Solution:

The angles of a hexagon sum to $720^\circ,$ so each of the four congruent angles measures $\frac{720 - 2 \cdot 90}{4} = 135$ degrees. Let $AB = x.$ The diagonals $BF$ and $CE$ cut off the right isosceles triangles $FAB$ and $CDE,$ each with legs $x$ and hypotenuse $x\sqrt{2},$ and the $135^\circ$ angles guarantee that the remaining piece $BCEF$ is a rectangle with sides $x\sqrt{2}$ and $x.$

Hence the area is $$2 \cdot \frac{1}{2}x^2 + x \cdot x\sqrt{2} = x^2(1 + \sqrt{2}) = 2116(\sqrt{2} + 1),$$ so $x^2 = 2116$ and $AB = x = 46.$

Solution:

The triangle inequality requires $\log n \lt \log 12 + \log 75 = \log 900$ and $\log 12 + \log n \gt \log 75,$ that is $\log n \gt \log 75 - \log 12 = \log \frac{25}{4}.$ (The third inequality is the first one again.)

So $\frac{25}{4} \lt n \lt 900,$ which for integers means $7 \le n \le 899.$ That gives $899 - 7 + 1 = 893$ possible values of $n.$

Solution:

$P = 1 \cdot 3 \cdot 5 \cdots 199,$ so $k$ is the total number of factors of $3$ among the odd numbers up to $199.$ The odd multiples of $3$ are $3 \cdot 1, 3 \cdot 3, \ldots, 3 \cdot 65,$ and there are $33$ of them. The odd multiples of $9$ are $9 \cdot 1, \ldots, 9 \cdot 21:$ $11$ of them. The odd multiples of $27$ are $27, 81, 135, 189:$ $4$ of them. The only odd multiple of $81$ at most $199$ is $81$ itself, and there are no multiples of $243.$

Each layer contributes one additional factor of $3,$ so $k = 33 + 11 + 4 + 1 = 49.$

Solution:

The term $a_6$ is smaller than every other term of the permutation, so $a_6 = 1.$ Now choose which five of the remaining $11$ numbers occupy positions $1$ through $5:$ they must appear in decreasing order, so their arrangement is forced, and the other six numbers must fill positions $7$ through $12$ in increasing order, which is also forced.

Every choice of the five numbers gives exactly one valid permutation, so the count is $\binom{11}{5} = 462.$

Solution:

Let the probability of face $F$ be $\frac{1}{6} + x,$ so the face opposite $F$ has probability $\frac{1}{6} - x$ (the six probabilities must sum to $1$). Since opposite faces sum to $7,$ a total of $7$ occurs exactly when the two dice show a pair of opposite faces. Of the six ordered pairs that sum to $7,$ four use only ordinary faces, and two pair $F$ with its opposite. Thus $$\frac{47}{288} = 4\left(\frac{1}{6}\right)^2 + 2\left(\frac{1}{6} + x\right)\left(\frac{1}{6} - x\right) = \frac{1}{6} - 2x^2.$$

Since $\frac{1}{6} = \frac{48}{288},$ this gives $2x^2 = \frac{1}{288},$ so $x = \frac{1}{24}.$ The probability of face $F$ is $\frac{1}{6} + \frac{1}{24} = \frac{5}{24},$ and $m + n = 5 + 24 = 29.$

Solution:

Place $A = (0, 0),$ $B = (1, 0),$ $C = (1, 1),$ $D = (0, 1).$ By the symmetry of the equilateral triangle across diagonal $\overline{AC},$ we have $BE = DF.$ Let $BE = t,$ so $CE = CF = 1 - t.$ Then $AE^2 = 1 + t^2$ and $EF^2 = 2(1 - t)^2,$ and setting them equal gives $t^2 - 4t + 1 = 0,$ so $t = 2 - \sqrt{3}$ (taking the root less than $1$).

Thus $E = (1,\, 2 - \sqrt{3}),$ and line $AE$ is $y = (2 - \sqrt{3})x.$ If the smaller square has side $q,$ its vertex opposite $B$ is $(1 - q,\, q),$ which must lie on line $AE:$ $$q = (2 - \sqrt{3})(1 - q) \quad \Longrightarrow \quad q = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} = \frac{(2 - \sqrt{3})(3 + \sqrt{3})}{6} = \frac{3 - \sqrt{3}}{6}.$$

So $a = 3,$ $b = 3,$ $c = 6,$ and $a + b + c = 12.$

Solution:

There are $999$ pairs in all ($a = 1, \ldots, 999$); count the forbidden ones. If $a$ has units digit $0,$ so does $b,$ and writing $a = 10r,$ $b = 10s$ gives $r + s = 100$ with $1 \le r \le 99:$ that is $99$ forbidden pairs.

Now suppose both units digits are nonzero. Then a number in the pair has a zero digit exactly when it is a three-digit number of the form $h0u$ with $h, u \in \{1, \ldots, 9\}$ (a one- or two-digit number with nonzero units digit has no zero digit). If $a = h0u,$ then $b = 1000 - a = 100(9 - h) + 90 + (10 - u)$ has tens digit $9,$ so $b$ is not also of that form. Hence the forbidden pairs here are those where exactly one of $a, b$ equals $h0u:$ $81 + 81 = 162$ pairs.

The total number of forbidden pairs is $99 + 162 = 261,$ so the answer is $999 - 261 = 738.$

Solution:

The rotations and reflections of the large triangle realize every permutation of the three corner triangles while fixing the center triangle. So two large triangles are indistinguishable exactly when they have the same center color and the same multiset of three corner colors.

Count the multisets of corner colors from six colors: all three the same ($6$ ways), exactly two the same ($6 \cdot 5 = 30$ ways, choosing the repeated color and then the different one), or all three different ($\binom{6}{3} = 20$ ways). That is $6 + 30 + 20 = 56$ multisets.

With $6$ independent choices for the center color, the total is $6 \cdot 56 = 336.$

Solution:

A common internal tangent meets the segment between the centers at the point dividing it in the ratio of the radii. For $\mathcal{C}_1$ and $\mathcal{C}_2$ that point is $(4, 0),$ at distance $4$ from $(0,0).$ If $t_1$ makes angle $\theta$ with the $x$-axis, then $\sin\theta = \frac{1}{4},$ so $\tan\theta = \frac{1}{\sqrt{15}}$ and $t_1$ is $y = \frac{1}{\sqrt{15}}(x - 4).$ For $\mathcal{C}_2$ and $\mathcal{C}_3$ the point is $(16, 0),$ at distance $4$ from $(12, 0);$ here $\sin\theta = \frac{2}{4} = \frac{1}{2},$ so the slope is $-\frac{1}{\sqrt{3}}$ and $t_2$ is $y = -\frac{1}{\sqrt{3}}(x - 16).$

Setting the two expressions equal and multiplying by $\sqrt{15}$ gives $x - 4 = -\sqrt{5}\,(x - 16),$ so $x(1 + \sqrt{5}) = 4 + 16\sqrt{5}$ and $$x = \frac{4 + 16\sqrt{5}}{1 + \sqrt{5}} = \frac{(4 + 16\sqrt{5})(\sqrt{5} - 1)}{4} = \frac{76 - 12\sqrt{5}}{4} = 19 - 3\sqrt{5}.$$

Thus $p + q + r = 19 + 3 + 5 = 27.$

Solution:

Teams $A$ and $B$ each have $5$ games left, none against each other, so all $2^5 \cdot 2^5 = 1024$ outcomes are equally likely. Since $A$ already leads by one point, $A$ finishes with more points exactly when $A$ wins at least as many remaining games as $B$ does.

The number of outcomes with equal win counts is $$\sum_{k=0}^{5} \binom{5}{k}^2 = \binom{10}{5} = 252.$$ By symmetry, the other $1024 - 252 = 772$ outcomes split evenly between $A$ winning more and $B$ winning more.

So the probability is $\frac{252 + 386}{1024} = \frac{638}{1024} = \frac{319}{512},$ and $m + n = 319 + 512 = 831.$

Solution:

Let $S_n = a_1 + \cdots + a_n.$ We claim $2S_n = a_{n+2} + a_n,$ which holds for $n = 1$ since $2 = 1 + 1.$ If it holds for $n,$ then $$2S_{n+1} = 2S_n + 2a_{n+1} = a_{n+2} + 2a_{n+1} + a_n = a_{n+3} + a_{n+1}$$ by the recurrence, completing the induction.

Therefore $S_{28} = \frac{a_{30} + a_{28}}{2} = \frac{20603361 + 6090307}{2} = 13346834,$ whose remainder upon division by $1000$ is $834.$

Solution:

By construction $ADFE$ is a parallelogram with $AD = 13,$ $DF = AE = 11,$ and $\angle ADF = 180^\circ - \angle DAE = 120^\circ.$ Hence $[ADF] = \frac{1}{2} \cdot 13 \cdot 11 \sin 120^\circ = \frac{143\sqrt{3}}{4},$ and by the law of cosines, $$AF^2 = 13^2 + 11^2 - 2 \cdot 13 \cdot 11 \cos 120^\circ = 169 + 121 + 143 = 433.$$

Since $G$ lies on the circle, inscribed angles give $\angle GCB = \angle GAB = \angle FAD$ (both subtend arc $GB$) and $\angle CBG = \angle CAG$ (both subtend arc $CG$); and $\angle CAG = \angle AFD$ because $\overline{AE} \parallel \overline{DF}.$ So $\triangle CBG \sim \triangle AFD$ with ratio $\frac{CB}{AF}.$ The side of an equilateral triangle inscribed in a circle of radius $2$ is $BC = 2\sqrt{3}.$

Therefore $$[CBG] = \left(\frac{2\sqrt{3}}{\sqrt{433}}\right)^2 \cdot \frac{143\sqrt{3}}{4} = \frac{12}{433} \cdot \frac{143\sqrt{3}}{4} = \frac{429\sqrt{3}}{433},$$ and $p + q + r = 429 + 3 + 433 = 865.$

Solution:

The sum of the $(k+1)$th through $m$th positive odd integers is $m^2 - k^2 = (m - k)(m + k).$ Writing $a = m - k$ and $b = m + k,$ the representations of $N$ correspond exactly to the factorizations $N = ab$ with $a \le b$ and $a, b$ of the same parity (then $m = \frac{a + b}{2},$ $k = \frac{b - a}{2}$). So we need $N$ to have exactly $5$ such factorizations.

If $N$ is odd, every divisor pair works, so $N$ needs $9$ or $10$ divisors, i.e. $N = p^8,$ $p^9,$ $p^2q^2,$ or $pq^4$ with $p, q$ distinct odd primes. Below $1000,$ $p^8$ and $p^9$ are impossible, $p^2 q^2$ gives $225$ and $441,$ and $pq^4$ gives $3^4 \cdot 5,$ $3^4 \cdot 7,$ $3^4 \cdot 11:$ five odd values.

If $N$ is even, both factors must be even, so $N = 4M$ and the factorizations correspond to divisor pairs of $M,$ with no parity restriction; we need $M \lt 250$ with $9$ or $10$ divisors. With $9$ divisors: $36,$ $100,$ $196,$ $225.$ With $10$ divisors ($pq^4$): $3 \cdot 2^4,$ $5 \cdot 2^4,$ $7 \cdot 2^4,$ $11 \cdot 2^4,$ $13 \cdot 2^4,$ $2 \cdot 3^4.$ That is $4 + 6 = 10$ even values, for a total of $5 + 10 = 15.$

Solution:

Write the integers from $0$ to $10^n - 1$ as $n$-digit strings with leading zeros. Each of the $n$ digit positions takes each digit value equally often, so each nonzero digit appears $n \cdot 10^{n-1}$ times. Adding the digit $1$ of $10^n$ itself, $$S_n = 1 + n \cdot 10^{n-1}\left(1 + \frac{1}{2} + \cdots + \frac{1}{9}\right) = 1 + \frac{7129}{2520}\, n \cdot 10^{n-1}.$$

Since $\gcd(7129, 2520) = 1,$ the sum is an integer exactly when $2520 \mid n \cdot 10^{n-1}.$ Now $2520 = 2^3 \cdot 3^2 \cdot 5 \cdot 7,$ and for $n \ge 4$ the factor $10^{n-1}$ supplies $2^3 \cdot 5,$ leaving the condition $63 \mid n$ (a power of $10$ has no factors of $3$ or $7$). For $n = 1, 2, 3$ the products $1, 20, 300$ are not multiples of $2520.$

The smallest solution is therefore $n = 63.$

Solution:

Each radical $\sqrt{y^2 - \frac{1}{16}}$ is the leg of a right triangle with hypotenuse $y$ and other leg $\frac{1}{4}.$ So the first equation says: in a triangle $XYZ$ with $x = YZ,$ $y = ZX,$ $z = XY,$ the altitude from $X$ has length $\frac{1}{4},$ and its foot splits $YZ$ into the two radical lengths. The other equations say the altitudes to sides $y$ and $z$ are $\frac{1}{5}$ and $\frac{1}{6}.$

If $K$ is the area of this triangle, then $K = \frac{1}{2} \cdot x \cdot \frac{1}{4}$ gives $x = 8K,$ and likewise $y = 10K$ and $z = 12K.$ These are proportional to $8, 10, 12,$ and $8^2 + 10^2 \gt 12^2,$ so the triangle is acute and the altitude feet do land inside the sides. Heron's formula with $s = 15K$ gives $$K^2 = 15K \cdot 7K \cdot 5K \cdot 3K = 1575K^4,$$ so $K^2 = \frac{1}{1575}$ and $K = \frac{1}{15\sqrt{7}}.$

Then $x + y + z = 30K = \frac{2}{\sqrt{7}},$ so $m + n = 2 + 7 = 9.$