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Difference between revisions of "2008 Indonesia MO Problems/Problem 1"

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Latest revision as of 16:50, 8 April 2020

Problem

Given triangle $ABC$. Points $D,E,F$ outside triangle $ABC$ are chosen such that triangles $ABD$, $BCE$, and $CAF$ are equilateral triangles. Prove that cicumcircles of these three triangles are concurrent.

Solution

Let $X$ be the intersection of the circumcircles of $\triangle ABD$ and $\triangle ACF$. Note that $AXBD$ and $AXCF$ are cyclic quadrilaterals. Thus, $\angle ADB + \angle AXB = 180^\circ$ and $\angle AFC + \angle AXC = 180^\circ$.


We know that $\triangle ABD$ and $\triangle ACF$ are equilateral, so $\angle ADB = \angle AFC = 60^\circ$. Therefore, $\angle AXB = \angle AXC = 120^\circ$, so $\angle BXC = 120^\circ$.


Since $\triangle BEC$ is equilateral as well, $\angle BEC = 60^\circ$. Note that $\angle BXC + \angle BEC = 180^\circ$, and since the circumcircle is the circle that passes through $B, E, C$, point $X$ must also be on the same circumcircle of $\triangle BEC$. Thus, the cicumcircles of these three triangles are concurrent.

See Also

2008 Indonesia MO (Problems)
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed by
Problem 2
All Indonesia MO Problems and Solutions
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