This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 5

1991 IMO, 2

Let $ \,ABC\,$ be a triangle and $ \,P\,$ an interior point of $ \,ABC\,$. Show that at least one of the angles $ \,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $ 30^{\circ }$.

2004 IMO Shortlist, 8

Given a cyclic quadrilateral $ABCD$, let $M$ be the midpoint of the side $CD$, and let $N$ be a point on the circumcircle of triangle $ABM$. Assume that the point $N$ is different from the point $M$ and satisfies $\frac{AN}{BN}=\frac{AM}{BM}$. Prove that the points $E$, $F$, $N$ are collinear, where $E=AC\cap BD$ and $F=BC\cap DA$. [i]Proposed by Dusan Dukic, Serbia and Montenegro[/i]

1992 Spain Mathematical Olympiad, 5

Given a triangle $ABC$, show how to construct the point $P$ such that $\angle PAB= \angle PBC= \angle PCA$. Express this angle in terms of $\angle A,\angle B,\angle C$ using trigonometric functions.

1991 IMO Shortlist, 4

Let $ \,ABC\,$ be a triangle and $ \,P\,$ an interior point of $ \,ABC\,$. Show that at least one of the angles $ \,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $ 30^{\circ }$.

1999 Brazil National Olympiad, 6

Given any triangle $ABC$, show how to construct $A'$ on the side $AB$, $B'$ on the side $BC$ and $C'$ on the side $CA$, such that $ABC$ and $A'B'C'$ are similar (with $\angle A = \angle A', \angle B = \angle B', \angle C = \angle C'$) and $A'B'C'$ has the least possible area.