Small Hint:

Consider $R(x) = P(x) + Q(x):$ the quadratic terms cancel, so $R$ is linear

Big Hint:

$R$ passes through $(16, 108)$ and $(20, 106);$ extend that line back to $x = 0$

Solution:

Let $R(x) = P(x) + Q(x).$ The leading coefficients $2$ and $-2$ cancel, so $R$ is a linear function. Since both graphs pass through $(16, 54)$ and $(20, 53),$ we get $R(16) = 108$ and $R(20) = 106.$

The slope of $R$ is $\frac{106 - 108}{20 - 16} = -\frac{1}{2},$ so $$\begin{aligned} P(0) + Q(0) &= R(0) \\ &= R(16) + 16 \cdot \frac{1}{2} \\ &= 108 + 8 = 116. \end{aligned}$$

Small Hint:

Write the condition as $100a + 10b + c = 81b + 9c + a$ and simplify

Big Hint:

From $99a = 71b + 8c,$ reduce modulo $8$ to pin down $b$ in terms of $a,$ then test small values of $a.$

Solution:

The condition says $100a + 10b + c = 81b + 9c + a,$ which simplifies to $99a = 71b + 8c.$ Since the digits also appear in a base-nine numeral, each is at most $8.$ Reducing modulo $8$ gives $3a \equiv -b \pmod 8,$ so $b \equiv 5a \pmod 8.$

For $a = 1,$ $b = 5$ makes $71b$ exceed $99;$ for $a = 2,$ $b = 2$ gives $8c = 198 - 142 = 56,$ so $c = 7.$ For each $a \ge 3,$ the required $b$ forces $99a - 71b$ outside the range $[0, 64],$ so there is no other solution.

The number is $227,$ and indeed $227 = 2 \cdot 81 + 7 \cdot 9 + 2$ $= 272_9.$

Small Hint:

Let the bisector of $\angle A$ meet $\overline{CD}$ at $A':$ alternate interior angles make triangle $ADA'$ isosceles with $DA' = 333$

Big Hint:

In isosceles triangle $ADA',$ the bisector from $D$ is also the median, so $P$ is the midpoint of $AA'.$ Do the same for $Q$ and use coordinates.

Solution:

Let the bisector of $\angle A$ meet $\overline{CD}$ at $A'.$ Since $\overline{AB} \parallel \overline{CD},$ we have $\angle DA'A = \angle A'AB = \angle A'AD,$ so triangle $ADA'$ is isosceles with $DA' = DA = 333.$ The bisector of $\angle D$ is then the median from $D$ in this triangle, so $P,$ which lies on both bisectors, is the midpoint of $\overline{AA'}.$ Symmetrically, $Q$ is the midpoint of $\overline{BB'},$ where $B'$ is on $\overline{CD}$ with $CB' = 333.$

Place $D = (0, 0)$ and $C = (650, 0),$ so $A = (75, h)$ and $B = (575, h)$ for the appropriate height $h.$ Then $A' = (333, 0)$ and $B' = (650 - 333, 0) = (317, 0),$ so $$\begin{aligned} P &= \left(\frac{75 + 333}{2}, \frac{h}{2}\right) \\ &= \left(204, \frac{h}{2}\right), \end{aligned}$$ $$\begin{aligned} Q &= \left(\frac{575 + 317}{2}, \frac{h}{2}\right) \\ &= \left(446, \frac{h}{2}\right). \end{aligned}$$

Therefore $PQ = 446 - 204 = 242.$

Small Hint:

Both numbers lie on the unit circle: $w$ has argument $30^\circ$ and $z$ has argument $120^\circ,$ while $\mathrm{i}$ has argument $90^\circ$

Big Hint:

Matching arguments gives $90 + 30r \equiv 120s \pmod{360},$ i.e. $r + 3 \equiv 4s \pmod{12};$ count how many $r \in [1, 100]$ hit each residue

Solution:

Both $w$ and $z$ have modulus $1:$ in polar form $w = \operatorname{cis} 30^\circ$ and $z = \operatorname{cis} 120^\circ,$ while $\mathrm{i} = \operatorname{cis} 90^\circ.$ The equation $\mathrm{i} \cdot w^r = z^s$ is therefore a statement about arguments: $$90 + 30r \equiv 120s \pmod{360},$$ i.e. $$r + 3 \equiv 4s \pmod{12}.$$

For each $s,$ this determines $r \pmod{12}:$ the residue $4s - 3$ is $1,$ $5,$ or $9$ modulo $12$ according as $s \equiv 1,$ $2,$ or $0 \pmod 3.$ Among $1 \le r \le 100$ there are $9$ values with $r \equiv 1 \pmod{12}$ and $8$ values each with $r \equiv 5$ or $r \equiv 9 \pmod{12}.$ Among $1 \le s \le 100$ there are $34$ values with $s \equiv 1 \pmod 3$ and $33$ values in each of the other two classes.

The count is $34 \cdot 9$ $+ 33 \cdot 8$ $+ 33 \cdot 8$ $= 306 + 264 + 264 = 834.$

Small Hint:

The meeting point is due north of the midpoint between them. Each swimmer’s velocity relative to the water is her ground velocity minus the current $(14, 0).$

Big Hint:

With $u = \frac{D}{2t},$ subtracting the two speed equations gives $(u + 14)^2 - (u - 14)^2$ $= 6400 - 3600$

Solution:

Put Sherry at the origin and Melanie at $(D, 0)$ on the south bank. A point on the north bank equidistant from both is $\left(\frac{D}{2}, 264\right).$ If both arrive at time $t,$ then each swimmer’s velocity relative to the water is her ground velocity minus the current $(14, 0),$ so $$\begin{aligned} &\left(\frac{D}{2t} - 14\right)^2 \\ &\quad {}+ \left(\frac{264}{t}\right)^2 = 60^2, \end{aligned}$$ $$\begin{aligned} &\left(-\frac{D}{2t} - 14\right)^2 \\ &\quad {}+ \left(\frac{264}{t}\right)^2 = 80^2. \end{aligned}$$

Subtracting, with $u = \frac{D}{2t}:$ $(u + 14)^2 - (u - 14)^2$ $= 56u$ $= 6400 - 3600$ $= 2800,$ so $u = 50.$ Substituting back, $(50 - 14)^2 + \left(\frac{264}{t}\right)^2 = 3600$ gives $\frac{264}{t} = 48,$ so $t = \frac{11}{2}.$

Therefore $D = 2ut = 100t = 550.$

Small Hint:

Count the bad pairs among the $\binom{24}{2}$ choices with $5 \lt a \lt b \lt 30:$ every four-term progression must use $a$ or $b$

Big Hint:

Beyond $a = 6$ and $20 \in \{a, b\},$ exactly three pairs $(a, b)$ complete a progression with two of the fixed terms. Watch for double counting.

Solution:

The sequence is increasing exactly when $5 \lt a \lt b \lt 30,$ giving $\binom{24}{2} = 276$ pairs. The six fixed terms contain no four-term arithmetic progression, so every progression must involve $a$ or $b.$ If only one of them is involved, three fixed terms must already be in progression: $3, 4, 5$ extends only by $6,$ and $30, 40, 50$ extends only by $20.$ So the single-variable violations are $a = 6$ ($23$ pairs) and $20 \in \{a, b\}$ ($23$ pairs), which overlap in the pair $(6, 20).$

If both $a$ and $b$ are involved, two fixed terms complete the progression. Checking the possible positions: $(4, 5, a, b)$ gives $(6, 7);$ $(3, 5, a, b)$ gives $(7, 9);$ $(3, a, b, 30)$ gives $(12, 21);$ $(4, a, b, 40)$ gives $(16, 28);$ $(5, a, b, 50)$ gives $(20, 35),$ out of range; and $(a, b, 30, 40)$ gives $(10, 20).$ Of these, $(6, 7)$ and $(10, 20)$ are already counted, so $(7, 9),$ $(12, 21),$ and $(16, 28)$ are the only new bad pairs.

The number of valid pairs is $276 - (23 + 23 - 1)$ $- 3 = 276 - 48 = 228.$

Small Hint:

Aim for a numerator of $1:$ look for two triples of distinct digits whose products differ by exactly $1$

Big Hint:

Save the three largest leftover digits for the denominator, then rule out bigger denominators by showing they force a numerator difference of at least $2$

Solution:

Try to make the numerator equal to $1$ while keeping large digits in the denominator. The products $2 \cdot 3 \cdot 6 = 36$ and $1 \cdot 5 \cdot 7 = 35$ differ by $1$ and leave $4, 8, 9$ for the denominator, giving the value $$\frac{36 - 35}{4 \cdot 8 \cdot 9} = \frac{1}{288}.$$

To beat this, a fraction would need numerator $1$ with denominator greater than $288.$ The denominators exceeding $288$ are $504,$ $432,$ $360,$ $378,$ $315,$ and $336,$ coming respectively from $\{7,8,9\},$ $\{6,8,9\},$ $\{5,8,9\},$ $\{6,7,9\},$ $\{5,7,9\},$ and $\{6,7,8\}.$ Splitting the remaining six digits into two triples, the smallest positive numerator differences are respectively $6,$ $2,$ $8,$ $2,$ $4,$ and $6.$ The resulting lower bounds $$\begin{aligned}&\frac{6}{504},\quad \frac{2}{432},\quad \frac{8}{360},\\&\frac{2}{378},\quad \frac{4}{315},\quad \frac{6}{336}\end{aligned}$$ all exceed $\frac{1}{288}.$

So the minimum positive value is $\frac{1}{288},$ and $m + n = 1 + 288 = 289.$

Small Hint:

Find $\omega_A$ first: its center lies on line $AO,$ its radius is half its distance from $A,$ and internal tangency to $\omega$ fixes everything

Big Hint:

The two intersection points of $\omega_B$ and $\omega_C$ lie on line $AO,$ and each triangle’s circumradius is the distance from that point to the center $O$

Solution:

Let $O$ be the center of $\omega.$ The center of $\omega_A$ lies on line $AO$ (the bisector of $\angle A$) at some distance $d$ from $A;$ since $\overline{AB}$ makes a $30^\circ$ angle with $AO,$ the radius is $r = d \sin 30^\circ = \frac{d}{2}.$ Internal tangency to $\omega$ requires the center to be $18 - r$ from $O,$ which forces the center past $O:$ $d - 18 = 18 - \frac{d}{2},$ so $d = 24,$ $r = 12,$ and the center is $6$ beyond $O.$

Place $O$ at the origin with $A = (0, 18).$ Then the three centers are $O_A = (0, -6)$ and $O_B, O_C = (\pm 3\sqrt{3}, 3),$ all with radius $12.$ The intersections of $\omega_B$ and $\omega_C$ lie on the $y$-axis: $27 + (y - 3)^2 = 144$ gives $y = 3 \pm \sqrt{117}.$ The point $(0, 3 + \sqrt{117})$ is closer to $A$ and belongs to the larger triangle, so the smaller triangle has vertex $(0, 3 - \sqrt{117}),$ at distance $\sqrt{117} - 3$ from $O.$

By symmetry the smaller triangle is equilateral with circumradius $\sqrt{117} - 3,$ so its side is $\sqrt{3}\left(\sqrt{117} - 3\right) = \sqrt{351} - \sqrt{27}.$ Thus $a + b = 351 + 27 = 378.$

Small Hint:

There is an even number of blocks between positions $i$ and $j$ exactly when $i$ and $j$ have opposite parity

Big Hint:

So each color must occupy one odd and one even position: the six odd slots hold all six colors once, as do the six even slots

Solution:

If a color occupies positions $i \lt j,$ the number of blocks between them is $j - i - 1,$ which is even exactly when $i$ and $j$ have opposite parity. So an arrangement is even precisely when every color occupies one odd position and one even position — that is, the six odd slots contain each color exactly once, and so do the six even slots.

Counting arrangements of the twelve blocks (blocks of the same color identical), there are $\frac{12!}{2^6}$ in total, and $6! \cdot 6!$ even ones (a permutation of the six colors in the odd slots and another in the even slots). The probability is $$\frac{6! \cdot 6! \cdot 2^6}{12!} = \frac{16}{231}.$$

Since $\gcd(16, 231) = 1,$ the answer is $m + n = 16 + 231 = 247.$

Small Hint:

If a sphere’s center sits at height $h$ above the plane, its circle has radius squared $r^2 - h^2;$ congruence ties the three heights together

Big Hint:

$A$ and $B$ are the feet of the centers, so $AB^2 = 24^2 - (h_2 - h_1)^2;$ combine with $h_2^2 - h_1^2 = 13^2 - 11^2$

Solution:

Let the sphere centers be at heights $h_1, h_2, h_3$ above the plane. Each circle’s center is the foot of the perpendicular from the sphere’s center, and the common circle radius $\rho$ satisfies $\rho^2 = 11^2 - h_1^2$ $= 13^2 - h_2^2$ $= 19^2 - h_3^2.$

The first two spheres are tangent, so their centers are $11 + 13 = 24$ apart, and projecting onto the plane, $AB^2 = 24^2 - (h_2 - h_1)^2.$ Thus $(h_2 - h_1)^2 = 576 - 560 = 16.$ Congruence gives $h_2^2 - h_1^2 = 169 - 121 = 48,$ so $h_2 - h_1 = 4$ and $h_2 + h_1 = 12$ (the other sign gives a negative sum), yielding $h_1 = 4,$ $h_2 = 8,$ and $\rho^2 = 121 - 16 = 105.$ Then $h_3^2 = 361 - 105 = 256,$ so $h_3 = 16.$

The first and third centers are $11 + 19 = 30$ apart, so $$\begin{aligned} AC^2 &= 30^2 - (h_3 - h_1)^2 \\ &= 900 - 144 = 756. \end{aligned}$$

Small Hint:

Power of a point: $AP \cdot AQ = 36$ and $CQ \cdot CP = 400$ give tangent lengths $6$ from $A$ and $20$ from $C$

Big Hint:

Equal tangents force $BC = AB + 14,$ and tangency to both parallel lines $AD$ and $BC$ gives $AB \cos^2\!\frac{\angle A}{2} = 6;$ finish with the law of cosines on $AC = 28$

Solution:

By power of a point, $AP \cdot AQ = 3 \cdot 12 = 36$ and $CQ \cdot CP = 16 \cdot 25 = 400,$ so the tangent lengths from $A$ and $C$ are $6$ and $20.$ The tangent point on $\overline{AB}$ is $6$ from $A,$ hence $AB - 6$ from $B;$ equal tangents from $B$ put the tangent point on $\overline{BC}$ at that same distance from $B,$ so its distance from $C$ is $BC - (AB - 6) = 20,$ giving $BC = AB + 14.$

Let $\angle BAD = 2\theta.$ The center lies on the bisector of $\angle A$ with the tangent length from $A$ equal to $6,$ so the radius is $\rho = 6\tan\theta.$ The circle is tangent to both parallel lines $AD$ and $BC,$ whose distance apart is $AB \sin 2\theta,$ so $AB \sin 2\theta = 2\rho = 12 \tan\theta,$ which simplifies to $AB \cos^2\theta = 6.$ In triangle $ABC,$ $\angle ABC = 180^\circ - 2\theta$ and $AC = 3 + 9 + 16 = 28,$ so the law of cosines gives $$\begin{aligned} 784 &= AB^2 + BC^2 \\ &\quad {}+ 2 \cdot AB \cdot BC \cos 2\theta. \end{aligned}$$ Substituting $BC = AB + 14$ and $\cos 2\theta = 2\cos^2\theta - 1,$ the $AB^2$ terms cancel and, using $AB\cos^2\theta = 6,$ the equation collapses to $24\,AB + 336 + 196 = 784,$ so $AB = \frac{21}{2}$ and $\cos^2\theta = \frac{4}{7}.$

Then $\sin 2\theta = 2\sqrt{\frac{3}{7}}\sqrt{\frac{4}{7}} = \frac{4\sqrt{3}}{7},$ and the area is $$\begin{aligned} &AB \cdot BC \sin 2\theta \\ &= \frac{21}{2} \cdot \frac{49}{2} \cdot \frac{4\sqrt{3}}{7} \\ &= 147\sqrt{3}, \end{aligned}$$ so $m + n = 147 + 3 = 150.$