Answer: 252

Solution:

Let $x$ be the number of students originally at the party, so $0.6x$ are girls and $0.4x$ like to dance. When the $20$ boys arrive, the number of girls is unchanged but the total becomes $x + 20,$ so $0.6x = 0.58(x + 20).$ Then $0.02x = 11.6,$ giving $x = 580.$

The number of students who now like to dance is $0.4 \cdot 580 + 20 = 232 + 20 = 252.$

Answer: 25

Solution:

Here $\overline{EM}$ is a side of the square. Let $h$ be the altitude of triangle $GEM.$ If $h \le 10,$ the triangle would lie entirely inside the square, and its area $\frac{1}{2} \cdot 10 \cdot h = 80$ would force $h = 16,$ a contradiction. So $h \gt 10$ and the apex $G$ lies outside the square; the opposite side $\overline{AI}$ cuts off a smaller triangle similar to $GEM$ with height $h - 10$ and base $\frac{10(h - 10)}{h}.$

The common region is triangle $GEM$ minus that small triangle: $$80 = 5h - \frac{1}{2} \cdot \frac{10(h - 10)}{h} \cdot (h - 10) = 5h - \frac{5(h - 10)^2}{h}.$$ Multiplying by $h$ gives $80h = 5h^2 - 5(h - 10)^2 = 5(20h - 100) = 100h - 500,$ so $20h = 500$ and $h = 25.$

Answer: 314

Solution:

Let $b,$ $j,$ and $s$ be the biking, jogging, and swimming rates. The two trips give $$2b + 3j + 4s = 74 \qquad \text{and} \qquad 4b + 2j + 3s = 91.$$ Doubling the first equation and subtracting the second yields $4j + 5s = 57,$ whose positive integer solutions are $(j, s) = (13, 1),$ $(8, 5),$ and $(3, 9).$

The corresponding values of $2b = 74 - 3j - 4s$ are $31,$ $30,$ and $29,$ so only $(j, s) = (8, 5)$ gives a whole-number rate, $b = 15.$ The sum of the squares is $15^2 + 8^2 + 5^2 = 225 + 64 + 25 = 314.$

Answer: 80

Solution:

Completing the square, $x^2 + 84x + 2008 = (x + 42)^2 + 244,$ so $y^2 - (x + 42)^2 = 244,$ which factors as $$(y - x - 42)(y + x + 42) = 244 = 2^2 \cdot 61.$$ The two factors have the same parity, and their product is even, so both are even: $y - x - 42 = 2$ and $y + x + 42 = 122.$

Adding gives $y = 62,$ and then $x = 18;$ indeed $18^2 + 84 \cdot 18 + 2008 = 3844 = 62^2.$ Therefore $x + y = 18 + 62 = 80.$

Answer: 14

Solution:

The contact point of the base stays at distance $\ell = \sqrt{r^2 + h^2}$ (the slant height) from the fixed vertex, so it traces a circle of radius $\ell.$ Rolling without slipping, the cone makes one rotation for each base circumference of arc, so returning after exactly $17$ rotations means $$2\pi\sqrt{r^2 + h^2} = 17 \cdot 2\pi r, \qquad \text{i.e.} \qquad \sqrt{r^2 + h^2} = 17r.$$

Squaring gives $h^2 = 288r^2,$ so $h/r = \sqrt{288} = 12\sqrt{2},$ and $m + n = 12 + 2 = 14.$

Answer: 17

Solution:

By induction, the $n$th entry of row $r$ is $2^{r-1}(r + 2n - 2):$ row $1$ gives $2^0(2n - 1),$ and summing two adjacent entries of row $r$ gives $2^{r-1}(r + 2n - 2) + 2^{r-1}(r + 2n) = 2^r\bigl((r + 1) + 2n - 2\bigr),$ the formula for row $r + 1.$ Row $r$ has $51 - r$ entries, so $1 \le n \le 51 - r.$

Since $67$ is odd, an entry is a multiple of $67$ exactly when $67 \mid r + 2n - 2.$ As $n$ runs through row $r,$ the quantity $r + 2n - 2$ takes the values $r, r + 2, \ldots, 100 - r,$ all with the same parity as $r$ and all less than $134.$ So the only possible multiple of $67$ is $67$ itself, which requires $r$ odd and $r \le 67 \le 100 - r,$ that is, $r \le 33.$

Each odd row $r = 1, 3, \ldots, 33$ contains exactly one such entry, for a total of $17.$

Answer: 708

Solution:

Consecutive squares $a^2$ and $(a + 1)^2$ differ by $2a + 1 \le 99$ for $a \le 49,$ so the squares from $1^2$ to $50^2 = 2500$ never skip a hundred-block: every set $S_0, S_1, \ldots, S_{25}$ contains a perfect square. For $a \ge 50$ the gap $2a + 1 \ge 101$ exceeds $100,$ so each of the sets $S_{26}, \ldots, S_{999}$ contains at most one square.

The largest number involved is $99999,$ and $316^2 = 99856 \le 99999 \lt 317^2.$ So the squares landing in $S_{26}, \ldots, S_{999}$ are $51^2, 52^2, \ldots, 316^2$ — that is, $266$ squares occupying $266$ distinct sets out of those $974.$

Therefore $974 - 266 = 708$ sets contain no perfect square.

Answer: 47

Solution:

For positive $x, y$ with $xy \lt 1,$ the tangent addition formula gives $\arctan x + \arctan y = \arctan\frac{x + y}{1 - xy}.$ Applying it twice: $$\arctan\frac{1}{3} + \arctan\frac{1}{4} = \arctan\frac{\frac{1}{3} + \frac{1}{4}}{1 - \frac{1}{12}} = \arctan\frac{7}{11}, \qquad \arctan\frac{7}{11} + \arctan\frac{1}{5} = \arctan\frac{\frac{7}{11} + \frac{1}{5}}{1 - \frac{7}{55}} = \arctan\frac{23}{24}.$$

The equation becomes $\arctan\frac{23}{24} + \arctan\frac{1}{n} = \arctan 1,$ so $\frac{23/24 + 1/n}{1 - 23/(24n)} = 1.$ Clearing denominators, $23n + 24 = 24n - 23,$ giving $n = 47.$

Answer: 190

Solution:

Each crate independently contributes height $3,$ $4,$ or $6,$ each with probability $\frac{1}{3},$ so there are $3^{10}$ equally likely stacks. If $x,$ $y,$ $z$ crates have heights $3,$ $4,$ $6,$ then $x + y + z = 10$ and $3x + 4y + 6z = 41;$ subtracting three times the first equation gives $y + 3z = 11,$ so $$(x, y, z) = (1, 8, 1), \quad (3, 5, 2), \quad (5, 2, 3).$$

These can be ordered in $\frac{10!}{1!\,8!\,1!} = 90,$ $\frac{10!}{3!\,5!\,2!} = 2520,$ and $\frac{10!}{5!\,2!\,3!} = 2520$ ways, for $5130$ stacks in all. The probability is $\frac{5130}{3^{10}} = \frac{190}{3^7},$ which is in lowest terms since $190 = 2 \cdot 5 \cdot 19.$ Thus $m = 190.$

Answer: 32

Solution:

By the triangle inequality, $30\sqrt{7} = DE \le DA + AE = DA + 10\sqrt{7},$ so $DA \ge 20\sqrt{7}.$ On the other hand, in triangle $ACD$ the angle at $D$ is $\frac{\pi}{3}$ and $AC = 10\sqrt{21},$ so the Law of Sines gives $$DA = \frac{AC \sin\angle DCA}{\sin\frac{\pi}{3}} = \frac{10\sqrt{21}}{\sqrt{3}/2}\sin\angle DCA = 20\sqrt{7}\sin\angle DCA \le 20\sqrt{7}.$$

Both bounds force $DA = 20\sqrt{7},$ so $\angle DCA = 90^\circ,$ and equality in the triangle inequality means $E$ lies on line $AD$ with $A$ between $D$ and $E.$ From the right triangle, $DC = \sqrt{DA^2 - AC^2} = \sqrt{2800 - 2100} = 10\sqrt{7},$ and since $\angle CDF = 60^\circ,$ the foot satisfies $DF = DC\cos 60^\circ = 5\sqrt{7}.$

Points $F$ and $E$ are on line $AD$ on the same side of $D,$ so $EF = DE - DF = 30\sqrt{7} - 5\sqrt{7} = 25\sqrt{7},$ and $m + n = 25 + 7 = 32.$

Answer: 172

Solution:

Let $a_n$ and $b_n$ count valid sequences of length $n$ beginning with $A$ and with $B.$ A sequence beginning with $A$ starts with $AA$ followed by any valid sequence of length $n - 2$ (possibly empty), so $a_{n+2} = a_n + b_n,$ where the empty sequence counts once. A sequence beginning with $B$ starts either with a single $B$ followed by a sequence beginning with $A,$ or with $BB$ followed by a sequence beginning with $B,$ so $b_{n+2} = a_{n+1} + b_n.$

Starting from $(a_1, b_1) = (0, 1)$ and $(a_2, b_2) = (1, 0),$ the pairs $(a_n, b_n)$ for $n = 3, 4, \ldots, 14$ are $$(1, 2),\ (1, 1),\ (3, 3),\ (2, 4),\ (6, 5),\ (6, 10),\ (11, 11),\ (16, 21),\ (22, 27),\ (37, 43),\ (49, 64),\ (80, 92).$$

The number of valid sequences of length $14$ is $80 + 92 = 172.$

Answer: 375

Solution:

Suppose the cars travel at $s$ kilometers per hour. The gap is $\lceil s/15 \rceil$ car lengths, so successive fronts are $4\lceil s/15 \rceil + 4$ meters apart, and in one hour a column of $1000s$ meters of traffic passes the eye — that is, $$N = \frac{1000s}{4\lceil s/15 \rceil + 4} = \frac{250s}{\lceil s/15 \rceil + 1}$$ gaps per hour.

For a fixed value $k = \lceil s/15 \rceil,$ the count $N$ is largest at $s = 15k,$ where it equals $\frac{3750k}{k + 1}.$ This is always less than $3750$ but approaches $3750$ as $k$ grows. Although the gap count never reaches $3750,$ the car count can: choose $k$ so large that more than $3749$ gaps pass, and start the hour with a car exactly at the eye. That car, plus one car for each of the $3749$ complete gaps that follow, makes $3750$ cars.

So $M = 3750,$ and the quotient when $M$ is divided by $10$ is $375.$

Answer: 40

Solution:

From $p(0,0) = 0$ we get $a_0 = 0.$ Adding and subtracting $p(1,0) = p(-1,0) = 0$ gives $a_3 = 0$ and $a_6 = -a_1;$ similarly $p(0,\pm 1) = 0$ give $a_5 = 0$ and $a_9 = -a_2.$ Then $p(1,1) = 0$ and $p(1,-1) = 0$ reduce to $a_4 + a_7 + a_8 = 0$ and $-a_4 - a_7 + a_8 = 0,$ so $a_8 = 0$ and $a_7 = -a_4.$ Now $p = a_1(x - x^3) + a_2(y - y^3) + a_4(xy - x^2y),$ and $p(2,2) = 0$ gives $-6a_1 - 6a_2 - 4a_4 = 0,$ i.e. $a_4 = -\frac{3}{2}(a_1 + a_2).$

Therefore $$p = a_1\left[x - x^3 - \tfrac{3}{2}xy(1 - x)\right] + a_2\left[y - y^3 - \tfrac{3}{2}xy(1 - x)\right],$$ and a point $(r, s)$ that is a zero for every choice of $a_1, a_2$ must kill both brackets. The first bracket factors as $r(1 - r)\left(1 + r - \tfrac{3}{2}s\right),$ so for a new point (with $r \ne 0, 1$) we need $s = \tfrac{2}{3}(r + 1).$ The second bracket is $\tfrac{1}{2}s(2 - 2s^2 - 3r + 3r^2);$ substituting $s^2 = \tfrac{4}{9}(r + 1)^2$ turns $2 - 2s^2 - 3r + 3r^2 = 0$ into $$\frac{19r^2 - 43r + 10}{9} = 0,$$ whose roots are $r = 2$ and $r = \frac{5}{19}.$

The root $r = 2$ reproduces the given point $(2, 2),$ so the new point has $r = \frac{5}{19}$ and $s = \frac{2}{3} \cdot \frac{24}{19} = \frac{16}{19}.$ Thus $(a, b, c) = (5, 16, 19)$ and $a + b + c = 40.$

Answer: 432

Solution:

Place the center $O$ at the origin with radius $9,$ so $A = (-9, 0)$ and $B = (9, 0).$ If the point of tangency is $T = (9\cos t, 9\sin t),$ the tangent line is $x\cos t + y\sin t = 9;$ it meets the $x$-axis at $C = (9/\cos t, 0),$ which lies beyond $A$ exactly when $-1 \lt \cos t \lt 0.$ Writing $u = \cos t,$ the signed distance from $A$ to the line is $-9u - 9,$ so the foot of the perpendicular is $P = A + 9(1 + u)(\cos t, \sin t).$

Then $P - B = \bigl(9(u^2 + u - 2),\ 9(1 + u)\sin t\bigr),$ and using $\sin^2 t = 1 - u^2:$ $$\frac{BP^2}{81} = (u^2 + u - 2)^2 + (1 + u)^2(1 - u^2) = 5 - 2u - 3u^2.$$ This quadratic in $u$ is maximized at $u = -\frac{1}{3},$ which is inside $(-1, 0)$ (there $C = (-27, 0)$), giving $\frac{BP^2}{81} = 5 + \frac{2}{3} - \frac{1}{3} = \frac{16}{3}.$

Therefore $m^2 = 81 \cdot \frac{16}{3} = 432.$

Answer: 871

Solution:

Put the corner at the origin $O$ with the two sides along the positive axes, and write $a = \sqrt{17}.$ The cut on the bottom edge starts at $P = (a, 0),$ and the two cuts meet at $R$ on the diagonal $y = x,$ each making a $30^\circ$ angle with the diagonal. In triangle $OPR,$ $\angle ROP = 45^\circ$ and $\angle ORP = 30^\circ,$ so the Law of Sines gives $PR = \frac{OP\sin 45^\circ}{\sin 30^\circ} = a\sqrt{2}.$ The fold lines are the horizontal and vertical lines through $R.$ Let $S$ be the point of the horizontal fold line directly above $P,$ and $T = (a, a)$ the point where the vertical line through $P$ meets the diagonal. Since $\angle OPR = 105^\circ,$ segment $PR$ makes a $75^\circ$ angle with the bottom edge, so $$SP = PR\sin 75^\circ = a\sqrt{2} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} = a\,\frac{\sqrt{3} + 1}{2}, \qquad ST = SP - PT = a\,\frac{\sqrt{3} + 1}{2} - a = a\,\frac{\sqrt{3} - 1}{2}.$$

When the bottom strip folds up along the horizontal line through $R,$ point $P$ stays at distance $SP$ from $S,$ moving in the vertical plane through $P$ perpendicular to that fold line. By symmetry the two taped cut edges meet above the diagonal, so $P$ lands at a point $P'$ directly above $T,$ and $P'T$ is the height of the tray. By the Pythagorean theorem, $$P'T^2 = P'S^2 - ST^2 = a^2\left(\frac{\sqrt{3} + 1}{2}\right)^2 - a^2\left(\frac{\sqrt{3} - 1}{2}\right)^2 = a^2\sqrt{3}.$$

So the height is $a \cdot 3^{1/4} = \sqrt{17} \cdot \sqrt[4]{3} = \sqrt[4]{17^2 \cdot 3} = \sqrt[4]{867},$ and $m + n = 867 + 4 = 871.$