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 $$P(0) + Q(0) = R(0) = R(16) + 16 \cdot \frac{1}{2} = 108 + 8 = 116.$$

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 $227 = 2 \cdot 81 + 7 \cdot 9 + 2 = 272_{\text{nine}}.$

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 $$P = \left(\frac{75 + 333}{2}, \frac{h}{2}\right) = \left(204, \frac{h}{2}\right), \qquad Q = \left(\frac{575 + 317}{2}, \frac{h}{2}\right) = \left(446, \frac{h}{2}\right).$$

Therefore $PQ = 446 - 204 = 242.$

Solution:

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

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

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 $$\left(\frac{D}{2t} - 14\right)^2 + \left(\frac{264}{t}\right)^2 = 60^2, \qquad \left(-\frac{D}{2t} - 14\right)^2 + \left(\frac{264}{t}\right)^2 = 80^2.$$

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

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 $276 - (23 + 23 - 1) - 3 = 276 - 48 = 228.$

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 $\{7,8,9\},$ $\{6,8,9\},$ $\{5,8,9\},$ $\{6,7,9\},$ $\{5,7,9\},$ and $\{6,7,8\}.$ Splitting the remaining six digits into two triples in each case, the closest product pairs are $30$ and $24,$ $30$ and $28,$ $36$ and $28,$ $30$ and $28,$ $36$ and $32,$ and $36$ and $30,$ respectively — differences of at least $2,$ and even $\frac{2}{432} = \frac{1}{216}$ exceeds $\frac{1}{288}.$

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

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 $a + b = 351 + 27 = 378.$

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 $m + n = 16 + 231 = 247.$

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 $$AC^2 = 30^2 - (h_3 - h_1)^2 = 900 - 144 = 756.$$

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 $$784 = AB^2 + BC^2 + 2 \cdot AB \cdot BC \cos 2\theta.$$ 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 $$AB \cdot BC \sin 2\theta = \frac{21}{2} \cdot \frac{49}{2} \cdot \frac{4\sqrt{3}}{7} = 147\sqrt{3},$$ so $m + n = 147 + 3 = 150.$

Solution:

Count element by element: $S_n$ equals the number of triples $(x, A, B)$ with $|A| = |B|$ and $x \in A \cap B.$ For a fixed $x$ and size $k,$ there are $\binom{n-1}{k-1}$ choices for each of $A$ and $B$ containing $x,$ so by the Vandermonde identity $$S_n = n \sum_{k=1}^{n} \binom{n-1}{k-1}^2 = n\binom{2n-2}{n-1}.$$

Therefore $$\frac{S_{2022}}{S_{2021}} = \frac{2022\binom{4042}{2021}}{2021\binom{4040}{2020}} = \frac{2022}{2021} \cdot \frac{4042 \cdot 4041}{2021^2} = \frac{2 \cdot 2022 \cdot 4041}{2021^2}.$$ Since $2021 = 43 \cdot 47$ divides neither $2022,$ $4041 = 3^2 \cdot 449,$ nor $2,$ this fraction is in lowest terms: $p = 2 \cdot 2022 \cdot 4041 = 16341804$ and $q = 2021^2 = 4084441.$

Then $p + q = 20426245,$ whose remainder modulo $1000$ is $245.$

Solution:

Every element of $S$ equals $\frac{k}{9999}$ for some $1 \le k \le 9999,$ where $9999 = 3^2 \cdot 11 \cdot 101.$ In lowest terms this is $\frac{m}{D}$ where $D \mid 9999,$ $m \le D,$ and $\gcd(m, D) = 1;$ conversely any such $\frac{m}{D}$ arises from $k = m \cdot \frac{9999}{D}.$ So $N$ counts the integers $m$ that are at most, and coprime to, some divisor $D$ of $9999.$

Classify $m$ by which of the primes $3, 11, 101$ divide it, always using the largest divisor $D$ coprime to $m.$ If $\gcd(m, 9999) = 1,$ take $D = 9999:$ there are $\varphi(9999) = 6000$ such $m.$ If $3 \mid m$ only, take $D = 11 \cdot 101 = 1111:$ multiples of $3$ up to $1111$ avoiding $11$ and $101$ number $370 - 33 - 3 = 334.$ If $11 \mid m$ only, take $D = 9 \cdot 101 = 909:$ that gives $82 - 27 = 55.$ If $101 \mid m$ only, then $D = 99 \lt 101$ admits none. If $33 \mid m$ but $101 \nmid m,$ take $D = 101:$ the values $33, 66, 99$ give $3$ more, and any $m$ divisible by $3 \cdot 101$ or $11 \cdot 101$ would need $D \le 11,$ which is impossible.

Therefore $N = 6000 + 334 + 55 + 3 = 6392,$ and the remainder modulo $1000$ is $392.$

Solution:

Write $a = BC = 219,$ $b = CA,$ $c = AB,$ and $s$ for the semiperimeter. The splitting line through $M$ meets $\overline{BC}$ at the point $X$ with $BX = s - \frac{c}{2}$ (then each piece has perimeter $s$). In triangle $BMX,$ the law of sines shows $\angle BXM = \frac{C}{2}:$ this needs $c \sin\left(B + \frac{C}{2}\right) = (a + b)\sin\frac{C}{2},$ which reduces via $a + b = 2R(\sin A + \sin B) = 4R\cos\frac{C}{2}\cos\frac{A - B}{2}$ and $c = 4R \sin\frac{C}{2}\cos\frac{C}{2}$ to $\sin\left(B + \frac{C}{2}\right) = \cos\frac{A - B}{2},$ true because those angles are complementary. Hence the splitting line through $M$ is parallel to the angle bisector from $C,$ and likewise the one through $N$ is parallel to the bisector from $B.$

The internal bisectors from $B$ and $C$ meet at $90^\circ + \frac{A}{2} \gt 90^\circ,$ so the acute angle between the two splitting lines is $90^\circ - \frac{A}{2} = 30^\circ,$ forcing $\angle A = 120^\circ.$ The law of cosines gives $$219^2 = b^2 + c^2 + bc = (b + c)^2 - bc.$$ Set $p = b + c,$ so $bc = p^2 - 219^2$ and $b, c$ are roots of $t^2 - pt + (p^2 - 219^2),$ requiring $4 \cdot 219^2 - 3p^2$ to be a perfect square $k^2.$ Then $3 \mid k$ and $3 \mid p;$ writing $p = 3r$ and $k = 3m$ turns the condition into $m^2 + 3r^2 = 146^2.$ The triangle inequality $p \gt 219$ and $4 \cdot 219^2 \ge 3p^2$ restrict $74 \le r \le 84,$ and checking these, only $r = 80$ works, with $m = 46.$

So $b + c = 240$ and $bc = 240^2 - 47961 = 9639,$ giving $\{b, c\} = \{51, 189\}$ — a valid triangle. The perimeter is $219 + 240 = 459.$

Solution:

Each radicand factors: $2x - xy = x(2 - y),$ and so on, so $0 \lt x, y, z \le 2.$ Substitute $x = 2\sin^2\alpha,$ $y = 2\sin^2\beta,$ $z = 2\sin^2\gamma$ with $\alpha, \beta, \gamma \in \left(0^\circ, 90^\circ\right].$ Then $\sqrt{x(2 - y)} = \sqrt{4\sin^2\alpha\cos^2\beta} = 2\sin\alpha\cos\beta,$ and each equation collapses by the sine addition formula: $$2\sin(\alpha + \beta) = 1, \qquad 2\sin(\beta + \gamma) = \sqrt{2}, \qquad 2\sin(\gamma + \alpha) = \sqrt{3}.$$

Taking $\alpha + \beta = 30^\circ,$ $\beta + \gamma = 45^\circ,$ $\gamma + \alpha = 60^\circ$ and solving, $\alpha = 22.5^\circ,$ $\beta = 7.5^\circ,$ $\gamma = 37.5^\circ.$ (The supplementary branch choices consistent with the angle ranges lead to the same value of the final square.) By the double-angle identity, $1 - x = \cos 2\alpha = \cos 45^\circ,$ $1 - y = \cos 15^\circ,$ and $1 - z = \cos 75^\circ.$

Therefore $$(1 - x)(1 - y)(1 - z) = \frac{\sqrt{2}}{2}\cos 15^\circ \sin 15^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sin 30^\circ}{2} = \frac{\sqrt{2}}{8},$$ whose square is $\frac{2}{64} = \frac{1}{32}.$ Thus $m + n = 1 + 32 = 33.$