Answer: 336

Solution:

Let the integers be $a,$ $b,$ and $c = a + b.$ Then $$N = abc = 6(a + b + c) = 6 \cdot 2c = 12c,$$ and cancelling $c$ from $abc = 12c$ leaves $ab = 12.$

The factorizations $(a, b) = (1, 12),$ $(2, 6),$ $(3, 4)$ give $c = 13,$ $8,$ $7$ and $N = 12c = 156,$ $96,$ $84.$ The sum of all possible values is $156 + 96 + 84 = 336.$

Answer: 120

Solution:

An integer is divisible by $8$ exactly when the number formed by its last three digits is. To make $N$ as large as possible, use all ten digits once each and put the largest digits first: the leading digits are $9876543,$ and the final three digits are some arrangement of $0,$ $1,$ $2$ — provided one of those arrangements is a multiple of $8.$

Checking $012,$ $021,$ $102,$ $120,$ $201,$ $210,$ the only multiple of $8$ is $120.$ So $N = 9{,}876{,}543{,}120,$ and the remainder upon division by $1000$ is $120.$

Answer: 192

Solution:

Each letter rules out exactly one successor ($A$ forbids $B,$ $B$ forbids $C,$ $C$ forbids $A$), so whatever letter has just been written, exactly $2$ of the $3$ letters may come next.

With $3$ choices for the first letter and $2$ for each of the remaining six positions, the number of seven-letter good words is $3 \cdot 2^6 = 192.$

Answer: 28

Solution:

Use position vectors, and let $G = \frac{A + B + C + D}{4}$ be the centroid of the tetrahedron. The center of the face opposite $A$ is $$\frac{B + C + D}{3} = \frac{4G - A}{3} = G - \frac{1}{3}(A - G),$$ so each face center is the image of the opposite vertex under the homothety centered at $G$ with ratio $-\frac{1}{3}.$

Hence the smaller tetrahedron is similar to the larger with ratio $\frac{1}{3},$ and its volume is $\left(\frac{1}{3}\right)^3 = \frac{1}{27}$ of the larger. Thus $m + n = 1 + 27 = 28.$

Answer: 216

Solution:

Take the first cut as horizontal. The line where the two cutting planes meet touches the log at exactly one point, so it is tangent to the circular cross-section of radius $6.$ The wedge therefore stands over the entire disk: its height is $0$ at the tangent point and, because the second cut is at $45^\circ,$ it rises linearly to $12$ at the diametrically opposite point.

Pair each point of the disk with its mirror image through the center: the wedge's heights over the two points add to exactly $12.$ So two copies of the wedge fit together into a cylinder of radius $6$ and height $12,$ and the wedge's volume is $$\frac{1}{2}\,\pi \cdot 6^2 \cdot 12 = 216\pi.$$ Thus $n = 216.$

Answer: 112

Solution:

A $180^\circ$ rotation takes each line to a parallel line, so $\triangle A'B'C'$ is congruent to $\triangle ABC$ with parallel sides. View $BC$ as horizontal and let $h$ be the height of $A$ above it. The centroid $G$ is at height $\frac{h}{3},$ so $A',$ the reflection of $A$ through $G,$ is at height $2 \cdot \frac{h}{3} - h = -\frac{h}{3},$ on the far side of line $BC,$ while $B'$ and $C'$ are at height $\frac{2h}{3}.$

Line $BC$ therefore slices off the corner of $\triangle A'B'C'$ at $A':$ the cut is parallel to $B'C',$ and the corner's height $\frac{h}{3}$ is one third of the triangle's full height $h,$ so the corner is similar with ratio $\frac{1}{3}$ and has area $\frac{1}{9}[ABC].$ The same happens at each side of $\triangle ABC,$ and these three corners are exactly the part of $\triangle A'B'C'$ outside $\triangle ABC.$ Hence the union has area $$[ABC] + 3 \cdot \tfrac{1}{9}[ABC] = \tfrac{4}{3}[ABC].$$

By Heron's formula with $s = 21,$ $[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84,$ so the union has area $\frac{4}{3} \cdot 84 = 112.$

Answer: 400

Solution:

Let $s$ be the side length and $\alpha = \angle BAC$ (the diagonal $AC$ bisects angle $A$). The diagonals then have lengths $AC = 2s\cos\alpha$ and $BD = 2s\sin\alpha.$ In triangle $ABD,$ side $BD$ subtends the angle $\angle BAD = 2\alpha,$ so the extended law of sines gives $$12.5 = R_1 = \frac{BD}{2\sin 2\alpha} = \frac{2s\sin\alpha}{4\sin\alpha\cos\alpha} = \frac{s}{2\cos\alpha}.$$ In triangle $ACD,$ side $AC$ subtends $\angle ADC = 180^\circ - 2\alpha,$ so similarly $25 = R_2 = \frac{s}{2\sin\alpha}.$

Dividing, $\tan\alpha = \frac{R_1}{R_2} = \frac{1}{2},$ so $\sin\alpha = \frac{1}{\sqrt{5}}$ and $\cos\alpha = \frac{2}{\sqrt{5}}.$ Then $s = 2R_2\sin\alpha = \frac{50}{\sqrt{5}} = 10\sqrt{5}.$

The area is half the product of the diagonals: $\frac{1}{2} \cdot 2s\cos\alpha \cdot 2s\sin\alpha = 2s^2\sin\alpha\cos\alpha = 2 \cdot 500 \cdot \frac{2}{5} = 400.$

Answer: 348

Solution:

The $n$th term of an arithmetic sequence is linear in $n,$ so the product of corresponding terms of two arithmetic sequences is a quadratic $t_n = an^2 + bn + c.$ Indexing the given terms by $n = 0, 1, 2:$ $$c = 1440, \qquad a + b + c = 1716, \qquad 4a + 2b + c = 1848,$$ which give $a + b = 276$ and $2a + b = 204,$ so $a = -72,$ $b = 348,$ $c = 1440.$

The eighth term is $t_7 = -72 \cdot 49 + 348 \cdot 7 + 1440 = 348.$ (Indeed $t_n = (180 - 24n)(8 + 3n),$ a product of two arithmetic sequences matching the given terms.)

Answer: 6

Solution:

Polynomial division gives $$P(x) = Q(x)\,(x^2 + 1) + x^2 - x + 1,$$ so $P(z_i) = z_i^2 - z_i + 1$ for each root $z_i$ of $Q.$

By Vieta's formulas for $Q(x) = x^4 - x^3 - x^2 - 1,$ we have $\sum z_i = 1$ and $\sum_{i \lt j} z_i z_j = -1,$ so $\sum z_i^2 = \left(\sum z_i\right)^2 - 2\sum_{i \lt j} z_i z_j = 1 + 2 = 3.$ Therefore $$\sum_{i=1}^4 P(z_i) = 3 - 1 + 4 = 6.$$

Answer: 156

Solution:

Let the integers be $x$ and $x + 60,$ and suppose $\sqrt{x} + \sqrt{x + 60} = \sqrt{y}.$ Squaring, $y = 2x + 60 + 2\sqrt{x(x + 60)},$ so $x(x + 60)$ must be a perfect square, say $z^2.$ Completing the square, $$(x + 30)^2 - z^2 = 900, \qquad \text{i.e.} \qquad (x + 30 + z)(x + 30 - z) = 900.$$ The two factors have the same parity and their product is even, so both are even.

The factor pairs $(450, 2),$ $(150, 6),$ $(90, 10),$ $(50, 18)$ give $x + 30 = 226,$ $78,$ $50,$ $34,$ so $x = 196,$ $48,$ $20,$ $4.$ For $x = 196$ the integers are $196$ and $256,$ both perfect squares, so $\sqrt{y} = 14 + 16 = 30$ and $y = 900$ is a perfect square — not allowed. For $x = 48$ the integers are $48$ and $108,$ with $\sqrt{48} + \sqrt{108} = 4\sqrt{3} + 6\sqrt{3} = \sqrt{300},$ and $300$ is not a perfect square.

The maximum possible sum is therefore $48 + 108 = 156.$

Answer: 578

Solution:

The hypotenuse is $AB = \sqrt{7^2 + 24^2} = 25,$ and the median to the hypotenuse gives $CM = \frac{25}{2}.$ Since $AD = BD,$ point $D$ lies on the perpendicular to $AB$ at $M,$ so $DM \perp AB$ and $$DM = \sqrt{15^2 - \left(\tfrac{25}{2}\right)^2} = \sqrt{\tfrac{275}{4}} = \tfrac{5\sqrt{11}}{2}.$$

Let $\beta = \angle AMC.$ In triangle $AMC$ with $AM = CM = \frac{25}{2}$ and $AC = 7,$ the law of cosines gives $$\cos\beta = \frac{\left(\frac{25}{2}\right)^2 + \left(\frac{25}{2}\right)^2 - 7^2} {2 \cdot \frac{25}{2} \cdot \frac{25}{2}} = \frac{527}{625}.$$ Since $C$ and $D$ are on the same side of $AB$ and $MD \perp AB,$ we have $\angle CMD = 90^\circ - \beta,$ so $\sin\angle CMD = \cos\beta.$

Therefore $[CDM] = \frac{1}{2} \cdot \frac{25}{2} \cdot \frac{5\sqrt{11}}{2} \cdot \frac{527}{625} = \frac{527\sqrt{11}}{40},$ and $m + n + p = 527 + 11 + 40 = 578.$

Answer: 134

Solution:

Let $t$ be the number of members. A candidate with $n$ votes has percentage $\frac{100n}{t},$ so the condition is $\frac{100n}{t} \le n - 1,$ which rearranges to $n(t - 100) \ge t.$ This forces $t \gt 100$ and $$n \ge \frac{t}{t - 100}.$$

If $t \le 133,$ then $\frac{t}{t - 100} \ge \frac{133}{33} \gt 4,$ so every candidate needs at least $5$ votes, and the total is at least $27 \cdot 5 = 135 \gt t$ — impossible.

For $t = 134,$ each candidate needs $n \ge \frac{134}{34},$ i.e. at least $4$ votes, and this is achievable: let $26$ candidates receive $4$ votes each and one receive $30.$ Indeed $\frac{400}{134} \approx 2.99 \le 3$ and $\frac{3000}{134} \approx 22.4 \le 29.$ So the smallest possible number of members is $134.$

Answer: 683

Solution:

Let $p_n$ be the probability that the bug is at its starting vertex after $n$ moves, so $p_0 = 1.$ The bug is home after move $n + 1$ exactly when it was elsewhere after move $n$ (probability $1 - p_n$) and then chose the starting vertex (probability $\frac{1}{2}$): $$p_{n+1} = \frac{1}{2}(1 - p_n).$$

The fixed point of this recurrence is $\frac{1}{3},$ and $p_{n+1} - \frac{1}{3} = -\frac{1}{2}\left(p_n - \frac{1}{3}\right),$ so $p_n = \frac{1}{3} + \frac{2}{3}\left(-\frac{1}{2}\right)^n.$

For $n = 10:$ $p_{10} = \frac{1}{3}\left(1 + \frac{2}{1024}\right) = \frac{1}{3} \cdot \frac{1026}{1024} = \frac{171}{512}.$ Since $171 = 9 \cdot 19$ and $512 = 2^9$ share no factor, $m + n = 171 + 512 = 683.$

Answer: 51

Solution:

Opposite sides are parallel, equal in length, and traversed in opposite directions, so $\overrightarrow{AB} = \overrightarrow{ED},$ $\overrightarrow{BC} = \overrightarrow{FE},$ $\overrightarrow{CD} = \overrightarrow{AF}:$ the hexagon is centrally symmetric, and opposite vertices' $y$-coordinates share a common sum, namely $\frac{0 + 2 + \cdots + 10}{3} = 10.$ From $y_A = 0$ and $y_B = 2$ we get $y_D = 10,$ $y_E = 8,$ and convexity puts $y_C = 6,$ $y_F = 4.$ Write $\overrightarrow{AB} = (b, 2),$ $\overrightarrow{BC} = (p, 4),$ $\overrightarrow{CD} = (q, 4).$ Equal side lengths give $s^2 = b^2 + 4 = p^2 + 16 = q^2 + 16,$ so $p = \pm q;$ since $p = q$ would make $B,$ $C,$ $D$ collinear, $p = -q.$

Since $\overrightarrow{AF} = \overrightarrow{CD},$ we have $F = (q, 4),$ and $\angle FAB = 120^\circ$ gives $$\overrightarrow{AB} \cdot \overrightarrow{AF} = bq + 8 = -\frac{s^2}{2} = -\frac{b^2 + 4}{2}.$$ Taking $b \gt 0$ forces $q \lt 0,$ so $q = -\sqrt{b^2 - 12},$ and the equation becomes $b\sqrt{b^2 - 12} = \frac{b^2 + 20}{2}.$ Squaring yields $3b^4 - 88b^2 - 400 = 0,$ so $b^2 = \frac{100}{3},$ giving $b = \frac{10}{\sqrt{3}},$ $q = -\frac{8}{\sqrt{3}},$ $p = \frac{8}{\sqrt{3}}.$

The vertices are $A = (0, 0),$ $B = \left(\frac{10}{\sqrt{3}}, 2\right),$ $C = (6\sqrt{3}, 6),$ $D = \left(\frac{10}{\sqrt{3}}, 10\right),$ $E = (0, 8),$ $F = \left(-\frac{8}{\sqrt{3}}, 4\right).$ The hexagon splits into the parallelogram $ABDE,$ with vertical side $AE = 8$ and horizontal offset $b$ (area $8b$), plus the two congruent triangles $BCD$ and $EFA,$ each with vertical base $8$ and horizontal height $\frac{8}{\sqrt{3}}.$ The total area is $$8 \cdot \frac{10}{\sqrt{3}} + 2 \cdot \frac{1}{2} \cdot 8 \cdot \frac{8}{\sqrt{3}} = \frac{144}{\sqrt{3}} = 48\sqrt{3},$$ so $m + n = 48 + 3 = 51.$

Answer: 15

Solution:

The coefficient of $x^k$ in $P(x)$ is $24 - |24 - k|$ for $1 \le k \le 47,$ and consecutive coefficients differ by $+1$ up through $x^{24}$ and by $-1$ afterwards. Multiplying by $1 - x$ therefore telescopes: $$(1 - x)P(x) = (x + x^2 + \cdots + x^{24}) - (x^{25} + \cdots + x^{48}) = (x + x^2 + \cdots + x^{24})(1 - x^{24}),$$ so for $x \ne 1,$ $$P(x) = x\left(\frac{x^{24} - 1}{x - 1}\right)^2.$$

The distinct zeros of $P$ are therefore $0$ together with the $24$th roots of unity other than $1:$ $z_k = \cos 15k^\circ + i \sin 15k^\circ$ for $k = 1, \ldots, 23.$ The zero $0$ contributes nothing, and $z_k^2 = \cos 30k^\circ + i \sin 30k^\circ,$ so $|b_k| = |\sin 30k^\circ|.$

As $k$ runs from $1$ to $12,$ the values $|\sin 30k^\circ|$ are $\frac{1}{2}, \frac{\sqrt{3}}{2}, 1, \frac{\sqrt{3}}{2}, \frac{1}{2}, 0$ repeated twice, summing to $4 + 2\sqrt{3};$ the terms for $k = 13, \ldots, 23$ repeat those for $k = 1, \ldots, 11$ and add another $4 + 2\sqrt{3}.$ The total is $8 + 4\sqrt{3},$ so $m + n + p = 8 + 4 + 3 = 15.$