For a pair of integers $a$ and $b$, with $0 < a < b < 1000$, set $S\subseteq \{ 1, 2, \dots , 2003\}$ is called a skipping set for $(a, b)$ if for any pair of elements $s_1, s_2 \in S$, $|s_1 - s_2|\not\in \{ a, b\}$. Let $f(a, b)$ be the maximum size of a skipping set for $(a, b)$. Determine the maximum and minimum values of $f$.

Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$, $PB$, $PC$ intersect sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Prove that \[ [PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC] \] if and only if $P$ lies on at least one of the medians of triangle $ABC$. (Here $[XYZ]$ denotes the area of triangle $XYZ$.)

Find all ordered triples of primes $(p, q, r)$ such that \[ p \mid q^r + 1, \quad q \mid r^p + 1, \quad r \mid p^q + 1. \] Reid Barton

Click for solution I remember seeing it at least twice on th forum, and solving it once before.. But you're right: it is nice. Suppose WLOG that $p<q,r$. We have $p|(q^{2r}-1,q^{p-1}-1)=q^{(2r,p-1)}-1|q^2-1$. This means that either $p|q-1$ or $p|q+1$. Assume first that all our primes are odd. Then we can't possibly have $p|q-1$, because then $p|q^r+1\equiv 2\pmod p\Rightarrow p=2$. We thus have $p|q+1$. From $q|r^{(2p,q-1)}-1$ and $p|q+1,p\ne 2$ we get $q|r^2-1\Rightarrow q|r-1$ or $r+1$. Again, we can't have $q|r-1$, so $q|r+1$. Just as above, from the equation $r|p^q+1$ and $q|r+1$ we deduce $r|p+1$. This is a contradiction: since $p,q,r>2$ and $p|q+1,q|r+1,r|p+1$, we must have $p<q<r<p$. This means that $p=2$. If $q$ is odd, we have $r|2^q+1\Rightarrow r|2^{(2q,r-1)}-1|2^2-1=3$ (the last divisibility takes place because $q$ is odd it cannot divide $r$, given that $q|r^2+1$). $r=3,q=5$ follows, and the only solutions $(p,q,r)$ are the cyclic permutations of $(2,5,3)$. If, on the other hand, $q=2$, then we get $r=5$, and we have the solution $(p,q,r)=(2,2,5)$ and all its cyclic permutations.

Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that \[ f(m+n)f(m-n) = f(m^2) \] for $m,n \in \mathbb{N}$.

Let $A, B, C$ be real numbers in the interval $\left(0,\frac{\pi}{2}\right)$. Let \begin{align*} X &= \frac{\sin A\sin (A-B)\sin (A-C)}{\sin (B+C)} \\ Y &= \frac{\sin B\sin(B-C)\sin (B-A)}{\sin (C+A)} \\ Z &= \frac{\sin C\sin (C-A)\sin (C-B)}{\sin (A+B)} . \end{align*}Prove that $X+Y+Z \geq 0$.

Let $\overline{AH_1}, \overline{BH_2}$, and $\overline{CH_3}$ be the altitudes of an acute scalene triangle $ABC$. The incircle of triangle $ABC$ is tangent to $\overline{BC}, \overline{CA},$ and $\overline{AB}$ at $T_1, T_2,$ and $T_3$, respectively. For $k = 1, 2, 3$, let $P_i$ be the point on line $H_iH_{i+1}$ (where $H_4 = H_1$) such that $H_iT_iP_i$ is an acute isosceles triangle with $H_iT_i = H_iP_i$. Prove that the circumcircles of triangles $T_1P_1T_2$, $T_2P_2T_3$, $T_3P_3T_1$ pass through a common point.

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