Answer: 146

Small Hint:

Let $s=x+y$ and $p=xy$, and rewrite both given equations using $s$ and $p$

Big Hint:

The two equations determine $s+p$ and $sp$, so $s$ and $p$ are roots of one quadratic

Solution:

Let $s=x+y$ and $p=xy.$ The equations become $s+p=71$ and $sp=880,$ so $s$ and $p$ are the roots of $$\begin{aligned}t^2-71t+880&=0,\\ (t-16)(t-55)&=0.\end{aligned}$$ Because $x$ and $y$ are positive integers, $s=16$ and $p=55$; indeed, $x=5$ and $y=11.$ Therefore $$\begin{aligned}x^2+y^2&=s^2-2p\\&=16^2-2(55)=146.\end{aligned}$$

Answer: 840

Small Hint:

Each segment $\overline{P_kQ_k}$ is parallel to the $3$-$4$-$5$ diagonal and is a fixed fraction of its length

Big Hint:

The two side constructions give two identical arithmetic sums; remember to include $\overline{AC}$

Solution:

Put $A=(0,0),$ $B=(4,0),$ and $C=(4,3).$ Then $$\begin{aligned}P_k&=\left(\frac{4k}{168},0\right),\\Q_k&=\left(4,3-\frac{3k}{168}\right),\end{aligned}$$ so the $3$-$4$-$5$ ratio gives $$P_kQ_k=5\left(1-\frac{k}{168}\right).$$ Hence one construction has total length $$5\sum_{k=1}^{167}\left(1-\frac{k}{168}\right)=\frac{5\cdot167}{2}.$$ The construction on the other two sides has the same total, and $AC=5.$ Thus the requested sum is $5(167)+5=840.$

Answer: 166

Small Hint:

Compare $A_{k+1}$ directly with $A_k$ instead of estimating the binomial coefficients

Big Hint:

Find the last $k$ for which $\frac{A_{k+1}}{A_k}$ is greater than $1$

Solution:

Consecutive terms satisfy $$\frac{A_{k+1}}{A_k}=\frac{1000-k}{k+1}\cdot\frac15.$$ This ratio exceeds $1$ exactly when $1000-k>5k+5,$ or $k<\frac{995}{6}.$ Thus the terms increase through $A_{166}$ and decrease afterward. Therefore the largest term is $A_{166}.$

Answer: 159

Small Hint:

The bound $\lvert\sin(5\pi x)\rvert\leq1$ restricts $x$ to a finite interval

Big Hint:

Separate the positive and negative half-waves of the sine function and count two crossings on each eligible full half-wave

Solution:

Let $h(x)=\sin(5\pi x)-\frac15\log_2x.$ Any root lies in $[2^{-5},2^5]=[\frac{1}{32},32].$

For $x<1,$ a root can occur only on a negative half-wave of the sine. There are two such half-waves, $(\frac{1}{5},\frac{2}{5})$ and $(\frac{3}{5},\frac{4}{5}).$ On each, $h$ is positive at both endpoints and negative at the midpoint, so there are two roots. Uniqueness on the descending half follows from monotonicity; on the ascending half, $$\begin{aligned}h''&=-25\pi^2\sin(5\pi x)\\&\quad+\frac1{5x^2\ln2}>0,\end{aligned}$$ so $h'$ increases from a negative value at the midpoint and vanishes once. The function $h$ therefore first decreases and then increases from a negative midpoint value to a positive endpoint value, giving exactly one root on this half. Thus there are $4$ roots below $1.$

Also $x=1$ is a root. For $1<x<32,$ roots can occur only on positive half-waves $(\frac{2m}{5},\frac{2m+1}{5})$ with $m=3,$ $4,$ $\ldots,$ $79.$ There are $77$ of these. The value of $h$ is negative at each endpoint and positive at the midpoint. The right half is strictly decreasing. On the left half, $$\begin{aligned}h'''&=-125\pi^3\cos(5\pi x)\\&\quad-\frac2{5x^3\ln2}<0,\end{aligned}$$ so $h'$ is strictly concave. It is positive at the left endpoint and negative at the midpoint, so it changes sign exactly once. Consequently, $h$ rises to one maximum and then falls to a still-positive midpoint, giving exactly one crossing on the left half. Hence every such half-wave contributes exactly two roots. Therefore the total number is $$4+1+2(77)=159.$$

Answer: 128

Small Hint:

If $\frac{a}{b}$ is in lowest terms and $ab=20!$, each full prime power of $20!$ must go entirely to one of $a$ or $b$

Big Hint:

Count ordered allocations of the distinct prime-power factors, then use the condition $a<b$

Solution:

The distinct primes dividing $20!$ are $2,$ $3,$ $5,$ $7,$ $11,$ $13,$ $17,$ and $19.$ If $\frac{a}{b}$ is in lowest terms and $ab=20!,$ the entire power of each of these eight primes must be assigned to either $a$ or $b.$ Thus there are $2^8$ ordered coprime factorizations $ab=20!.$ Since $a\neq b,$ exactly half have $0<\frac{a}{b}<1.$ The number sought is $2^7=128.$

Answer: 743

Small Hint:

Write $r=n+f$ with integer $n$ and $0\leq f<1$

Big Hint:

After removing the common integer part, count how many of the $73$ fractional terms cross $1$

Solution:

Write $r=n+f,$ where $n$ is an integer and $0\leq f<1.$ There are $73$ summands. Since $546=73\cdot7+35,$ we must have $n=7,$ and exactly $35$ of the numbers $f+\frac{19}{100},$ $\ldots,$ $f+\frac{91}{100}$ have floor $1.$ These must be the terms with numerators $57,$ $58,$ $\ldots,$ $91.$ Hence $$f+\frac{56}{100}<1\leq f+\frac{57}{100},$$ so $43\leq100f<44.$ Therefore $\lfloor100r\rfloor=700+43=743.$

Answer: 383

Small Hint:

Define $f(t)=\sqrt{19}+\frac{91}{t}$; the equation says that $f$ applied five times returns $x$

Big Hint:

Use the two fixed points of $f$ and track the ratio $\frac{f(t)-\alpha}{f(t)-\beta}$

Solution:

Let $f(t)=\sqrt{19}+\frac{91}{t},$ and let $\alpha>0>\beta$ be its fixed points. They satisfy $$t^2-\sqrt{19}\,t-91=0.$$ A direct subtraction using $91=\alpha(\alpha-\sqrt{19})=\beta(\beta-\sqrt{19})$ gives $$\frac{f(t)-\alpha}{f(t)-\beta}=\frac{\beta}{\alpha}\,\frac{t-\alpha}{t-\beta}.$$ The given equation is $f^5(x)=x.$ If $x$ were neither $\alpha$ nor $\beta,$ iterating the displayed ratio five times would force $(\frac{\beta}{\alpha})^5=1,$ which is impossible because $\frac{\beta}{\alpha}<0.$ Thus the only roots are $\alpha$ and $\beta.$

Their absolute values sum to $\alpha-\beta,$ the difference of the roots of the quadratic. Hence $$A=\sqrt{(\sqrt{19})^2+4(91)}=\sqrt{383},$$ so $A^2=383.$

Answer: 10

Small Hint:

Let the two integer roots be $r$ and $s$, and eliminate $a$ using Vieta’s formulas

Big Hint:

Complete a product after obtaining $rs=-6(r+s)$

Solution:

Let the integer roots be $r$ and $s.$ Vieta’s formulas give $r+s=-a$ and $rs=6a,$ so $rs=-6(r+s).$ Therefore $$(r+6)(s+6)=36.$$ Conversely, every ordered integer factorization $uv=36$ gives integer roots $r=u-6,$ $s=v-6,$ and $a=12-u-v.$ Unordered positive factor pairs of $36$ have sums $37,$ $20,$ $15,$ $13,$ $12,$ and the corresponding negative factor pairs have their negatives as sums. These ten sums are distinct, so they give $10$ distinct values of $a.$

Answer: 44

Small Hint:

If $u=\sec x+\tan x$, then $\sec x-\tan x=\frac{1}{u}$

Big Hint:

Rewrite $\csc x+\cot x$ as $\frac{\sec x+1}{\tan x}$

Solution:

Let $u=\sec x+\tan x=\frac{22}{7}.$ Since $\sec x+\tan x$ and $\sec x-\tan x$ have product $1,$ $$\begin{aligned}\sec x&=\frac{u+u^{-1}}2,\\\tan x&=\frac{u-u^{-1}}2.\end{aligned}$$ Also $$\begin{aligned}\csc x+\cot x&=\frac{1+\cos x}{\sin x}\\&=\frac{\sec x+1}{\tan x}\\&=\frac{u+1}{u-1}.\end{aligned}$$ Substituting $u=\frac{22}{7}$ gives $\frac{29}{15}.$ Thus $m+n=29+15=44.$

Answer: 532

Small Hint:

The ordering is decided at the first position where the two received strings differ

Big Hint:

At one position, compute the probabilities of equality and of receiving $a$ in $S_a$ and $b$ in $S_b$

Solution:

At any position, the received letters agree with probability $$2\left(\frac23\right)\left(\frac13\right)=\frac49.$$ The favorable first difference, $S_a$ receiving $a$ and $S_b$ receiving $b,$ has probability $(\frac{2}{3})^2=\frac{4}{9}.$ It can occur in the first, second, or third position after zero, one, or two agreements. Hence $$\begin{aligned}p&=\frac49\left(1+\frac49+\left(\frac49\right)^2\right)\\&=\frac{532}{729}.\end{aligned}$$ This fraction is in lowest terms, so its numerator is $532.$

Answer: 135

Small Hint:

Join the center of $C$ to the centers and tangency point of two neighboring disks

Big Hint:

The resulting right triangle has angle $15^\circ$, adjacent leg $1$, and opposite leg equal to a disk radius

Solution:

Let $O$ be the center of $C,$ let $U$ and $V$ be the centers of two neighboring disks, and let $T$ be their tangency point. By the $12$-fold symmetry, $\angle UOV=30^\circ,$ and $OT$ bisects that angle. Also $T$ is the midpoint of $\overline{UV},$ so triangle $OUT$ is right at $T.$ Since $T$ lies on $C,$ $OT=1,$ while $UT$ is the disk radius $r.$ Thus $$r=\tan15^\circ=2-\sqrt3.$$ The total area is $$\begin{aligned}12\pi r^2&=12\pi(7-4\sqrt3)\\&=\pi(84-48\sqrt3).\end{aligned}$$ Therefore $a+b+c=84+48+3=135.$

Answer: 677

Small Hint:

The center of the rhombus is also the center of the rectangle, and its half-diagonals have lengths $15$ and $20$

Big Hint:

Use coordinates for $P$ and $Q$; their vectors from the common center are perpendicular

Solution:

Let the rectangle have width $w$ and height $h,$ with $A=(0,0)$ and $B=(w,0).$ Then $P=(w-15,0)$ and $Q=(w,20).$ The diagonals of the rhombus bisect each other at the rectangle’s center $O=(\frac{w}{2},\frac{h}{2}).$ Put $p=\overrightarrow{OP}=(\frac{w}{2}-15,-\frac{h}{2}).$ Since $OP=15,$ $OQ=20,$ and the rhombus diagonals are perpendicular, while $\overrightarrow{PQ}=(15,20),$ we have $$\begin{aligned}|p|&=15,\\p\mathbin{\cdot}(15,20)&=-225.\end{aligned}$$ Solving these two equations, with the second coordinate negative because $h>0,$ gives $p=(\frac{21}{5},-\frac{72}{5}).$ Hence $$\begin{aligned}w&=2\left(15+\frac{21}{5}\right)=\frac{192}{5},\\h&=\frac{144}{5}.\end{aligned}$$ The perimeter is $2(w+h)=\frac{672}{5},$ so $m+n=672+5=677.$

Answer: 990

Small Hint:

If there are $r$ red and $b$ blue socks, it is equivalent to require probability $\frac{1}{2}$ of drawing one of each color

Big Hint:

Use $n=r+b$ and $d=r-b$ to turn the probability equation into a square condition

Solution:

Let $r$ and $b$ be the two color counts and $n=r+b.$ The probability of drawing different colors is also $\frac{1}{2},$ so $$\begin{aligned}\frac{rb}{\binom n2}&=\frac12,\\4rb&=n(n-1).\end{aligned}$$ Since $4rb=(r+b)^2-(r-b)^2,$ this becomes $(r-b)^2=n.$ Thus $n=k^2$ and, choosing red as the more numerous color, $$r=\frac{k^2+k}{2}.$$ The largest square at most $1991$ is $44^2=1936,$ giving $r=\frac{1936+44}{2}=990.$

Answer: 384

Small Hint:

Let $2u$ be the central angle subtended by each side of length $81$, and set $t=2\cos u$

Big Hint:

Express $\frac{\sin(5u)}{\sin u}$ and the three diagonal ratios in terms of $t$

Solution:

Let $2u$ be the central angle subtended by each $81$-side, and put $t=2\cos u.$ The remaining arc has half-angle $\pi-5u,$ so the chord ratio gives $$\begin{aligned}\frac{31}{81}&=\frac{\sin5u}{\sin u}\\&=t^4-3t^2+1.\end{aligned}$$ This yields $t^2=\frac{25}{9}$ or $\frac{2}{9}.$ Because $5u<\pi,$ we have $t>2\cos36^\circ,$ so $t=\frac{5}{3}.$

The three diagonals from $A$ subtend the same minor angles as $2u,$ $3u,$ and $4u.$ Relative to an $81$-side, their length ratios are $$\begin{aligned}\frac{\sin2u}{\sin u}&=t,\\\frac{\sin3u}{\sin u}&=t^2-1,\\\frac{\sin4u}{\sin u}&=t^3-2t.\end{aligned}$$ Their sum is therefore $$\begin{aligned}81(t^3+t^2-t-1)&=81\left(\frac{128}{27}\right)\\&=384.\end{aligned}$$

Answer: 12

Small Hint:

Interpret each radical as the length of a vector $(2k-1,a_k)$ and apply the triangle inequality

Big Hint:

After finding $S_n$, factor the difference of two squares that results from requiring it to be an integer

Solution:

The two component sums are $\sum_{k=1}^n(2k-1)=n^2$ and $\sum_{k=1}^na_k=17.$ Therefore the triangle inequality for vectors gives $$\begin{aligned}&\sum_{k=1}^n\sqrt{(2k-1)^2+a_k^2}\\&\quad\geq\sqrt{(n^2)^2+17^2}\\&\quad=\sqrt{n^4+289}.\end{aligned}$$ Equality is attainable by taking $a_k$ proportional to $2k-1,$ so $S_n=\sqrt{n^4+289}.$

If this is the integer $m,$ then $$(m-n^2)(m+n^2)=289=17^2.$$ The factor pair $17,17$ gives $n=0,$ while the pair $1,289$ gives $m=145$ and $n^2=144.$ Thus the unique positive $n$ is $12.$