Small Hint:

Imagine pouring all three jars together: the result is $10$ liters that must be $50$% acid, which determines $k$

Big Hint:

With $k$ known, let $x$ be the amount added to jar $A$ and set jar $A$’s acid equal to half of its new volume

Solution:

If all three jars were combined, the result would be $10$ liters of $50$% acid, since both final jars are $50$% acid. The total acid is therefore $5$ liters, so $4(0.45) + 5(0.48) + 0.01k = 5,$ which gives $k = 80.$

Now let $x$ be the number of liters poured from jar $C$ into jar $A$. Jar $A$ then holds $4 + x$ liters containing $1.8 + 0.8x$ liters of acid, so $1.8 + 0.8x = 0.5(4 + x),$ giving $0.3x = 0.2,$ so $x = \frac{2}{3}.$

Thus $m + n = 2 + 3 = 5,$ and $k + m + n = 80 + 5 = 85.$

Small Hint:

Since $\overline{BE} \parallel \overline{DF},$ write $E = B - 9(u, v)$ and $F = D + 8(u, v)$ for one unit vector $(u, v)$

Big Hint:

$\overline{EF} \parallel \overline{AB}$ forces $E$ and $F$ to the same height, giving $v = \frac{10}{17};$ then $EF$ is a difference of $x$-coordinates

Solution:

Place $D = (0, 0),$ $C = (12, 0),$ $B = (12, 10),$ $A = (0, 10).$ Since $\overline{BE} \parallel \overline{DF},$ there is a unit vector $(u, v)$ with $u, v \gt 0$ such that $E = B - 9(u, v)$ and $F = D + 8(u, v):$ line $BE$ heads down and to the left so that it can cross $\overline{AD},$ while $\overline{DF}$ points up and to the right into the rectangle.

Because $\overline{EF} \parallel \overline{AB}$ is horizontal, $E$ and $F$ have equal heights: $10 - 9v = 8v,$ so $v = \frac{10}{17}$ and $u = \sqrt{1 - v^2} = \frac{3\sqrt{21}}{17}.$

Then $E$ and $F$ have $x$-coordinates $12 - 9u$ and $8u,$ so $EF = |12 - 17u|$ $= |12 - 3\sqrt{21}|$ $= 3\sqrt{21} - 12,$ since $3\sqrt{21} \gt 12.$ Thus $m + n + p = 3 + 21 + 12 = 36.$

Small Hint:

The new coordinates of $P$ are its signed distances to the new axes: $\alpha$ is the distance to line $M$ and $\beta$ the distance to line $L$

Big Hint:

Use the point-to-line distance formula, fixing each sign by checking whether $P$ is on the same side as $A$ (for $\alpha$) or as $B$ (for $\beta$)

Solution:

Line $L$ is $5x - 12y - 132 = 0$ and line $M$ is $12x + 5y - 90 = 0.$ The new $x$-coordinate of a point is its signed distance to line $M,$ counted positive on the side containing $A,$ and the new $y$-coordinate is its signed distance to line $L,$ positive on the side containing $B.$

Substituting $P = (-14, 27)$ into $12x + 5y - 90$ gives $-168 + 135 - 90 = -123,$ while $A$ gives $193 \gt 0;$ dividing by $\sqrt{12^2 + 5^2} = 13,$ we get $\alpha = -\frac{123}{13}.$ Substituting $P$ into $5x - 12y - 132$ gives $-70 - 324 - 132 = -526,$ and $B$ gives $-179,$ so $P$ lies on the same side of $L$ as $B$ and $\beta = \frac{526}{13}.$

Therefore $\alpha + \beta = \frac{-123 + 526}{13} = \frac{403}{13} = 31.$

Small Hint:

Extend $\overline{CM}$ and $\overline{CN}$ until they meet $\overline{AB}$

Big Hint:

$BM$ is both a bisector and an altitude of triangle $BCP,$ so that triangle is isosceles and $M$ is a midpoint; then $MN$ is a midline

Solution:

Extend $\overline{CM}$ and $\overline{CN}$ to meet $\overline{AB}$ at $P$ and $Q,$ respectively. In triangle $BCP,$ the segment $BM$ is both an angle bisector and an altitude, so the triangle is isosceles with $BP = BC = 120,$ and $M$ is the midpoint of $\overline{CP}.$ Similarly, triangle $ACQ$ is isosceles with $AQ = AC = 117,$ and $N$ is the midpoint of $\overline{CQ}.$

Hence $\overline{MN}$ is a midline of triangle $CPQ,$ so $MN = \frac{PQ}{2}.$ Since $$\begin{aligned} PQ &= BP + AQ - AB \\ &= 120 + 117 - 125 \\ &= 112, \end{aligned}$$ we conclude $MN = 56.$

Small Hint:

Compare two overlapping triples of consecutive vertices: their sums differ by the labels three apart, so labels three apart are congruent mod $3$

Big Hint:

Each of the residue classes $\{3,6,9\},$ $\{1,4,7\},$ $\{2,5,8\}$ occupies one of the three position classes; count assignments and orderings, then divide by $9$ rotations

Solution:

Two overlapping triples of consecutive vertices share two labels, so their sums differ by labels three positions apart. Since all triple sums are multiples of $3,$ labels three apart are congruent mod $3.$ Thus the position classes $\{1, 4, 7\},$ $\{2, 5, 8\},$ $\{3, 6, 9\}$ each carry a single residue class of digits, and the digits $1$ through $9$ consist of exactly three digits from each residue class mod $3.$

Every triple of consecutive positions then contains one digit from each residue class, with sum $\equiv 0 + 1 + 2 \equiv 0 \pmod 3,$ so all $3!$ assignments of residue classes to position classes are acceptable, and within each position class the three digits can be arranged in $3!$ ways. That gives $3! \cdot (3!)^3 = 6^4 = 1296$ acceptable labelings.

Because the digits are distinct, no nontrivial rotation fixes a labeling, so the $1296$ labelings split into rotation classes of size $9,$ giving $\frac{1296}{9} = 144$ distinguishable arrangements.

Small Hint:

Write the parabola in vertex form and note that $a + b + c$ is the value of $y$ at $x = 1$

Big Hint:

So $\frac{9(a-2)}{16}$ must be an integer; choose the integer that makes $a$ as small as possible while keeping $a \gt 0$

Solution:

In vertex form the parabola is $y = a\left(x - \frac{1}{4}\right)^2 - \frac{9}{8}.$ Since $a + b + c$ equals the value of $y$ at $x = 1,$ $$\begin{aligned} a + b + c &= a\left(\frac{3}{4}\right)^2 - \frac{9}{8} \\ &= \frac{9(a - 2)}{16}. \end{aligned}$$

If this equals the integer $n,$ then $a = 2 + \frac{16n}{9}.$ The condition $a \gt 0$ requires $16n \gt -18,$ that is $n \ge -1,$ and $a$ is smallest when $n = -1,$ giving $a = 2 - \frac{16}{9} = \frac{2}{9}.$

Thus $p + q = 2 + 9 = 11.$

Small Hint:

Reduce both sides mod $m - 1:$ every power of $m$ is congruent to $1$

Big Hint:

So $m - 1$ must divide $2010.$ Conversely, expand $m^n$ by repeatedly splitting one power into $m$ copies of the next-lower power; each split adds $m - 1$ terms.

Solution:

The value $m = 1$ fails, since the right side would be $2011$ while the left side is $1.$ For $m \ge 2,$ reduce mod $m - 1:$ every power of $m$ is $\equiv 1,$ so the equation forces $1 \equiv 2011 \pmod{m - 1},$ that is, $m - 1$ divides $2010.$

Conversely, suppose $2010 = (m - 1)n.$ Take $x_0 = n,$ let $m$ of the $x_k$ equal $0,$ and for each $r = 1, 2, \ldots, n - 1$ let $m - 1$ of the $x_k$ equal $r.$ This uses $m + (m - 1)(n - 1)$ $= n(m - 1) + 1$ $= 2011$ terms, and the sum telescopes: $$\begin{aligned} &m \\ &\quad {}+ (m - 1) \\ &\quad {}\cdot (m + m^2 + \cdots + m^{n-1}) \\ &\quad {}= m + (m^n - m) \\ &\quad {}= m^n = m^{x_0}. \end{aligned}$$

So the equation is solvable exactly when $m - 1$ divides $2010 = 2 \cdot 3 \cdot 5 \cdot 67,$ which has $2^4 = 16$ divisors. There are $16$ such $m.$

Small Hint:

A flap folded down from a vertex hangs to a depth equal to the distance from that vertex to its fold line, so all three fold lines must be at distance $h$ from their vertices

Big Hint:

Two folds cutting the same side must not overlap: for each side, multiplying $h$ by the sum of the other two side lengths gives at most twice the triangle’s area; the largest such sum is the binding constraint

Solution:

Write $a = BC = 23,$ $b = CA = 27,$ $c = AB = 30,$ and let $K$ be the area of $\triangle ABC.$ By Heron’s formula with semiperimeter $40,$ $K = \sqrt{40 \cdot 17 \cdot 13 \cdot 10}$ $= 20\sqrt{221}.$ When the corner at a vertex is folded down at a right angle, the flap hangs to a depth equal to the distance from that vertex to the fold line, so for a level tabletop of height $h,$ each fold line must lie at distance $h$ from its vertex.

The flap at $A$ is similar to $\triangle ABC$ with ratio $\frac{h}{\frac{2K}{a}} = \frac{ha}{2K}$ (dividing $h$ by the distance from $A$ to $\overline{BC}$), so it uses up $AU = c \cdot \frac{ha}{2K}$ of side $\overline{AB};$ likewise the flap at $B$ uses $BZ = c \cdot \frac{hb}{2K}$ of the same side. The two folds fit without crossing exactly when $AU + BZ \le c,$ that is, $h(a + b) \le 2K.$ The other two sides give $h(b + c) \le 2K$ and $h(c + a) \le 2K.$

The binding constraint comes from the largest sum, $b + c = 57,$ so the maximum height is $$h = \frac{2K}{57} = \frac{40\sqrt{221}}{57},$$ and $k + m + n = 40 + 221 + 57$ $= 318.$

Small Hint:

Rewrite the equation in exponential form and square both sides to get $\cos^2 x = 24 \sin^3 x$

Big Hint:

Replace $\cos^2 x$ by $1 - \sin^2 x$ and look for a rational root of the resulting cubic in $\sin x$

Solution:

In exponential form the equation says $(24 \sin x)^{\frac{3}{2}} = 24 \cos x.$ Squaring gives $24^3 \sin^3 x = 24^2 \cos^2 x,$ so $\cos^2 x = 24 \sin^3 x.$

Writing $s = \sin x$ and using $\cos^2 x = 1 - s^2,$ we get $24s^3 + s^2 - 1 = 0,$ which factors as $(3s - 1)(8s^2 + 3s + 1) = 0.$ The quadratic factor has negative discriminant, so $\sin x = \frac{1}{3}.$

Then $$\begin{aligned} 24 \cot^2 x &= 24 \cdot \frac{1 - \sin^2 x}{\sin^2 x} \\ &= 24 \cdot \frac{\frac{8}{9}}{\frac{1}{9}} \\ &= 24 \cdot 8 = 192. \end{aligned}$$

Small Hint:

A triangle inscribed in a circle is obtuse exactly when its three vertices lie strictly inside some semicircle

Big Hint:

Count obtuse triangles by their clockwise-first vertex, and handle even and odd $n$ separately (only even $n$ has right triangles)

Solution:

By the inscribed angle theorem, an inscribed triangle is obtuse exactly when its three vertices lie strictly within some semicircle. Count obtuse triangles by their “first” vertex, the vertex from which the other two are reached going clockwise within half the circle. If $n = 2k,$ the open semicircle clockwise of a vertex contains $k - 1$ vertices, giving $n\binom{k-1}{2}$ obtuse triangles; if $n = 2k + 1,$ it contains $k$ vertices, giving $n\binom{k}{2}.$

For $n = 2k$ the probability is $$\frac{2k\binom{k-1}{2}}{\binom{2k}{3}} = \frac{3(k - 2)}{2(2k - 1)} = \frac{93}{125},$$ so $375(k - 2) = 186(2k - 1),$ giving $3k = 564,$ $k = 188,$ and $n = 376.$

For $n = 2k + 1$ the probability is $\frac{3(k - 1)}{2(2k - 1)} = \frac{93}{125},$ so $375(k - 1) = 186(2k - 1),$ giving $3k = 189,$ $k = 63,$ and $n = 127.$ The sum of all possible values is $376 + 127 = 503.$

Small Hint:

From $2^3$ on, every remainder is a multiple of $8,$ and the powers of $2$ repeat mod $125$ with period $100$

Big Hint:

Show $2^{50} \equiv -1 \pmod{125},$ so remainders $50$ apart in the cycle sum to exactly $1000$

Solution:

The remainders $2^0 = 1,$ $2^1 = 2,$ and $2^2 = 4$ occur, and for $n \ge 3$ every $2^n$ is divisible by $8.$ Modulo $125$ the powers of $2$ for $n \ge 3$ repeat with period $100,$ so $R$ consists of $1,$ $2,$ $4,$ and the $100$ distinct remainders of $2^3, 2^4, \ldots, 2^{102}.$

The key fact is $2^{50} \equiv -1 \pmod{125}:$ indeed $2^{50} + 1 = (2^{10} + 1)$ $\cdot (2^{40} - 2^{30} + 2^{20} - 2^{10} + 1),$ where $2^{10} + 1 = 1025$ is divisible by $25$ and the second factor is $\equiv 1 + 1 + 1 + 1 + 1$ $\equiv 0 \pmod 5$ because $2^{10} \equiv -1 \pmod 5.$ Hence for $n \ge 3,$ the sum $2^{n+50} + 2^n$ is divisible by $125$ and by $8,$ so by $1000.$

Pairing each remainder in the cycle with the one $50$ steps later therefore gives $50$ pairs of distinct remainders, each pair summing to exactly $1000,$ so those $100$ remainders contribute a multiple of $1000$ to $S.$ Thus $S \equiv 1 + 2 + 4 = 7 \pmod{1000}.$

Small Hint:

Every man next to a man means the men split into blocks of sizes $2+2+2,$ $2+4,$ $4+2,$ $3+3,$ or $6;$ place the blocks into gaps between the women

Big Hint:

The conditional probability simplifies to $\frac{6(n+1)}{n^2 + 8n + 6};$ require it to be at most $\frac{1}{100}$

Solution:

Let $n$ be the number of women; only the pattern of men’s and women’s positions matters. If every man stands next to another man, the men form maximal blocks whose sizes, in order, are $2+2+2,$ $2+4,$ $4+2,$ $3+3,$ or $6.$ A pattern with $j$ blocks amounts to choosing $j$ of the $n + 1$ gaps determined by the women, so there are $\binom{n+1}{3}$ patterns for $2+2+2,$ $\binom{n+1}{2}$ for each of the three two-block orders, and $n + 1$ for a single block.

At least four men stand together in the orders $2+4,$ $4+2,$ and $6,$ so $$\begin{aligned} p &= \frac{2\binom{n+1}{2} + (n+1)}{\binom{n+1}{3} + 3\binom{n+1}{2} + (n+1)} \\ &= \frac{(n+1)^2}{\frac{(n+1)(n^2 + 8n + 6)}{6}} \\ &= \frac{6(n+1)}{n^2 + 8n + 6}. \end{aligned}$$

The condition $p \le \frac{1}{100}$ becomes $f(n) = n^2 - 592n - 594 \ge 0.$ Since $f(593) = 593 - 594 = -1 \lt 0$ and $f(594) = 2 \cdot 594 - 594$ $= 594 \gt 0,$ the least number of women is $594.$