Small Hint:

Let the integers be $a,$ $b,$ and $c = a + b,$ and substitute into $abc = 6(a + b + c)$

Big Hint:

The equation reduces to $ab = 12;$ each way of factoring $12$ gives one value of $N = 12c$

Solution:

Let the integers be $a,$ $b,$ and $c = a + b.$ Then $$\begin{aligned} N &= abc = 6(a + b + c) \\ &= 6 \cdot 2c = 12c, \end{aligned}$$ 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 $156 + 96 + 84 = 336.$

Small Hint:

A number is a multiple of $8$ exactly when the number formed by its last three digits is a multiple of $8$

Big Hint:

Use all ten digits, putting $9876543$ up front, and arrange the digits $0,$ $1,$ $2$ at the end to form a multiple of $8$

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 $120.$

Small Hint:

Each letter forbids exactly one letter from coming immediately after it

Big Hint:

There are $3$ choices for the first letter, and every later letter has exactly $2$ legal choices

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 $3 \cdot 2^6 = 192.$

Small Hint:

The center of the face opposite vertex $A$ is $\frac{B + C + D}{3};$ compare it with the centroid $\frac{A + B + C + D}{4}$

Big Hint:

Each face center is the opposite vertex scaled by $-\frac{1}{3}$ about the centroid, so the two tetrahedra are similar; volume scales as the cube

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 $$\begin{aligned} \frac{B + C + D}{3} &= \frac{4G - A}{3} \\ &= G - \frac{1}{3}(A - G), \end{aligned}$$ 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 $m + n = 1 + 27 = 28.$

Small Hint:

The line where the two cutting planes meet touches the log at one point, so it is tangent to the circular cross-section

Big Hint:

Glue the wedge to a $180^\circ$-rotated copy of itself: the two heights above each point of the disk add up to $12$

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 $n = 216.$

Small Hint:

The $180^\circ$ rotation takes each line to a parallel line, so the two triangles overlap in a central hexagon

Big Hint:

The centroid sits at $\frac{1}{3}$ of each height, so $A'$ lands at height $\frac{h}{3}$ beyond line $BC;$ each protruding corner is similar to the triangle with ratio $\frac{1}{3}$

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 $$\begin{aligned} &[ABC] + 3 \cdot \tfrac{1}{9}[ABC] \\ &= \tfrac{4}{3}[ABC]. \end{aligned}$$

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

Small Hint:

Let $s$ be the side and $\alpha = \angle BAC.$ The diagonals have lengths $2s\cos\alpha$ and $2s\sin\alpha$

Big Hint:

The extended law of sines in triangles $ABD$ and $ACD$ gives $R_1 = \frac{s}{2\cos\alpha}$ and $R_2 = \frac{s}{2\sin\alpha},$ so $\tan\alpha = \frac{R_1}{R_2}$

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 $$\begin{aligned} 12.5 = R_1 &= \frac{BD}{2\sin 2\alpha} \\ &= \frac{2s\sin\alpha}{4\sin\alpha\cos\alpha} \\ &= \frac{s}{2\cos\alpha}. \end{aligned}$$ 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: $$\begin{aligned} &\frac{1}{2} \cdot 2s\cos\alpha \cdot 2s\sin\alpha \\ &= 2s^2\sin\alpha\cos\alpha \\ &= 2 \cdot 500 \cdot \frac{2}{5} = 400. \end{aligned}$$

Small Hint:

The product of two linear functions of $n$ is a quadratic function of $n$

Big Hint:

Fit $an^2 + bn + c$ through the three given terms at $n = 0,$ $1,$ $2,$ then evaluate at $n = 7$

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 $n = 0,$ $1,$ $2:$ $$\begin{aligned} c &= 1440, \\ a + b + c &= 1716, \\ 4a + 2b + c &= 1848, \end{aligned}$$ 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.)

Small Hint:

Divide $P(x)$ by $Q(x);$ at the roots of $Q,$ only the remainder survives

Big Hint:

$P(x) = Q(x)(x^2 + 1)$ $+ x^2 - x + 1,$ so sum $z_i^2 - z_i + 1$ using Vieta’s formulas on $Q$

Solution:

Polynomial division gives $$\begin{aligned} &P(x) = Q(x)\,(x^2 + 1) \\ &\quad {}+ x^2 - x + 1, \end{aligned}$$ 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 $$\sum_{i=1}^4 P(z_i) = 3 - 1 + 4 = 6.$$

Small Hint:

Write the integers as $x$ and $x + 60.$ If $\sqrt{x} + \sqrt{x + 60} = \sqrt{y}$ with $y$ an integer, then $x(x + 60)$ must be a perfect square

Big Hint:

Set $x(x + 60) = z^2$ and factor $(x + 30 + z)(x + 30 - z) = 900;$ discard the case where both integers are perfect squares

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, $$(x + 30)^2 - z^2 = 900,$$ i.e. $$(x + 30 + z)(x + 30 - z) = 900.$$ 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 $48 + 108 = 156.$

Small Hint:

The median to the hypotenuse gives $CM = \frac{25}{2},$ and $D$ lies on the perpendicular to $AB$ at $M,$ so $DM = \sqrt{15^2 - \left(\frac{25}{2}\right)^2}$

Big Hint:

Use area $= \frac{1}{2} \cdot CM \cdot DM \cdot \sin\angle CMD,$ where $\angle CMD = 90^\circ - \angle AMC,$ and get $\cos\angle AMC$ from the law of cosines in triangle $AMC$

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 $$\begin{aligned} DM &= \sqrt{15^2 - \left(\tfrac{25}{2}\right)^2} \\ &= \sqrt{\tfrac{275}{4}} = \tfrac{5\sqrt{11}}{2}. \end{aligned}$$

Let $\beta = \angle AMC.$ In triangle $AMC$ with $AM = CM = \frac{25}{2}$ and $AC = 7,$ the law of cosines gives $$\begin{aligned} \cos\beta &= \frac{\left(\frac{25}{2}\right)^2 + \left(\frac{25}{2}\right)^2 - 7^2} {2 \cdot \frac{25}{2} \cdot \frac{25}{2}} \\ &= \frac{527}{625}. \end{aligned}$$ 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 $$\begin{aligned} [CDM] &= \frac{1}{2} \cdot \frac{25}{2} \cdot \frac{5\sqrt{11}}{2} \cdot \frac{527}{625} \\ &= \frac{527\sqrt{11}}{40}, \end{aligned}$$ and $m + n + p = 527 + 11 + 40$ $= 578.$

Small Hint:

If a candidate got $n$ votes out of $t,$ the condition says $\frac{100n}{t} \le n - 1,$ i.e. $n \ge \frac{t}{t - 100}$

Big Hint:

Since $n$ is an integer, each candidate needs $\lceil \frac{t}{t-100} \rceil$ votes, so $t \ge 27 \lceil \frac{t}{t-100} \rceil;$ find the smallest such $t$ and a vote distribution

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 $134.$

Small Hint:

Let $p_n$ be the probability the bug is home after $n$ moves; then $p_{n+1} = \frac{1}{2}(1 - p_n)$

Big Hint:

Subtract the fixed point: $p_n - \frac{1}{3}$ is geometric with ratio $-\frac{1}{2}$

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

Small Hint:

Equal, parallel opposite sides make the hexagon centrally symmetric, so the $y$-coordinates of opposite vertices all have the same sum

Big Hint:

The $y$-coordinates in order are $0,$ $2,$ $6,$ $10,$ $8,$ $4.$ Express $\angle FAB = 120^\circ$ as a dot product and solve for the side length

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 $$\begin{aligned} \overrightarrow{AB} \cdot \overrightarrow{AF} &= bq + 8 \\ &= -\frac{s^2}{2} = -\frac{b^2 + 4}{2}. \end{aligned}$$ 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 $b = \frac{10}{\sqrt{3}},$ $q = -\frac{8}{\sqrt{3}},$ $p = \frac{8}{\sqrt{3}}.$