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Problem 1214

AIME late

Geometry Difficulty 5.2 Prove it Iberoamerican Mathematical Olympiad · Ibero-American Mathematical Olympiad

$ABC$ is an acute-angled triangle. $P$ is a point inside its circumcircle. The rays $AP$ , $BP$ , $CP$ intersect the circle again at $D$ , $E$ , $F$ . Find $P$ so that $DEF$ is equilateral.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

$PAB$ and $PED$ are similar, so $DE / AB = PD / PB$ . Similarly, $DF / AC = PD / PC$ , so $DE / DF = (AB / AC)(PC / PB)$ . Thus we need $PB / PC = AB / AC$ . So $P$ must lie on the circle of Apollonius, which is the circle we constructed with center $X$ . Similarly, it must lie on the circle of Apollonius with center $Y$ and hence be one of their points of intersection. It also lies on the third circle and hence we choose the point of intersection inside the triangle.

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