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# Difference between revisions of "2001 IMO Problems/Problem 1"

## Problem

Consider an acute triangle $\triangle ABC$. Let $P$ be the foot of the altitude of triangle $\triangle ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $\triangle ABC$. Assume that $\angle C \geq \angle B+30^{\circ}$. Prove that $\angle A+\angle COP < 90^{\circ}$.

## Solution

Take $D$ on the circumcircle with $AD \parallel to BC$. Notice that $\angle CBD = \angle BCA$, so $\angle ABD \ge 30^\circ$. Hence $\angle AOD \ge 60^\circ$. Let $Z$ be the midpoint of $AD$ and $Y$ the midpoint of $BC$. Then $AZ \ge R/2$, where $R$ is the radius of the circumcircle. But $AZ = YX$ (since $AZYX$ is a rectangle).

Now $O$ cannot coincide with $Y$ (otherwise $\angle A$ would be $90^\circ$ and the triangle would not be acute-angled). So $OX > YX \ge R/2$. But $XC = YC - YX < R - YX \le R/2$. So $OX > XC$.

Hence $\angle COX < \angle OCX$. Let $CE$ be a diameter of the circle, so that $\angle OCX = \angle ECB$. But $\angle ECB = \angle EAB$ and $\angle EAB + \angle BAC = \angle EAC = 90^\circ$, since $EC$ is a diameter. Hence $\angle COX + \angle BAC < 90^\circ$.