Small Hint:

Equal leftovers mean the red used exceeds the blue used by $164 - 130,$ and similarly for white

Big Hint:

Those excesses all went into the pink stripe, so one stripe’s worth of paint is $34 + 58$ ounces

Solution:

Say each stripe used $s$ ounces of paint. Blue was used only on the blue stripe, so $s$ ounces of blue were used. Since the three leftovers are equal and the colors started $34$ and $58$ ounces apart, red use exceeded blue use by $164 - 130 = 34$ ounces and white use exceeded blue use by $188 - 130 = 58$ ounces. That extra red and white is exactly the pink stripe, so $s = 34 + 58 = 92.$

Bill therefore had $130 - 92 = 38$ ounces of each color left, for a total of $3 \cdot 38 = 114$ ounces.

Small Hint:

Write $a^{(\log_3 7)^2}$ as $\left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7}$

Big Hint:

$27^{\log_3 7} = \left(3^{\log_3 7}\right)^3 = 7^3.$ Handle the other two terms the same way.

Solution:

By the power rule for exponents, $$\begin{aligned} a^{(\log_3 7)^2} &= \left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7} \\ &= \left(3^{\log_3 7}\right)^3 = 7^3 = 343. \end{aligned}$$

In the same way, $$\begin{aligned} b^{(\log_7 11)^2} &= 49^{\log_7 11} = \left(7^{\log_7 11}\right)^2 \\ &= 11^2 = 121, \end{aligned}$$ and $$\begin{aligned} c^{(\log_{11} 25)^2} &= \left(\sqrt{11}\right)^{\log_{11} 25} \\ &= \left(11^{\log_{11} 25}\right)^{\frac{1}{2}} \\ &= 25^{\frac{1}{2}} = 5. \end{aligned}$$

The sum is $343 + 121 + 5 = 469.$

Small Hint:

Place coordinates: $A = (0, 0),$ $B = (100, 0),$ $D = (0, h),$ so $E = (0, \frac{h}{2})$ and $C = (100, h)$

Big Hint:

Perpendicular lines have slopes whose product is $-1:$ $\frac{h}{100} \cdot \left(-\frac{h}{200}\right) = -1$

Solution:

Let $AD = h,$ and place $A = (0, 0),$ $B = (100, 0),$ $C = (100, h),$ $D = (0, h),$ so $E = \left(0, \frac{h}{2}\right).$ Line $AC$ has slope $\frac{h}{100}$ and line $BE$ has slope $\frac{\frac{h}{2}}{-100} = -\frac{h}{200}.$ Perpendicularity gives $$\frac{h}{100} \cdot \left(-\frac{h}{200}\right) = -1,$$ so $h^2 = 20000$ and $h = 100\sqrt{2} \approx 141.42.$

The greatest integer less than $AD$ is $141.$

Small Hint:

With $c$ children the grape counts are an arithmetic sequence, and the total is $c(n + 1 - c) = 2009 = 7^2 \cdot 41$

Big Hint:

The last-place child ate $n + 2 - 2c \ge 0$ grapes; check which divisors $c$ of $2009$ survive that condition

Solution:

Let $c$ be the number of children. The grape counts $n,$ $n - 2,$ $\ldots,$ $n + 2 - 2c$ form an arithmetic sequence, so the total is $c$ times the average of the first and last terms: $$\begin{aligned} &c \cdot \frac{n + (n + 2 - 2c)}{2} \\ &= c(n + 1 - c) \\ &= 2009 = 7^2 \cdot 41. \end{aligned}$$ Thus $2009$ is divisible by $c$ and $n = \frac{2009}{c} + c - 1.$

The last-place child ate $n + 2 - 2c = \frac{2009}{c} + 1 - c \ge 0$ grapes, which forces $c(c - 1) \le 2009,$ ruling out $c = 49,$ $287,$ and $2009.$ The remaining divisors give $n = 2009$ for $c = 1,$ $n = 287 + 6 = 293$ for $c = 7,$ and $n = 49 + 40 = 89$ for $c = 41.$

The smallest possible value is $n = 89.$

Small Hint:

Centers of internally tangent circles lie along radii of circle $A;$ put the center of $A$ at the origin with the radius-$3$ vertex on top

Big Hint:

By symmetry $E$’s center is on the vertical axis. Tangency to $B$ puts it at height $4 - r;$ then tangency to $C$ gives an equation in $r.$

Solution:

Place the center of circle $A$ at the origin with the triangle’s vertices at $(0, 10)$ and $\left(\pm 5\sqrt{3}, -5\right).$ A circle internally tangent to $A$ at a vertex has its center on the radius to that vertex, so circle $B$ has center $(0, 7)$ and circles $C$ and $D$ have centers $\left(\mp 4\sqrt{3}, -4\right)$ (at distance $10 - 2 = 8$ from the origin).

By symmetry the center of circle $E,$ of radius $r,$ lies on the $y$-axis at $(0, y).$ External tangency to $B$ gives $7 - y = r + 3,$ so $y = 4 - r.$ External tangency to $C$ gives $$\begin{aligned} &\left(4\sqrt{3}\right)^2 \\ &\quad {}+ (4 - r + 4)^2 = (r + 2)^2, \end{aligned}$$ that is, $48 + (8 - r)^2 = (r + 2)^2,$ which simplifies to $112 - 16r = 4r + 4,$ so $r = \frac{27}{5}.$

Then $m + n = 27 + 5 = 32.$

Small Hint:

Count the complement: five-element subsets with no two consecutive numbers

Big Hint:

Subtracting $0,$ $1,$ $2,$ $3,$ $4$ from the five numbers in increasing order matches no-two-consecutive subsets of $\{1, \ldots, 14\}$ with arbitrary subsets of $\{1, \ldots, 10\}$

Solution:

Count the complement: subsets $a_1 \lt a_2 \lt a_3 \lt a_4 \lt a_5$ with no two consecutive. Setting $b_i = a_i - (i - 1)$ turns each such subset into five distinct numbers $b_1 \lt b_2 \lt \cdots \lt b_5$ in $\{1, \ldots, 10\},$ and this map is reversible, so there are $\binom{10}{5} = 252$ subsets with no two consecutive numbers.

Therefore $m = \binom{14}{5} - \binom{10}{5}$ $= 2002 - 252 = 1750,$ and the remainder upon division by $1000$ is $750.$

Small Hint:

$(2i)!! = 2^i \cdot i!,$ and since $\binom{2i}{i} = \frac{2^i (2i-1)!!}{i!}$ is an integer, all odd factors of $i!$ cancel into $(2i-1)!!$

Big Hint:

Term $i$ in lowest terms has denominator $2^{i + e_i},$ $e_i$ the exponent of $2$ in $i!.$ These strictly increase, so the last term controls.

Solution:

The $i$th term is $\frac{(2i-1)!!}{(2i)!!}$ with odd numerator, and $(2i)!! = 2^i \cdot i!.$ Because $\binom{2i}{i} = \frac{(2i)!}{i!\,i!} = \frac{2^i (2i-1)!!}{i!}$ is an integer, every odd prime power dividing $i!$ also divides $(2i-1)!!.$ Hence in lowest terms the $i$th term has denominator exactly $2^{a_i}$ where $a_i = i + e_i$ and $e_i$ is the exponent of $2$ in $i!.$ The $a_i$ strictly increase, so over the common denominator $2^{a_{2009}}$ every term except the last contributes an even numerator while the last contributes an odd one. The sum in lowest terms therefore has denominator exactly $2^{a_{2009}},$ so $b = 1.$

By Legendre’s formula, $$\begin{aligned} e_{2009} &= 1004 + 502 + 251 \\ &\quad {}+ 125 + 62 + 31 + 15 \\ &\quad {}+ 7 + 3 + 1 \\ &= 2001, \end{aligned}$$ so $a = 2009 + 2001 = 4010.$ Then $\frac{ab}{10} = \frac{4010 \cdot 1}{10} = 401.$

Small Hint:

The chance the first six appears on roll $k$ is $p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}$

Big Hint:

Compute $\sum p_k^2$ for a tie and $\sum p_k p_{k+1}$ for a difference of one; both are geometric series with ratio $\frac{25}{36}$

Solution:

The probability that a player’s first six appears on roll $k$ is $p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}.$ The probability of a tie is $$\sum_{k=1}^{\infty} p_k^2 = \frac{1}{36} \cdot \frac{1}{1 - \frac{25}{36}} = \frac{1}{11}.$$

The probability that Linda needs exactly one more roll than Dave is $\sum_{k=1}^{\infty} p_k p_{k+1}$ $= \frac{5}{6} \sum_{k=1}^{\infty} p_k^2$ $= \frac{5}{66},$ and by symmetry the same holds with the players swapped.

The total probability is $\frac{1}{11} + 2 \cdot \frac{5}{66} = \frac{6 + 10}{66} = \frac{8}{33},$ so $m + n = 8 + 33 = 41.$

Small Hint:

Subtracting $1$ from each variable matches positive solutions of the $2009$ equation with nonnegative solutions of the $2000$ equation, since $4 + 3 + 2 = 9$

Big Hint:

So $m - n$ counts nonnegative solutions of $4x + 3y + 2z = 2000$ in which some variable is $0.$ Count the three cases and correct for double counting.

Solution:

If $(x, y, z)$ is a positive solution of $4x + 3y + 2z = 2009,$ then $(x - 1, y - 1, z - 1)$ is a nonnegative solution of $4x + 3y + 2z = 2000,$ and conversely, since $4 + 3 + 2 = 9.$ So $m$ equals the number of nonnegative solutions of $4x + 3y + 2z = 2000,$ and $m - n$ counts the nonnegative solutions of that equation in which at least one variable is $0.$

If $x = 0:$ $3y + 2z = 2000$ forces $y$ even, $0 \le y \le 666,$ giving $334$ solutions. If $y = 0:$ $2x + z = 1000$ with $0 \le x \le 500$ gives $501.$ If $z = 0:$ $4x + 3y = 2000$ forces $y \equiv 0 \pmod 4,$ $0 \le y \le 664,$ giving $167.$ The solutions $(0, 0, 1000)$ and $(500, 0, 0)$ are each counted twice, so $$\begin{aligned} m - n &= 334 + 501 + 167 - 2 \\ &= 1000. \end{aligned}$$

The remainder upon division by $1000$ is $0.$

Small Hint:

The triangle is right-angled at $B.$ The conditions say $AD$ bisects angle $BAC$ and $CB$ bisects angle $ACD.$

Big Hint:

With $A = (0,0),$ $B = (5,0),$ $C = (5,12):$ ray $CD$ is the mirror of ray $CA$ over the vertical line $CB,$ so it hits $(10, 0),$ and $\tan\frac{\angle BAC}{2} = \frac{2}{3}$

Solution:

Since $5^2 + 12^2 = 13^2,$ angle $B$ is right. Place $A = (0, 0),$ $B = (5, 0),$ $C = (5, 12).$ The condition at $A$ says $\angle BAD = \angle CAD,$ so $D$ lies on the bisector of angle $BAC.$ Using the half-angle formula with $\tan \angle BAC = \frac{12}{5},$ $$\begin{aligned} \tan \frac{\angle BAC}{2} &= \frac{\sin \angle BAC}{1 + \cos \angle BAC} \\ &= \frac{\frac{12}{13}}{1 + \frac{5}{13}} = \frac{2}{3}, \end{aligned}$$ so $D$ lies on the line $y = \frac{2}{3}x.$

The condition at $C$ says $CB$ bisects angle $ACD,$ so ray $CD$ is the reflection of ray $CA$ over line $CB,$ which is the vertical line $x = 5.$ The reflection of $A$ is $(10, 0),$ so $D$ lies on the line through $C = (5, 12)$ and $(10, 0),$ namely $5y = 120 - 12x.$

Solving $y = \frac{2}{3}x$ and $5y = 120 - 12x$ gives $x = \frac{180}{23},$ $y = \frac{120}{23}.$ Then $$AD = \frac{60}{23}\sqrt{3^2 + 2^2} = \frac{60\sqrt{13}}{23},$$ so $p + q + r = 60 + 23 + 13 = 96.$

Small Hint:

The inequality says exactly that $\frac{m}{n} \lt k \lt mn$

Big Hint:

Write $m = nq + r$ with $0 \le r \lt n;$ the count of such $k$ is $mn - q - 1 = 50,$ so $q(n^2 - 1) + nr = 51$ leaves only a few cases

Solution:

The inequality $|\log m - \log k| \lt \log n$ is equivalent to $\frac{m}{n} \lt k \lt mn.$ Write $m = nq + r$ with $0 \le r \lt n;$ since $m \ge n \ge 2$ (for $n = 1$ no $k$ works), $q \ge 1.$ The integers $k$ in the interval are $q + 1,$ $q + 2,$ $\ldots,$ $mn - 1,$ so there are $mn - q - 1 = 50$ of them, that is, $mn - q = 51,$ or $$q(n^2 - 1) + nr = 51.$$

For $n \ge 8$ the left side is at least $63,$ so $2 \le n \le 7.$ Checking each case, only $n = 2,$ $r = 0,$ $q = 17$ (so $m = 34$) and $n = 3,$ $r = 1,$ $q = 6$ (so $m = 19$) work. These give $mn = 68$ and $mn = 57;$ indeed $17 \lt k \lt 68$ and $\frac{19}{3} \lt k \lt 57$ each contain exactly $50$ integers.

The sum of all possible values of $mn$ is $68 + 57 = 125.$

Small Hint:

Add up all $k$ sums in two ways: the $2k$ elements are distinct, so the total is at least $1 + 2 + \cdots + 2k$

Big Hint:

The $k$ sums are distinct integers at most $2009,$ so the total is at most $2009 + 2008 + \cdots.$ Compare the bounds, then build a pairing that achieves the resulting $k.$

Solution:

Let $S = \sum_{i=1}^{k} (a_i + b_i).$ The $2k$ chosen elements are distinct positive integers, so $S \ge 1 + 2 + \cdots + 2k$ $= k(2k + 1).$ The $k$ sums are distinct integers at most $2009,$ so $S \le 2009 + 2008 + \cdots$ ${}+ (2010 - k)$ $= \frac{k(4019 - k)}{2}.$ Combining, $$\begin{aligned} &k(2k + 1) \le \frac{k(4019 - k)}{2} \\ &\implies 4k + 2 \le 4019 - k \\ &\implies k \le \frac{4017}{5} = 803.4, \end{aligned}$$ so $k \le 803.$

To achieve $k = 803,$ take the pairs $(i,\, 1206 + i)$ for $1 \le i \le 401,$ whose sums are the even numbers $1208,$ $1210,$ $\ldots,$ $2008,$ together with the pairs $(a,\, a + 403)$ for $402 \le a \le 803,$ whose sums are the odd numbers $1207,$ $1209,$ $\ldots,$ $2009.$ The elements used are $1$–$803,$ $805$–$1607$ with no repeats, and all $803$ sums are distinct and at most $2009.$

The maximum is $k = 803.$

Small Hint:

Model the circle in the complex plane with $A = -2,$ $B = 2,$ and $C_i = 2\omega^i$ where $\omega = e^{\frac{\pi \mathrm{i}}{7}}$

Big Hint:

$AC_i \cdot BC_i = 4\,|\omega^{2i} - 1|,$ and the $\omega^{2i}$ are exactly the nontrivial $7$th roots of unity, whose product of distances to $1$ is $7$

Solution:

Put the circle in the complex plane with center $0,$ $A = -2,$ $B = 2,$ and $C_i = 2\omega^i$ for $i = 1, \ldots, 6,$ where $\omega = e^{\frac{\pi \mathrm{i}}{7}}.$ Then $AC_i = 2\,|\omega^i + 1|$ and $BC_i = 2\,|\omega^i - 1|,$ so $$AC_i \cdot BC_i = 4\,\bigl|\omega^{2i} - 1\bigr|.$$

As $i$ runs over $1, \ldots, 6,$ the numbers $\omega^{2i}$ run over all six nontrivial $7$th roots of unity. Let $\zeta = \omega^2.$ Since $\prod_{j=1}^{6} (x - \zeta^j)$ $= 1 + x + \cdots + x^6,$ plugging in $x = 1$ gives $\prod_{j=1}^{6} \bigl|1 - \zeta^j\bigr| = 7.$ Therefore $$\begin{aligned} n &= \prod_{i=1}^{6} 4\,\bigl|\omega^{2i} - 1\bigr| \\ &= 4^6 \cdot 7 = 28672. \end{aligned}$$