# 2008 IMO Problems/Problem 1

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## Problem

An acute-angled triangle $ABC$ has orthocentre $H$. The circle passing through $H$ with centre the midpoint of $BC$ intersects the line $BC$ at $A_1$ and $A_2$. Similarly, the circle passing through $H$ with centre the midpoint of $CA$ intersects the line $CA$ at $B_1$ and $B_2$, and the circle passing through $H$ with centre the midpoint of $AB$ intersects the line $AB$ at $C_1$ and $C_2$. Show that $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ lie on a circle.

## Solution

Let $M_A$, $M_B$, and $M_C$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. It's not hard to see that $M_BM_C\parallel BC$. We also have that $AH\perp BC$, so $AH \perp M_BM_C$. Now note that the radical axis of two circles is perpendicular to the line connecting their centers. We know that $H$ is on the radical axis of the circles centered at $M_B$ and $M_C$, so $A$ is too. We then have $AC_1\cdot AC_2=AB_2\cdot AB_1\Rightarrow \frac{AB_2}{AC_1}=\frac{AC_2}{AB_1}$. This implies that $\triangle AB_2C_1\sim \triangle AC_2B_1$, so $\angle AB_2C_1=\angle AC_2B_1$. Therefore $\angle C_1B_2B_1=180^{\circ}-\angle AB_2C_1=180^{\circ}-\angle AC_2B_1$. This shows that quadrilateral $C_1C_2B_1B_2$ is cyclic. Note that the center of its circumcircle is at the intersection of the perpendicular bisectors of the segments $C_1C_2$ and $B_1B_2$. However, these are just the perpendicular bisectors of $AB$ and $CA$, which meet at the circumcenter of $ABC$, so the circumcenter of $C_1C_2B_1B_2$ is the circumcenter of triangle $ABC$. Similarly, the circumcenters of $A_1A_2B_1B_2$ and $C_1C_2A_1A_2$ are coincident with the circumcenter of $ABC$. The desired result follows.