2018 USA ELMO Shortlist #G2

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Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be the midpoint of $\overline{AH}$ and let $T$ be on line $BC$ with $\angle TAO=90^{\circ}$. Let $X$ be the foot of the altitude from $O$ onto line $PT$. Prove that the midpoint of $\overline{PX}$ lies on the nine-point circle* of $\triangle ABC$.

*The nine-point circle of $\triangle ABC$ is the unique circle passing through the following nine points: the midpoint of the sides, the feet of the altitudes, and the midpoints of $\overline{AH}$, $\overline{BH}$, and $\overline{CH}$




Let $M$ be the midpoint of $BC$ and $\Gamma$ be the circle that has diameter $BC$. Because $$AP^2=OM^2=OC^2-MC^2=PM^2-MC^2,$$ we know that $P$ lies on the radical axis of $A, \Gamma$. As $T$ lies on that radical axis as well, we have $TP \perp AM$, which means $TP \cup AM=N$ lies on the nine-point circle. But $OX || AM$, so $MN$ is the $P$-midline in $\triangle{XPO}$, as desired. $\square$



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