Given a triangle $ABC$, let $P$ and $Q$ be points on segments $\overline{AB}$ and $\overline{AC}$, respectively, such that $AP=AQ$. Let $S$ and $R$ be distinct points on segment $\overline{BC}$ such that $S$ lies between $B$ and $R$, $\angle BPS=\angle PRS$, and $\angle CQR=\angle QSR$. Prove that $P,Q,R,S$ are concyclic (in other words, these four points lie on a circle).

Find all integers $n \geq 3$ such that among any $n$ positive real numbers $a_1, a_2, \hdots, a_n$ with $\text{max}(a_1,a_2,\hdots,a_n) \leq n \cdot \text{min}(a_1,a_2,\hdots,a_n)$, there exist three that are the side lengths of an acute triangle.

Let $a,b,c$ be positive real numbers. Prove that $\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \geq \frac{2}{3}(a^2+b^2+c^2)$.

Let $\alpha$ be an irrational number with $0<\alpha < 1$, and draw a circle in the plane whose circumference has length $1$. Given any integer $n\ge 3$, define a sequence of points $P_1, P_2, \ldots , P_n$ as follows. First select any point $P_1$ on the circle, and for $2\le k\le n$ define $P_k$ as the point on the circle for which the length of arc $P_{k-1}P_k$ is $\alpha$, when travelling counterclockwise around the circle from $P_{k-1}$ to $P_k$. Suppose that $P_a$ and $P_b$ are the nearest adjacent points on either side of $P_n$. Prove that $a+b\le n$.

For distinct positive integers $a, b<2012$, define $f(a, b)$ to be the number of integers $k$ with $1\le k<2012$ such that the remainder when $ak$ divided by $2012$ is greater than that of $bk$ divided by $2012$. Let $S$ be the minimum value of $f(a, b)$, where $a$ and $b$ range over all pairs of distinct positive integers less than $2012$. Determine $S$.

Let $P$ be a point in the plane of $\triangle ABC$, and $\gamma$ a line passing through $P$. Let $A', B', C'$ be the points where the reflections of lines $PA, PB, PC$ with respect to $\gamma$ intersect lines $BC, AC, AB$ respectively. Prove that $A', B', C'$ are collinear.

These problems are copyright $\copyright$ Mathematical Association of America.