Answer: 25

Small Hint:

Factor everything into powers of $2$ and $3:$ $12^{12} = 2^{24} 3^{12},$ $6^6 = 2^6 3^6,$ $8^8 = 2^{24}$

Big Hint:

The lcm of $6^6$ and $8^8$ already supplies $2^{24}$ but only $3^6,$ so $k$ must contribute exactly $3^{12}$

Solution:

Since $12^{12} = 2^{24} 3^{12},$ $6^6 = 2^6 3^6,$ and $8^8 = 2^{24},$ the number $k$ can involve no primes other than $2$ and $3,$ so write $k = 2^a 3^b.$ The least common multiple of the three numbers is then $2^{\max(24,\,a)} \, 3^{\max(6,\,b)}.$

Matching this to $2^{24} 3^{12}$ requires $\max(24, a) = 24,$ i.e. $0 \le a \le 24,$ and $\max(6, b) = 12,$ i.e. $b = 12.$ That gives $25$ choices for $a$ and one for $b,$ so there are $25$ values of $k.$

Answer: 480

Small Hint:

All four constraints together confine $(x, y)$ to the square $1 \le x, y \le 30$

Big Hint:

Inside that square, subtract the pairs with $y \gt 2x$ and those with $x \gt 2y;$ the two bad sets are disjoint and have equal size

Solution:

The chains unpack into four conditions: $x \le 2y,$ $2y \le 60,$ $y \le 2x,$ and $2x \le 60.$ So $(x, y)$ lies in the square $1 \le x, y \le 30,$ and within it we must avoid $x \gt 2y$ and $y \gt 2x,$ which cannot both happen.

Pairs with $y \gt 2x:$ for each $x$ from $1$ to $14$ the values $y = 2x + 1, \ldots, 30$ work, giving $\sum_{x=1}^{14} (30 - 2x)$ $= 420 - 210$ $= 210$ pairs. By the symmetry swapping $x$ and $y,$ there are also $210$ pairs with $x \gt 2y.$

The answer is $30 \cdot 30 - 210 - 210 = 480.$

Answer: 800

Small Hint:

For $x \ge 0$ the equation rearranges to $2x(y + 20) = 400 - y^2,$ and the right side factors

Big Hint:

Each sign case yields two lines; the four rays bound a parallelogram with horizontal sides on $y = 20$ and $y = -20$

Solution:

For $x \ge 0$ rewrite the equation as $2x(y + 20)$ $= 400 - y^2$ $= (20 - y)(20 + y),$ so either $y = -20$ or $y = 20 - 2x.$ For $x \le 0$ it becomes $2x(y - 20) = -(y - 20)(y + 20),$ so either $y = 20$ or $y = -20 - 2x.$ The graph therefore consists of two horizontal rays and two rays of slope $-2.$

These rays bound a parallelogram: the top edge runs from $(-20, 20)$ to $(0, 20)$ along $y = 20,$ the bottom edge from $(0, -20)$ to $(20, -20)$ along $y = -20,$ and the two slanted edges of slope $-2$ connect them.

The parallelogram has horizontal base $20$ and height $40$ between the lines $y = 20$ and $y = -20,$ so its area is $20 \cdot 40 = 800.$

Answer: 17

Small Hint:

A player’s sum is odd exactly when the player holds an odd number of odd tiles; there are five odd tiles in all

Big Hint:

The odd tiles must split $3,$ $1,$ $1$ among the players. Count the ways to deal odds and evens separately.

Solution:

A player’s three tiles have an odd sum exactly when the player holds an odd number of odd tiles — one or three. The nine tiles include five odd and four even, and the only way to split five odd tiles into three groups of size one or three is $3 + 1 + 1.$

Count favorable deals: choose which player gets three odd tiles ($3$ ways), choose that player’s odd tiles ($\binom{5}{3} = 10$ ways), give one of the two remaining odd tiles to each other player ($2$ ways), then split the four even tiles two and two between those players ($\binom{4}{2} = 6$ ways), for $3 \cdot 10 \cdot 2 \cdot 6 = 360$ deals. The total number of deals is $\binom{9}{3}\binom{6}{3} = 84 \cdot 20 = 1680.$

The probability is $\frac{360}{1680} = \frac{3}{14},$ so $m + n = 3 + 14 = 17.$

Answer: 40

Small Hint:

$\frac{k(k-1)}{2}$ is an integer, so each cosine is $\pm 1,$ with sign depending only on $k \bmod 4$

Big Hint:

Group the $80$ terms into consecutive blocks of four starting at $k = 19;$ consecutive triangular numbers differ by $k,$ so each block collapses

Solution:

Since $k(k-1)$ is even, $n_k = \frac{k(k-1)}{2}$ is an integer and $\cos\frac{k(k-1)\pi}{2}$ $= \cos(n_k \pi)$ $= (-1)^{n_k}.$ The parity of the triangular number $n_k$ depends only on $k \bmod 4:$ it is even for $k \equiv 0, 1 \pmod 4$ and odd for $k \equiv 2, 3 \pmod 4.$ So $A_k = n_k$ when $k \equiv 0, 1 \pmod 4$ and $A_k = -n_k$ when $k \equiv 2, 3 \pmod 4.$

Group the $80$ terms into $20$ consecutive blocks of four starting at $k = 19 \equiv 3 \pmod 4.$ Using $n_{j+1} - n_j = j,$ each block with $k \equiv 3 \pmod 4$ collapses: $$\begin{aligned} &A_k + A_{k+1} \\ &\quad {}+ A_{k+2} + A_{k+3} \\ &= (n_{k+1} - n_k) \\ &\quad {}- (n_{k+3} - n_{k+2}) \\ &= k - (k + 2) \\ &= -2. \end{aligned}$$

The total is $20 \cdot (-2) = -40,$ so the requested absolute value is $40.$

Answer: 308

Small Hint:

Two pairs of similar triangles: $AQ \parallel DC$ gives $\frac{PQ}{PC} = \frac{PA}{PD},$ and $BC \parallel PD$ gives $\frac{RC}{RP} = \frac{BC}{PD}$

Big Hint:

With $a = \frac{PA}{AD},$ the ratios give $PQ : QR : RC$ $= a(a+2) : 1 : (a+1);$ use $\frac{PQ}{QR} = \frac{735}{112}$ to find $a$

Solution:

Let $a = \frac{PA}{AD}.$ Since $AQ \parallel DC,$ triangles $PAQ$ and $PDC$ are similar, so $\frac{PQ}{PC} = \frac{PA}{PD} = \frac{a}{a+1}.$ Since $BC \parallel AD,$ i.e. $BC \parallel PD,$ triangles $RBC$ and $RDP$ are similar, so $\frac{RC}{RP} = \frac{BC}{PD} = \frac{1}{a+1},$ which gives $\frac{RC}{PC} = \frac{1}{a+2}.$

Writing $PC = L,$ we get $PQ = \frac{a}{a+1}L,$ $RC = \frac{L}{a+2},$ and $$\begin{aligned} QR &= L - PQ - RC \\ &= \frac{L}{(a+1)(a+2)}. \end{aligned}$$ Hence $\frac{PQ}{QR} = a(a+2) = \frac{735}{112} = \frac{105}{16},$ so $16a^2 + 32a - 105 = 0,$ which factors as $(4a - 7)(4a + 15) = 0,$ giving $a = \frac{7}{4}.$

Finally $RC = (a + 1)\,QR$ $= \frac{11}{4} \cdot 112$ $= 308.$

Answer: 196

Small Hint:

Substitute $x_i = 2y_i - 1$ to turn the odd variables into arbitrary positive integers

Big Hint:

The new equation is $y_1 + y_2 + y_3 + y_4 = 51;$ count its positive solutions with stars and bars

Solution:

Write $x_i = 2y_i - 1$ where each $y_i$ is a positive integer. Then $x_1 + x_2 + x_3 + x_4 = 98$ becomes $2(y_1 + y_2 + y_3 + y_4) - 4 = 98,$ so $y_1 + y_2 + y_3 + y_4 = 51.$

By stars and bars, the number of solutions in positive integers is $\binom{50}{3} = 19600.$ Therefore $\frac{n}{100} = 196.$

Answer: 618

Small Hint:

Write out terms: $1000 - x,$ $2x - 1000,$ $2000 - 3x,$ $5x - 3000, \ldots$ — the coefficients are Fibonacci numbers

Big Hint:

Keeping the twelfth and thirteenth terms nonnegative pins $x$ between $\frac{55000}{89}$ and $\frac{89000}{144},$ an interval containing exactly one integer

Solution:

Computing terms, $a_3 = 1000 - x,$ $a_4 = 2x - 1000,$ $a_5 = 2000 - 3x,$ $a_6 = 5x - 3000,$ and in general $$\begin{aligned} a_{2k+1} &= 1000 F_{2k-1} - x F_{2k}, \\ a_{2k+2} &= x F_{2k+1} - 1000 F_{2k}, \end{aligned}$$ where $F_1 = F_2 = 1,$ $F_3 = 2, \ldots$ are the Fibonacci numbers. The sequence keeps going exactly as long as its terms stay nonnegative, so a long sequence requires $\frac{x}{1000}$ to be squeezed between the ratios $\frac{F_{2k}}{F_{2k+1}}$ and $\frac{F_{2k-1}}{F_{2k}}$ for larger and larger $k.$

For the first $13$ terms to be nonnegative we need $a_{12} = 89x - 55000 \ge 0$ and $a_{13} = 89000 - 144x \ge 0,$ i.e. $617.9\ldots \le x \le 618.05\ldots,$ so $x = 618.$ If $x \le 617$ the sequence turns negative by $a_{12},$ and if $x \ge 619$ it turns negative by $a_{13},$ so every other integer gives a shorter sequence.

Indeed $x = 618$ yields $1000, 618, 382, 236, 146,$ $90, 56, 34, 22, 12,$ $10, 2, 8, -6,$ a sequence of $14$ terms, the maximum possible. The answer is $618.$

Answer: 87

Small Hint:

Plot the two arrival times as a point in a $60 \times 60$ square; they meet exactly when $|x - y| \lt m$

Big Hint:

The complement consists of two right triangles that fit together into a square of side $60 - m,$ so $(60 - m)^2 = \frac{3}{5} \cdot 3600$

Solution:

Let the arrival times be $x$ and $y$ minutes after $9$ a.m., so $(x, y)$ is uniform in a $60 \times 60$ square. The two people meet exactly when $|x - y| \lt m.$

The non-meeting region $|x - y| \ge m$ consists of two right triangles with legs $60 - m,$ with total area $(60 - m)^2.$ Meeting with probability $40\%$ means $$(60 - m)^2 = 0.6 \cdot 3600 = 2160,$$ so $60 - m = \sqrt{2160} = 12\sqrt{15}.$

Thus $m = 60 - 12\sqrt{15},$ and $a + b + c = 60 + 12 + 15 = 87.$

Answer: 152

Small Hint:

All eight centers lie at height $100$ at the vertices of an octagon of side $200;$ the ninth center is at height $r$ above the octagon’s center

Big Hint:

Tangency gives $R^2 + (r - 100)^2 = (r + 100)^2,$ where $R$ is the circumradius and $200 = 2R\sin 22.5^\circ$

Solution:

The eight centers are at height $100,$ at the vertices of a regular octagon of side $200$ (adjacent spheres are tangent). If the ninth sphere has radius $r,$ it rests on the surface with its center at height $r$ directly above the octagon’s center, and tangency to each sphere gives $\sqrt{R^2 + (r - 100)^2} = r + 100,$ where $R$ is the octagon’s circumradius. Hence $$\begin{aligned} R^2 &= (r + 100)^2 - (r - 100)^2 \\ &= 400r. \end{aligned}$$

A side of a regular octagon subtends $45^\circ$ at the center, so $200 = 2R \sin 22.5^\circ$ and, using $\sin^2 22.5^\circ = \frac{1 - \cos 45^\circ}{2} = \frac{2 - \sqrt{2}}{4},$ $$\begin{aligned} R^2 &= \frac{10000}{\sin^2 22.5^\circ} \\ &= \frac{40000}{2 - \sqrt{2}} \\ &= 20000\,(2 + \sqrt{2}). \end{aligned}$$

Then $r = \frac{R^2}{400}$ $= 50\,(2 + \sqrt{2})$ $= 100 + 50\sqrt{2},$ so $a + b + c = 100 + 50 + 2 = 152.$

Answer: 525

Small Hint:

Set $B = (0,0,0),$ $A = (20,0,0),$ $C = (0,20,0),$ $D = (0,20,20),$ and find the plane through $P,$ $Q,$ $R$

Big Hint:

The plane $2x + 2y - z = 30$ cuts a hexagon. Projecting to the $xy$-plane scales area by $\frac{1}{3},$ the vertical component of the unit normal.

Solution:

The cube has side $20.$ Take $B = (0,0,0),$ $A = (20,0,0),$ $C = (0,20,0),$ and $D = (0,20,20),$ so $\overline{AD}$ is an interior diagonal. Then $P = (15, 0, 0),$ $Q = (0, 15, 0),$ $R = (0, 20, 10),$ and the plane through them is $2x + 2y - z = 30.$

Evaluating $2x + 2y - z$ at the cube’s vertices and checking all twelve edges, the plane also crosses the edges at $(5, 20, 20),$ $(20, 5, 20),$ and $(20, 0, 10),$ so the cross-section is the hexagon with vertices $(15,0,0),$ $(0,15,0),$ $(0,20,10),$ $(5,20,20),$ $(20,5,20),$ $(20,0,10)$ in order. Its projection onto the $xy$-plane is the hexagon $(15,0),$ $(0,15),$ $(0,20),$ $(5,20),$ $(20,5),$ $(20,0),$ whose area by the shoelace formula is $175.$

The plane’s unit normal $\frac{1}{3}(2, 2, -1)$ has vertical component of magnitude $\frac{1}{3},$ so projecting onto the $xy$-plane multiplies area by $\frac{1}{3}.$ The cross-section therefore has area $3 \cdot 175 = 525.$

Answer: 83

Small Hint:

Parameterize $P,$ $Q,$ and $R$ by separate fractions $x,$ $y,$ and $z$ along $DE,$ $EF,$ and $FD$; translate the three collinearities into equations

Big Hint:

The equations are $x(1+y)=1,$ $y(1+z)=1,$ and $z(1+x)=1;$ show they force $x=y=z,$ then compare squared distances from the common center of the two equilateral triangles

Solution:

Place $A = (0, \sqrt{3}),$ $B = (-1, 0),$ $C = (1, 0),$ so $D = (0, 0),$ $E = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $F = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right).$ Write $P=D+x(E-D),$ $Q=E+y(F-E),$ and $R=F+z(D-F),$ where $0\le x,y,z\le1.$ Computing the three two-dimensional cross products, the collinearities $C,P,Q;$ $A,Q,R;$ and $B,R,P$ give, respectively, $$\begin{gathered} x(1+y)=1,\\ y(1+z)=1,\\ z(1+x)=1. \end{gathered}$$

Let $f(u)=\frac{1}{1+u}.$ The equations say $x=f(y),$ $y=f(z),$ and $z=f(x),$ so $$x=f(f(f(x)))=\frac{x+2}{2x+3}.$$ Hence $x^2+x-1=0,$ and positivity gives $x=\frac{\sqrt5-1}{2}.$ The equations then force $y=z=x;$ call this common value $t.$ Thus the desired configuration really is the symmetric one, rather than merely being assumed to be.

With $P = \left(\frac{t}{2}, \frac{t\sqrt{3}}{2}\right)$ and $Q = \left(\frac{1}{2} - t, \frac{\sqrt{3}}{2}\right),$ both triangles are equilateral with center $G = \left(0, \frac{\sqrt{3}}{3}\right).$ Therefore the area ratio is $\frac{GA^2}{GP^2}.$ Using $t^2 = 1 - t,$ $$\begin{aligned} GP^2 &= \frac{t^2}{4} + 3\left(\frac{t}{2} - \frac{1}{3}\right)^2 \\ &= t^2 - t + \frac{1}{3} \\ &= \frac{7}{3} - \sqrt{5}, \\ GA^2 &= \frac{4}{3}. \end{aligned}$$

Hence $$\begin{aligned} \frac{[ABC]}{[PQR]} &= \frac{\frac{4}{3}}{\frac{7}{3} - \sqrt{5}} \\ &= \frac{4}{7 - 3\sqrt{5}} \\ &= 7 + 3\sqrt{5}, \end{aligned}$$ so $a = 7,$ $b = 3,$ $c = 5,$ and $a^2 + b^2 + c^2 = 49 + 9 + 25 = 83.$