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 $k + m + n = 80 + 5 = 85.$

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 $(p, q)$ with $p, q \gt 0$ such that $E = B - 9(p, q)$ and $F = D + 8(p, q):$ 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: $10 - 9q = 8q,$ so $q = \frac{10}{17}$ and $p = \sqrt{1 - q^2} = \frac{3\sqrt{21}}{17}.$

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

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

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 $$PQ = BP + AQ - AB = 120 + 117 - 125 = 112,$$ we conclude $MN = 56.$

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.

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 $x = 1,$ $$a + b + c = a\left(\frac{3}{4}\right)^2 - \frac{9}{8} = \frac{9(a - 2)}{16}.$$

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

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: $$m + (m - 1)(m + m^2 + \cdots + m^{n-1}) = m + (m^n - m) = m^n = m^{x_0}.$$

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 $m.$

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 $\frac{h}{2K/a} = \frac{ha}{2K}$ (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 $k + m + n = 40 + 221 + 57 = 318.$

Solution:

In exponential form the equation says $(24 \sin x)^{3/2} = 24 \cos x.$ 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 $$24 \cot^2 x = 24 \cdot \frac{1 - \sin^2 x}{\sin^2 x} = 24 \cdot \frac{8/9}{1/9} = 24 \cdot 8 = 192.$$

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 $376 + 127 = 503.$

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 $2^{50} + 1 = (2^{10} + 1)(2^{40} - 2^{30} + 2^{20} - 2^{10} + 1),$ 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 $S \equiv 1 + 2 + 4 = 7 \pmod{1000}.$

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 $$p = \frac{2\binom{n+1}{2} + (n+1)}{\binom{n+1}{3} + 3\binom{n+1}{2} + (n+1)} = \frac{(n+1)^2}{\frac{(n+1)(n^2 + 8n + 6)}{6}} = \frac{6(n+1)}{n^2 + 8n + 6}.$$

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 $594.$

Solution:

Let $h$ be the height of $A,$ and let $e_1,$ $e_2,$ $e_3$ be unit vectors along the three mutually perpendicular edges at $A.$ If $u$ is the upward unit normal of the plane, the height of the vertex along edge $i$ is $h + 10(e_i \cdot u),$ so $10(e_1 \cdot u) = 10 - h,$ $10(e_2 \cdot u) = 11 - h,$ and $10(e_3 \cdot u) = 12 - h.$ Because $e_1, e_2, e_3$ form an orthonormal basis, $(e_1 \cdot u)^2 + (e_2 \cdot u)^2 + (e_3 \cdot u)^2 = 1.$

Therefore $$(10 - h)^2 + (11 - h)^2 + (12 - h)^2 = 100,$$ which simplifies to $3h^2 - 66h + 265 = 0,$ so $h = \frac{33 \pm \sqrt{33^2 - 3 \cdot 265}}{3} = \frac{33 \pm \sqrt{294}}{3}.$

Since $A$ is the closest vertex to the plane, $h \lt 10,$ forcing $h = \frac{33 - \sqrt{294}}{3},$ and $r + s + t = 33 + 294 + 3 = 330.$

Solution:

Scale so that $A_1A_2 = 2.$ A $90^\circ$ rotation about the center carries the whole configuration to itself, so $B_1B_3B_5B_7$ is a square and the distances $a = M_iB_i$ and $b = M_iB_{i-2}$ do not depend on $i.$ Both $B_3$ and $B_1$ lie on ray $R_3$ (at distances $a$ and $b$ from $M_3$), so $b - a = B_1B_3 = 2.$ Also, $R_1 \perp R_3$ makes triangle $M_1B_1M_3$ right-angled at $B_1,$ so $a^2 + b^2 = M_1M_3^2.$

Lines $A_1A_2$ and $A_3A_4$ meet at a point $C$ at a right angle, and triangle $A_2CA_3$ is an isosceles right triangle with legs $A_2C = A_3C = \sqrt{2},$ so $M_1C = M_3C = 1 + \sqrt{2}$ and $a^2 + b^2 = M_1M_3^2 = 2(1 + \sqrt{2})^2.$ Then $$(a + b)^2 = 2(a^2 + b^2) - (b - a)^2 = 4(1 + \sqrt{2})^2 - 4 = 8 + 8\sqrt{2}.$$

Since triangle $M_1CM_3$ is an isosceles right triangle and $A_3$ lies on segment $\overline{M_3C},$ we have $\angle A_3M_3M_1 = 45^\circ,$ while $\tan \angle M_1M_3B_1 = \frac{a}{b}$ from the right triangle. The tangent addition formula gives $$\tan \angle A_3M_3B_1 = \frac{1 + \frac{a}{b}}{1 - \frac{a}{b}} = \frac{a + b}{b - a}, \qquad \text{so} \quad \tan^2 \angle A_3M_3B_1 = \frac{8 + 8\sqrt{2}}{4} = 2 + 2\sqrt{2}.$$ Therefore $$\cos 2\angle A_3M_3B_1 = \frac{1 - \tan^2 \angle A_3M_3B_1}{1 + \tan^2 \angle A_3M_3B_1} = \frac{-1 - 2\sqrt{2}}{3 + 2\sqrt{2}} = -(1 + 2\sqrt{2})(3 - 2\sqrt{2}) = 5 - 4\sqrt{2} = 5 - \sqrt{32},$$ and $m + n = 5 + 32 = 37.$

Solution:

By Vieta's formulas, $a + b + c = 0$ and $ab + bc + ca = -2011.$ Negating all three roots replaces $m$ by $-m,$ so we may assume two roots, say $a \ge b,$ are nonnegative. Substituting $c = -(a + b)$ into the second equation gives $ab - (a + b)^2 = -2011,$ that is, $$a^2 + ab + b^2 = 2011.$$

Multiplying by $4$ and completing the square, $(2a + b)^2 + 3b^2 = 8044.$ Since $a \ge b \ge 0,$ we have $3b^2 \le a^2 + ab + b^2 = 2011,$ so $0 \le b \le 25.$ Checking these values, $8044 - 3b^2$ is a perfect square only for $b = 10,$ where $8044 - 300 = 7744 = 88^2.$

Then $2a + b = 88$ gives $a = 39,$ and $c = -(a + b) = -49.$ (Indeed $39 \cdot 10 - 49^2 = -2011.$) Therefore $|a| + |b| + |c| = 39 + 10 + 49 = 98.$