Small Hint:

Work in rooms per minute: Abe’s rate is $\frac{1}{900},$ so Bea’s is $\frac{1}{600}$ and Coe’s is $\frac{1}{450}$

Big Hint:

After $90$ minutes alone, $\frac{1}{10}$ of the room is done. Divide each remaining portion of the room by the combined rate of the painters working then.

Solution:

Abe paints $\frac{1}{900}$ of the room per minute, so Bea paints $\frac{3}{2} \cdot \frac{1}{900} = \frac{1}{600}$ per minute and Coe paints $\frac{2}{900} = \frac{1}{450}$ per minute. In the first $90$ minutes Abe paints $\frac{90}{900} = \frac{1}{10}$ of the room.

Abe and Bea together paint $\frac{1}{900} + \frac{1}{600} = \frac{1}{360}$ per minute, and they must bring the total from $\frac{1}{10}$ up to $\frac{1}{2},$ which takes $\frac{2}{5} \cdot 360 = 144$ minutes. All three together paint $\frac{1}{360} + \frac{1}{450} = \frac{1}{200}$ per minute, so the remaining half of the room takes $\frac{1}{2} \cdot 200 = 100$ minutes.

The total time is $90 + 144 + 100 = 334$ minutes.

Small Hint:

Take a population of $100$ and fill in a Venn diagram: each exactly-one region holds $10$ men and each exactly-two region holds $14$

Big Hint:

If $x$ men have all three factors, then $\frac{x}{x + 14} = \frac{1}{3}.$ Then count the men outside $A$ and the men with no factor at all.

Solution:

Take a population of $100$ men and fill in a Venn diagram. Each of the three exactly-one regions contains $10$ men, and each of the three exactly-two regions contains $14.$ If $x$ men have all three factors, then the men with both $A$ and $B$ number $x + 14,$ so the given conditional probability says $\frac{x}{x + 14} = \frac{1}{3},$ giving $x = 7.$

The union of the three sets therefore contains $3 \cdot 10 + 3 \cdot 14 + 7 = 79$ men, leaving $21$ with no risk factor. The men with risk factor $A$ number $10 + 14 + 14 + 7 = 45,$ so $55$ men do not have $A.$

The desired probability is $\frac{21}{55},$ which is in lowest terms, so $p + q = 21 + 55 = 76.$

Small Hint:

Each half of a $36$-side is a hinged bar of length $18$ whose vertical extent in the hexagon is $12;$ the Pythagorean theorem gives its horizontal reach

Big Hint:

Cut the hexagon along the line through the two midpoint hinges into two congruent trapezoids and set the total area equal to $36a$

Solution:

In the hexagon, each side of length $36$ has folded at its midpoint into two bars of length $18.$ The two sides of length $a$ are $24$ apart, so each bar spans a vertical distance of $12$ and hence a horizontal distance of $\sqrt{18^2 - 12^2} = \sqrt{180} = 6\sqrt{5}.$

The line through the two midpoint hinges splits the hexagon into two congruent trapezoids with parallel sides $a$ and $a + 12\sqrt{5}$ and height $12,$ so the hexagon has area $$\begin{aligned} &2 \cdot \frac{a + (a + 12\sqrt{5})}{2} \cdot 12 \\ &= 24a + 144\sqrt{5}. \end{aligned}$$

Setting this equal to the rectangle’s area $36a$ gives $12a = 144\sqrt{5},$ so $a = 12\sqrt{5}$ and $a^2 = 720.$

Small Hint:

Write the decimals as $\frac{ab}{99}$ and $\frac{abc}{999},$ then clear denominators using $99 = 9 \cdot 11$ and $999 = 27 \cdot 37$

Big Hint:

Reducing the equation modulo $11$ forces $a = b;$ then a short linear equation pins down the digits

Solution:

Writing $ab$ and $abc$ for the two- and three-digit numbers, the decimals equal $\frac{ab}{99}$ and $\frac{abc}{999}.$ Since $99 = 9 \cdot 11$ and $999 = 27 \cdot 37,$ the common denominator is $27 \cdot 37 \cdot 11 = 10989,$ and multiplying the equation by it gives $$\begin{aligned} &111 \cdot ab + 11 \cdot abc \\ &= \frac{33}{37} \cdot 10989 \\ &= 9801. \end{aligned}$$

Modulo $11,$ since $9801 = 11 \cdot 891$ and $111 \equiv 1,$ this forces $ab$ to be divisible by $11,$ so $a = b.$ Then $ab = 11a,$ and dividing the equation by $11$ gives $111a + abc = 891.$ Since $abc = 110a + c,$ this is $221a + c = 891,$ which requires $a = 4$ and $c = 7.$

Thus $a = b = 4,$ $c = 7,$ and the three-digit number $abc$ is $447.$

Small Hint:

Neither cubic has an $x^2$ term, so the third roots are $t = -r - s$ and $t - 1.$ Equate the coefficients of $x$ in the two cubics.

Big Hint:

Comparing constant terms and substituting $t = 4r - 3s + 13$ yields $(r-s)^2 + 7(r-s) - 8 = 0;$ finish each case using $r + s + t = 0.$

Solution:

Both cubics have zero $x^2$ coefficient, so their roots sum to $0:$ the third root of $p$ is $t = -r - s,$ and the third root of $q$ is $-(r+4) - (s-3) = t - 1.$ The coefficient of $x$ is $a$ in both, so $$\begin{aligned} &rs + st + tr \\ &= (r+4)(s-3) \\ &\quad {}+ (s-3)(t-1) \\ &\quad {}+ (t-1)(r+4), \end{aligned}$$ which simplifies to $t = 4r - 3s + 13.$

The constant terms give $b = -rst$ and $b + 240 =$ $-(r+4)(s-3)(t-1),$ so $240 =$ $rst - (r+4)(s-3)(t-1),$ i.e. $rs - 4st + 3tr - 3r$ $+ 4s + 12t - 252 = 0.$ Substituting $t = 4r - 3s + 13$ reduces this to $12\left[(r-s)^2 + 7(r-s) - 8\right] = 0,$ so $r - s = 1$ or $r - s = -8.$

If $r - s = 1,$ then $t = 4r - 3s + 13 = r + 16$ and $t = -r - s = -2r + 1,$ so $r = -5:$ the roots are $-5,$ $-6,$ $11,$ and $b = -rst = -330.$ If $r - s = -8,$ then $t = r - 11 = -2r - 8,$ so $r = 1:$ the roots are $1,$ $9,$ $-10,$ and $b = 90.$ The requested sum is $330 + 90 = 420.$

Small Hint:

The answer is the probability of three sixes divided by the probability of two sixes, each averaged over the two equally likely dice

Big Hint:

Two sixes in a row make the biased die much more likely: weight each die by its chance of producing the observed two sixes before predicting the third roll

Solution:

The desired conditional probability is $$\begin{aligned} &\frac{\Pr(\text{three sixes})}{\Pr(\text{first two are sixes})} \\ &= \frac{\frac{1}{2}\left(\frac{2}{3}\right)^3 + \frac{1}{2}\left(\frac{1}{6}\right)^3} {\frac{1}{2}\left(\frac{2}{3}\right)^2 + \frac{1}{2}\left(\frac{1}{6}\right)^2}, \end{aligned}$$ since each die is chosen with probability $\frac{1}{2}$ and the fair die shows a six with probability $\frac{1}{6}.$

The numerator is $\frac{1}{2}\left(\frac{8}{27} + \frac{1}{216}\right) = \frac{65}{432}$ and the denominator is $\frac{1}{2}\left(\frac{4}{9} + \frac{1}{36}\right) = \frac{17}{72},$ so the probability is $\frac{65}{432} \cdot \frac{72}{17} = \frac{65}{102}.$

Since $65 = 5 \cdot 13$ and $102 = 2 \cdot 3 \cdot 17$ share no factor, $p + q = 65 + 102 = 167.$

Small Hint:

Since $\cos(\pi k) = (-1)^k,$ the product of the $f(k)$ telescopes: even $k$ contribute $(k+1)(k+2)$ and odd $k$ contribute its reciprocal

Big Hint:

For even $n$ the product is $\frac{n+2}{2};$ for odd $n$ it is $\frac{1}{2(n+2)}.$ The sum of logs is $\pm 1$ when the product is $10$ or $\frac{1}{10}.$

Solution:

Since $\cos(\pi k) = (-1)^k$ and $k^2 + 3k + 2 = (k+1)(k+2),$ we have $f(k) = \left[(k+1)(k+2)\right]^{(-1)^k}.$ The sum of the logarithms is the log of the product $\prod_{k=1}^n f(k),$ which telescopes: consecutive factors $\frac{1}{(k+1)(k+2)}$ and $(k+2)(k+3)$ leave only boundary terms.

For even $n$ the product is $\frac{3 \cdot 4 \cdots (n+2)}{2 \cdot 3 \cdots (n+1)} = \frac{n+2}{2},$ and for odd $n$ it is $\frac{3 \cdot 4 \cdots (n+1)}{2 \cdot 3 \cdots (n+2)} = \frac{1}{2(n+2)}.$ The absolute value of the log equals $1$ exactly when the product is $10$ or $\frac{1}{10}.$

For even $n,$ $\frac{n+2}{2} = 10$ gives $n = 18;$ for odd $n,$ $2(n+2) = 10$ gives $n = 3.$ The requested sum is $18 + 3 = 21.$

Small Hint:

Let $s$ be the radius of circle $E$ and drop its center to a foot $F$ on $\overline{AB};$ express $CF$ and $DF$ with the Pythagorean theorem

Big Hint:

$DF = DC + CF$ gives $s\sqrt{15} = (2 - 3s) + \sqrt{4 - 4s};$ isolating one radical and squaring twice leaves a quadratic in $s$

Solution:

Let $C,$ $D,$ $E$ also name the circles’ centers, let $s$ be the radius of circle $E,$ so circle $D$ has radius $3s,$ and let $F$ be the foot of $E$ on $\overline{AB}.$ Tangency gives $CE = 2 - s,$ $DE = 3s + s = 4s,$ and $EF = s,$ while $D$ lies on $\overline{AB}$ with $DC = 2 - 3s.$

Right triangles $CEF$ and $DEF$ give $CF = \sqrt{(2-s)^2 - s^2}$ $= \sqrt{4 - 4s}$ and $DF = \sqrt{(4s)^2 - s^2} = s\sqrt{15}.$ Since $F$ is on the opposite side of $C$ from $A,$ we have $DF = DC + CF,$ so $$s\sqrt{15} = (2 - 3s) + \sqrt{4 - 4s}.$$

Moving $2 - 3s$ to the left and squaring gives $24s^2 - 8s = 2\sqrt{15}\,s\,(2 - 3s),$ i.e. $12s - 4 = \sqrt{15}\,(2 - 3s);$ squaring again yields $9s^2 + 84s - 44 = 0,$ so $s = \frac{-14 + 4\sqrt{15}}{3}.$ The radius of circle $D$ is $3s = 4\sqrt{15} - 14 = \sqrt{240} - 14,$ and $m + n = 240 + 14 = 254.$

Small Hint:

Count subsets other than the full set by locating an empty chair followed clockwise by three chairs of the subset

Big Hint:

Each such empty-full-full-full pattern leaves $2^6$ choices for the remaining chairs; subtract the double-counted subsets containing two separate runs of three or more

Solution:

The full set of $10$ chairs qualifies; count the others by locating each maximal run of at least three adjacent chosen chairs at its clockwise start. Any such subset contains a block of four consecutive chairs that is empty-chosen-chosen-chosen. There are $10$ positions for this block, and the remaining $6$ chairs are free, giving $10 \cdot 2^6 = 640.$

This counts once for each maximal run of length at least $3.$ Two such runs require at least $3 + 3$ chosen chairs plus two gaps, so three runs are impossible, and subsets with exactly two runs are counted twice. To have two runs, place two disjoint empty-chosen-chosen-chosen blocks: $\frac{10 \cdot 3}{2} = 15$ ways (the second block fits in $3$ positions among the remaining $6$ chairs), with the last $2$ chairs free, for $15 \cdot 2^2 = 60$ subsets.

The total is $1 + 640 - 60 = 581.$

Small Hint:

Clear denominators in $\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}$ to get $z^2 + zw + w^2 = 0$

Big Hint:

Multiplying by $z - w$ gives $w^3 = z^3,$ so the vertices are $z$ times the cube roots of unity: an equilateral triangle with circumradius $2014$

Solution:

Multiplying $\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}$ by $zw(z+w)$ gives $zw = (z+w)^2,$ i.e. $z^2 + zw + w^2 = 0.$ Multiplying by $z - w$ yields $z^3 - w^3 = 0,$ so $w = \omega z$ or $w = \omega^2 z,$ where $\omega$ is a primitive cube root of unity (and both indeed satisfy the original equation).

Thus $P$ is the equilateral triangle with vertices $z,$ $\omega z,$ $\omega^2 z,$ inscribed in the circle of radius $2014.$ Its area is $$\frac{3\sqrt{3}}{4}\,(2014)^2 = 3 \cdot 1007^2 \sqrt{3},$$ so $n = 3 \cdot 1007^2 = 3042147.$

The remainder when $n$ is divided by $1000$ is $147.$

Small Hint:

Place $D$ at the origin and $E$ on the positive $x$-axis; then $R = (\cos 60^\circ, \sin 60^\circ)$ and the law of sines gives $DE = \frac{\sqrt{3}+1}{2}$

Big Hint:

Perpendicular slopes locate $C$ on the $x$-axis, and $CA = AR$ is a linear equation in the $x$-coordinate of $A$

Solution:

Since $\angle RDE = 180^\circ - 75^\circ - 45^\circ$ $= 60^\circ,$ place $D = (0,0)$ with $E$ on the positive $x$-axis, so $R = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).$ The law of sines gives $DE = \frac{\sin 75^\circ}{\sin 45^\circ} = \frac{\sqrt{3}+1}{2},$ and $M = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right).$

The slope of $EM$ is $\frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} - \frac{\sqrt{3}+1}{2}} = \frac{-\sqrt{3}}{1 + 2\sqrt{3}},$ so line $RC$ has slope $\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Descending from $R$ by $\frac{\sqrt{3}}{2}$ to the $x$-axis moves us left by $\frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{6\sqrt{3} - 3}{22},$ so $C = (c, 0)$ with $c = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{7 - 3\sqrt{3}}{11}.$

For $A = (t, 0),$ the condition $CA = AR$ reads $(t - c)^2 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4},$ which is linear in $t:$ $t = \frac{1 - c^2}{1 - 2c} = \frac{9 + 4\sqrt{3}}{11}.$ Then $$\begin{aligned} AE &= t - \frac{\sqrt{3}+1}{2} \\ &= \frac{18 + 8\sqrt{3} - 11\sqrt{3} - 11}{22} \\ &= \frac{7 - 3\sqrt{3}}{22} = \frac{7 - \sqrt{27}}{22}, \end{aligned}$$ so $a + b + c = 7 + 27 + 22 = 56.$

Small Hint:

Write $1 - \cos 3A = 2\sin^2 \frac{3A}{2}$ and combine $\cos 3B + \cos 3C$ by sum-to-product; the whole condition factors

Big Hint:

One angle must equal $120^\circ.$ The remaining side is longest when that angle lies between the sides of lengths $10$ and $13.$

Solution:

Using $1 - \cos 3A = 2\sin^2\frac{3A}{2}$ and $\cos 3B + \cos 3C =$ $2\cos\frac{3(B+C)}{2}\cos\frac{3(B-C)}{2},$ together with $\frac{3(B+C)}{2} = 270^\circ - \frac{3A}{2}$ so that $\cos\frac{3(B+C)}{2} = -\sin\frac{3A}{2},$ the condition becomes $$\begin{aligned} 0 &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\sin\tfrac{3A}{2} + \cos\tfrac{3(B-C)}{2}\right) \\ &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\cos\tfrac{3(B-C)}{2} - \cos\tfrac{3(B+C)}{2}\right) \\ &= 4\sin\tfrac{3A}{2}\sin\tfrac{3B}{2}\sin\tfrac{3C}{2}. \end{aligned}$$

For an angle $X$ of a triangle, $\frac{3X}{2}$ lies strictly between $0^\circ$ and $270^\circ,$ so $\sin\frac{3X}{2} = 0$ exactly when $X = 120^\circ.$ Hence one angle of the triangle is $120^\circ.$

The remaining side is longest when the $120^\circ$ angle sits between the sides of lengths $10$ and $13$ (if $120^\circ$ were opposite one of them, the remaining side would be shorter than that side). By the law of cosines its length is $\sqrt{10^2 + 13^2 + 10 \cdot 13} = \sqrt{399},$ so $m = 399.$

Small Hint:

Pairing left shoe $j$ with right shoe $\pi(j)$ defines a random permutation; a collection of $k$ pairs using exactly $k$ adults’ shoes is a union of cycles

Big Hint:

The condition says every cycle of $\pi$ has length at least $5:$ count $10$-cycles and products of two $5$-cycles

Solution:

The child’s pairing matches left shoe $j$ with right shoe $\pi(j)$ for a uniformly random permutation $\pi$ of $\{1, \ldots, 10\}.$ A collection of $k$ pairs uses $k$ left and $k$ right shoes, so it involves exactly $k$ adults precisely when those adults’ indices are closed under $\pi$ — that is, when the collection is a union of cycles of $\pi.$ The condition therefore says $\pi$ has no cycle of length less than $5.$

The cycle lengths must partition $10$ into parts of size at least $5:$ either one $10$-cycle or two $5$-cycles. There are $9!$ ten-cycles, and $\frac{1}{2}\binom{10}{5}(4!)^2 = \frac{9!}{5}$ permutations that are products of two $5$-cycles.

The probability is $$\frac{9! + \frac{1}{5} \cdot 9!}{10!} = \frac{1 + \frac{1}{5}}{10} = \frac{3}{25},$$ so $m + n = 3 + 25 = 28.$