Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Prove it Taiwan

Let the orthocenter of triangle $ABC$ be $H$ , and let its circumcircle be $\Gamma$ . Take a point $P$ on $\Gamma$ different from $A$ , $B$ , $C$ , and let $M$ be the midpoint of segment $HP$ . Take points $D$ , $E$ , $F$ on lines $BC$ , $CA$ , $AB$ respectively such that $AP \parallel HD$ , $BP \parallel HE$ , $CP \parallel HF$ . Prove that: $D$ , $E$ , $F$ , $M$ are collinear.

Solution

(∠ denotes directed angles.)

Let $A'$ , $P'$ be the antipodal points of $A$ , $P$ with respect to $\Gamma$ respectively, let $M_a$ , $M'$ be the midpoints of $\overline{HA'}$ , $\overline{HP'}$ respectively, and let $\Omega$ be the nine-point circle of $\triangle ABC$ . Then $M$ , $M_a$ , $M'$ lie on $\Omega$ . Let $M'H$ meet $\Omega$ again at a point $X$ , and let $D'$ be the intersection of $AH$ and $BC$ . Then
$\angle XD'D = \angle XD'M_a = \angle XM'M_a = \angle HM'M_a.$

$\angle XHD = \angle (HM', AP) = \angle (HM', A'P') = \angle HM'M_a.$

Therefore $D$ , $D'$ , $H$ , $X$ are concyclic, i.e., $HP' \perp XD$ . Note that $\overline{MM'}$ is a diameter of $\Omega$ , so
$\angle (HP', XM) = \angle M'XM = 90^\circ,$
hence $X$ , $M$ , $D$ are collinear and $HP' \perp DM$ . By the same reasoning, $HP' \perp EM$ , $HP' \perp FM$ , and therefore $D$ , $E$ , $F$ , $M$ are collinear. This completes the proof.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.