Solution:

Since $36 = 4 \cdot 9,$ the number $N$ must be divisible by both $4$ and $9.$ Its digits are distinct members of $\{0, 2, 4, 6, 8\},$ whose total is $20,$ so $N$ cannot use all five. The digit sum must be a multiple of $9,$ and being even it must be $18;$ the only such digit sets are $\{4, 6, 8\}$ and $\{0, 4, 6, 8\}.$

The largest number formed from $\{0, 4, 6, 8\}$ is $8640,$ which ends in $40,$ a multiple of $4.$ So $N = 8640,$ and the remainder upon division by $1000$ is $640.$

Solution:

The points with $d(P) \ge t$ form a concentric square of side $1 - 2t.$ So $d(P) \ge \frac{1}{5}$ puts $P$ inside the concentric square of side $\frac{3}{5},$ and $d(P) \le \frac{1}{3}$ keeps $P$ outside the open concentric square of side $\frac{1}{3}.$

Since the unit square has area $1,$ the probability is the area between those two squares: $$\left(\frac{3}{5}\right)^2 - \left(\frac{1}{3}\right)^2 = \frac{9}{25} - \frac{1}{9} = \frac{81 - 25}{225} = \frac{56}{225}.$$ Thus $m + n = 56 + 225 = 281.$

Solution:

For each value $v = b - a,$ the pairs $(a, b) = (1, v+1), (2, v+2), \ldots, (20-v, 20)$ show that $v$ occurs exactly $20 - v$ times, so $K = \prod_{v=1}^{19} v^{20-v}.$ The exponent of $2$ in $K$ is therefore $\sum_v (20 - v)\,e(v),$ where $e(v)$ is the exponent of $2$ in $v.$

Only even $v$ contribute: $v = 2, 6, 10, 14, 18$ give $e = 1;$ $v = 4, 12$ give $e = 2;$ $v = 8$ gives $e = 3;$ and $v = 16$ gives $e = 4.$ The total is $$18 + 16 \cdot 2 + 14 + 12 \cdot 3 + 10 + 8 \cdot 2 + 6 + 4 \cdot 4 + 2 = 150,$$ so $n = 150.$

Solution:

Number the gates $1$ through $12.$ All $12 \cdot 11 = 132$ ordered pairs of distinct (old, new) gates are equally likely, and Dave walks $400$ feet or less exactly when the gate numbers differ by at most $4.$

A gate $i$ has $\min(i + 4, 12) - \max(i - 4, 1)$ qualifying new gates: gates $1$ and $12$ have $4$ each, gates $2$ and $11$ have $5,$ gates $3$ and $10$ have $6,$ gates $4$ and $9$ have $7,$ and gates $5$ through $8$ have $8$ each. The total is $2(4 + 5 + 6 + 7) + 4 \cdot 8 = 76.$

The probability is $\frac{76}{132} = \frac{19}{33},$ so $m + n = 19 + 33 = 52.$

Solution:

Let $a = \log_{10} x,$ $b = \log_{10} y,$ and $c = \log_{10} z.$ Taking logs of $xyz = 10^{81}$ gives $a + b + c = 81.$ Since $\log_{10} yz = b + c,$ the second condition is $a(b + c) + bc = ab + ac + bc = 468.$

Squaring the sum, $$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) = 81^2 - 2 \cdot 468 = 6561 - 936 = 5625,$$ so the requested value is $\sqrt{5625} = 75.$

Solution:

If there is a linear factor, then some integer $b$ is a root, so $b^4 - nb + 63 = 0$ and $n = b^3 + \frac{63}{b},$ forcing $b \mid 63$ and $b \gt 0.$ The smallest value is $48,$ at $b = 3.$

Otherwise the polynomial splits into two quadratics, which we may take monic; since the $x^3$ coefficient vanishes, they have the form $$(x^2 + px + q)(x^2 - px + r) = x^4 + (q + r - p^2)x^2 + p(r - q)x + qr.$$ Matching coefficients gives $q + r = p^2,$ $qr = 63,$ and $n = p(q - r).$ The factor pairs of $63$ with square sum are $\{7, 9\}$ (sum $16,$ so $p = 4$) and $\{1, 63\}$ (sum $64,$ so $p = 8$), giving $n = 4 \cdot 2 = 8$ or $n = 8 \cdot 62 = 496.$

The smallest positive value overall is $n = 8;$ indeed $(x^2 + 4x + 9)(x^2 - 4x + 7) = x^4 - 8x + 63.$

Solution:

Write $w = x + yi$ with $x, y$ real. The sum of the roots is $4w + 12i - 4 = -a,$ which is real, so $4y + 12 = 0$ and $y = -3.$ The roots are then $x,$ $x + 6i,$ and $2x - 4 - 6i.$ Because the coefficients are real, the two non-real roots must be conjugates, so $2x - 4 = x,$ giving $x = 4.$ The roots are $4,$ $4 + 6i,$ and $4 - 6i.$

Now $$1 + a + b + c = P(1) = (1 - 4)\bigl(1 - (4 + 6i)\bigr)\bigl(1 - (4 - 6i)\bigr) = (-3)(9 + 36) = -135,$$ so $a + b + c = -136$ and $|a + b + c| = 136.$

Solution:

Let $k = |\mathcal{A}|,$ so $|\mathcal{B}| = 12 - k$ with $1 \le k \le 11.$ Since every element lies in exactly one set, $k \notin \mathcal{A}$ means $k \in \mathcal{B},$ and $12 - k \notin \mathcal{B}$ means $12 - k \in \mathcal{A}.$ If $k = 6,$ then $6$ would have to belong to both sets, which is impossible, so $k \ne 6.$

For each other $k,$ the elements $k$ and $12 - k$ are already placed, and the remaining $k - 1$ elements of $\mathcal{A}$ can be chosen from the other $10$ numbers in $\binom{10}{k-1}$ ways, with $\mathcal{B}$ taking the rest. Hence $$N = \sum_{k=1}^{11} \binom{10}{k-1} - \binom{10}{5} = 2^{10} - 252 = 772.$$

Solution:

Center the hexagon at the origin with circumradius $1:$ $A = (1, 0),$ $B = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $C = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ and $F = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right).$ Then $H = \left(0, \frac{\sqrt{3}}{2}\right)$ and $G = \left(\frac{3}{4}, \frac{\sqrt{3}}{4}\right).$ Rotation by $60^\circ$ permutes the six segments, so the smaller hexagon is regular and concentric, and the area ratio is the square of the ratio of distances from the center to a vertex.

One vertex is the intersection of $AH$ and $FG.$ Line $AH$ is $x + \frac{2}{\sqrt{3}}\,y = 1,$ and line $FG$ is $y = 3\sqrt{3}\,x - 2\sqrt{3}.$ Substituting gives $x + 6x - 4 = 1,$ so $x = \frac{5}{7}$ and $y = \frac{\sqrt{3}}{7}.$

That vertex has squared distance $\frac{25}{49} + \frac{3}{49} = \frac{4}{7}$ from the center, while $A$ is at distance $1.$ The ratio of areas is $\frac{4}{7},$ and $m + n = 4 + 7 = 11.$

Solution:

Write $f(x) = a(x - r)(x - s)$ with integer roots $r, s;$ such a polynomial is determined by $a$ and the unordered pair $\{r, s\}.$ The condition $f(0) = 2010$ says $a \cdot rs = 2010 = 2 \cdot 3 \cdot 5 \cdot 67.$ Since $2010$ is squarefree, each of the four primes goes entirely to one of $|a|,$ $|r|,$ $|s|.$

First suppose $|r| \ne |s|.$ Choosing which $k \ge 1$ of the four primes divide the roots ($\binom{4}{k}$ ways) and splitting those primes between the two roots ($2^{k-1}$ unordered ways) gives $\sum_{k=1}^{4} \binom{4}{k} 2^{k-1} = 4 + 12 + 16 + 8 = 40$ choices of magnitudes. For each, the four sign patterns $(+,+),$ $(+,-),$ $(-,+),$ $(-,-)$ of $(r, s)$ are distinct and each forces the sign of $a,$ giving $4 \cdot 40 = 160$ polynomials.

If $|r| = |s|,$ squarefreeness forces $|r| = |s| = 1,$ so $|a| = 2010:$ the options are roots $1, 1$ or $-1, -1$ with $a = 2010,$ or roots $1, -1$ with $a = -2010,$ adding $3$ more. In total $160 + 3 = 163.$

Solution:

There are $\binom{9}{5} = 126$ matrices satisfying (1); we subtract those with two or more constant lines. Two lines of $0$'s are impossible (they would need at least $5$ zeros), and a line of $1$'s and a line of $0$'s cannot cross, so they must be parallel rows or parallel columns; likewise two lines of $1$'s cannot be parallel ($6$ ones), so they must cross, using exactly $3 + 3 - 1 = 5$ ones.

Case 1: a line of $1$'s and a parallel line of $0$'s. There are $6$ choices for the all-$1$ row or column, $2$ for the parallel all-$0$ line, and $3$ ways to fill the remaining parallel line with two $1$'s and one $0:$ $6 \cdot 2 \cdot 3 = 36$ matrices. Every perpendicular line then contains both a $1$ and a $0,$ so no third constant line appears and nothing is double-counted.

Case 2: two crossing lines of $1$'s and $0$'s elsewhere. The pair can be a row and a column ($3 \cdot 3 = 9$), a row or column with a diagonal ($6 \cdot 2 = 12$), or the two diagonals ($1$), for $22$ matrices; one checks the four remaining $0$'s never form a constant line. So $126 - 36 - 22 = 68.$

Solution:

Since the integer bases are in ratio $8 : 7,$ they are $8a$ and $7a$ for a positive integer $a.$ Equal areas make the corresponding altitudes inversely proportional to the bases, say $7h$ and $8h.$ The legs are then $\sqrt{16a^2 + 49h^2}$ and $\sqrt{\frac{49}{4}a^2 + 64h^2},$ and equal perimeters give $$8a + 2\sqrt{16a^2 + 49h^2} = 7a + 2\sqrt{\tfrac{49}{4}a^2 + 64h^2}.$$

Moving $7a$ to the left and squaring yields $a\sqrt{16a^2 + 49h^2} = 15h^2 - 4a^2;$ squaring again and simplifying leaves $225h^4 = 169a^2h^2,$ so $h = \frac{13a}{15}.$ The legs become $$\sqrt{16a^2 + 49 \cdot \tfrac{169a^2}{225}} = \frac{109a}{15} \qquad \text{and} \qquad \sqrt{\tfrac{49}{4}a^2 + 64 \cdot \tfrac{169a^2}{225}} = \frac{233a}{30}.$$

For all sides to be integers, $30$ must divide $a.$ Taking $a = 30$ gives the triangles $(218, 218, 240)$ and $(233, 233, 210),$ each with perimeter $676$ and area $21840.$ The minimum common perimeter is $676.$

Solution:

Condition on Alex and Dylan holding $a$ and $a + 9.$ Blair and Corey then draw $2$ of the remaining $50$ cards, and Alex and Dylan are teammates exactly when both of those cards are below $a$ (Alex and Dylan are the high team) or both are above $a + 9$ (the low team). There are $a - 1$ cards below and $52 - (a + 9) = 43 - a$ cards above, so $$p(a) = \frac{\binom{a-1}{2} + \binom{43-a}{2}}{\binom{50}{2}}.$$

The numerator is $\frac{(a-1)(a-2) + (43-a)(42-a)}{2} = a^2 - 44a + 904,$ so $p(a) \ge \frac{1}{2}$ becomes $a^2 - 44a + 904 \ge \frac{25 \cdot 49}{2},$ that is, $(a - 22)^2 \ge \frac{385}{2}.$ Since $a$ is an integer, $|a - 22| \ge 14,$ so $a \le 8$ or $a \ge 36.$

The parabola is smallest at the admissible points closest to $a = 22:$ $p(8) = p(36) = \frac{\binom{7}{2} + \binom{35}{2}}{\binom{50}{2}} = \frac{616}{1225} = \frac{88}{175},$ which is indeed at least $\frac{1}{2}.$ Thus $m + n = 88 + 175 = 263.$

Solution:

Because the right angle is at $C,$ segment $AB$ is a diameter of the circumcircle; let $O$ be its center, so the radius is $2.$ Let $\alpha = \angle ACP$ and extend $\overline{CP}$ to meet the circle again at $D.$ The central angle over arc $AD$ is $\angle AOD = 2\angle ACD = 2\alpha,$ while vertical angles give $\angle DPB = \angle APC = 2\alpha.$ So $\overline{OD}$ and $\overline{PD}$ make equal angles with line $AB,$ and triangle $ODP$ is isosceles with $DP = DO = 2.$

By the power of the point $P,$ $$AP \cdot PB = CP \cdot PD = 1 \cdot 2 = 2, \qquad AP + PB = 4,$$ so $AP$ and $PB$ are the roots of $t^2 - 4t + 2,$ namely $2 \pm \sqrt{2}.$ Since $\angle BAC \lt 45^\circ,$ we have $BC \lt AC$ with $AC^2 + BC^2 = 16,$ so $AC \gt 2\sqrt{2},$ and the triangle inequality gives $AP \ge AC - CP \gt 2\sqrt{2} - 1 \gt 2 - \sqrt{2}.$ Hence $AP = 2 + \sqrt{2}.$

Therefore $$\frac{AP}{BP} = \frac{2 + \sqrt{2}}{2 - \sqrt{2}} = \frac{(2 + \sqrt{2})^2}{2} = 3 + 2\sqrt{2},$$ and $p + q + r = 3 + 2 + 2 = 7.$

Solution:

By the angle bisector theorem, $AE = \frac{13}{27} \cdot 15$ and $CD = \frac{14}{29} \cdot 13,$ so $E$ lies on $\overline{AN}$ and $D$ lies on $\overline{MC},$ with $$NE = AN - AE = \frac{15}{2} - \frac{65}{9} = \frac{5}{18}, \qquad MD = CM - CD = \frac{13}{2} - \frac{182}{29} = \frac{13}{58}.$$

Since $AMPN$ is cyclic, $\angle ENP = \angle ANP = 180^\circ - \angle AMP = \angle DMP,$ and since $AEPD$ is cyclic, $\angle NEP = 180^\circ - \angle AEP = \angle ADP = \angle MDP.$ Hence triangles $ENP$ and $DMP$ are similar, so $\frac{NP}{MP} = \frac{NE}{MD}.$ By the law of sines in triangles $ANP$ and $AMP,$ whose angles $\angle ANP$ and $\angle AMP$ are supplementary, $$\frac{\sin\angle BAQ}{\sin\angle CAQ} = \frac{\sin\angle NAP}{\sin\angle MAP} = \frac{NP}{MP} = \frac{5/18}{13/58} = \frac{145}{117}.$$

Comparing the areas of triangles $ABQ$ and $ACQ,$ which share the cevian $\overline{AQ},$ $$\frac{BQ}{CQ} = \frac{[ABQ]}{[ACQ]} = \frac{AB \sin\angle BAQ}{AC \sin\angle CAQ} = \frac{15}{13} \cdot \frac{145}{117} = \frac{725}{507},$$ which is in lowest terms since $507 = 3 \cdot 13^2$ and $725 = 5^2 \cdot 29.$ Thus $m - n = 725 - 507 = 218.$