ so $(a, b, c) = (4, 5, 8).$ The box minus four corner tetrahedra of volume $\frac{abc}{6}$ each leaves $$V = abc - 4 \cdot \frac{abc}{6} = \frac{abc}{3} = \frac{160}{3}.$$

All four faces are congruent triangles with sides $\sqrt{41},$ $\sqrt{80},$ $\sqrt{89}.$ By Heron's formula in the form $16F^2 = 2(a^2b^2 + b^2c^2 + c^2a^2) - (a^4 + b^4 + c^4)$ applied to the squared sides $41, 80, 89,$ we get $16F^2 = 28098 - 16002 = 12096,$ so $F = \sqrt{756} = 6\sqrt{21}.$

The point equidistant from all four faces is the insphere center, and decomposing the tetrahedron into four pyramids over the faces gives $V = \frac{1}{3} r \cdot 4F.$ Hence $$r = \frac{3V}{4F} = \frac{160}{24\sqrt{21}} = \frac{20}{3\sqrt{21}} = \frac{20\sqrt{21}}{63},$$ and $m + n + p = 20 + 21 + 63 = 104.$

Solution:

For a box with dimensions $a, b, c,$ the conditions are $2(ab + bc + ca) = 54$ and $abc = 23,$ so $ab + bc + ca = 27.$ The smallest sphere containing a box has the box's space diagonal as a diameter, so $$r^2 = \max_{\mathcal{B}} \frac{a^2 + b^2 + c^2}{4} = \max_{\mathcal{B}} \frac{(a + b + c)^2 - 54}{4}.$$

With $ab + bc + ca$ and $abc$ fixed, $s = a + b + c$ ranges over an interval, and at an endpoint the cubic $t^3 - st^2 + 27t - 23$ has a double root, meaning two dimensions coincide. Setting $b = c:$ $2ab + b^2 = 27$ and $ab^2 = 23,$ so eliminating $a$ gives $\frac{b(27 - b^2)}{2} = 23,$ i.e. $b^3 - 27b + 46 = 0,$ which factors as $(b - 2)(b^2 + 2b - 23) = 0.$ The roots are $b = 2$ and $b = 2\sqrt{6} - 1.$

For $b = 2,$ $a = \frac{23}{4}$ and $s = \frac{23}{4} + 4 = \frac{39}{4} = 9.75;$ for $b = 2\sqrt{6} - 1,$ $s \approx 9.31$ is smaller. So the maximum of $a^2 + b^2 + c^2$ is $\left(\frac{39}{4}\right)^2 - 54 = \frac{657}{16},$ giving $r^2 = \frac{657}{64}$ and $p + q = 657 + 64 = 721.$