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Estonian Mathematical Olympiad

Geometry Difficulty 4.6 AIME Prove it Estonia

A point $X$ is chosen on the median $AD$ of triangle $ABC$ . The circumcircle of triangle $ABX$ intersects the median $BE$ of triangle $ABC$ at point $Y \neq B$ . The circumcircle of triangle $EXY$ intersects the line $DE$ at point $K \neq E$ . Prove that the location of $K$ does not depend on $X$ .

Solution

We will use directed angles (Fig. 3 and 4 depict both possible configurations). As $DE$ is a midline in $ABC$ , we have $DE \parallel AB$ . Thus
$\angle DEY = \angle DEB = \angle ABE = \angle ABY = \angle AXY = \angle DXY.$
Therefore points $D$ , $E$ , $X$ , $Y$ are concyclic. This means that the circumcircle of $EXY$ intersects $DE$ at $D$ . Thus $K = D$ , i.e. the midpoint of $BC$ , regardless of the choice of $X$ .

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