Given an acute $\vartriangle ABC$ whose orthocenter is denoted by $H$. A line $\ell$ passes $H$ and intersects $AB,AC$ at $P ,Q$ such that $H$ is the mid-point of $P,Q$. Assume the other intersection of the circumcircle of $\vartriangle ABC$ with the circumcircle of $\vartriangle APQ$ is $X$. Let $C'$ is the symmetric point of $C$ with respect to $X$ and $Y$ is the another intersection of the circumcircle of $\vartriangle ABC$ and $AO$, where O is the circumcenter of $\vartriangle APQ$. Show that $CY$ is tangent to circumcircle of $\vartriangle BCC'$.
Problem
Source: https://artofproblemsolving.com/community/c6h1740825p11314688
Tags: geometry, tangent, circumcircle, orthocenter
parmenides51
22.03.2020 14:48
Let $M$ be the midpoint of $BC$, let $D, E, F$ be the feet of the altitudes, let $Z$ be the second meeting of $(AH)$ and $(ABC)$. It is well-known by orthocenter reflection or something that $M, H, Z$ collinear, and by radical axis $N = AZ \cap EF \cap BC$.
First claim: $PQ \perp MH$
Proof: Maybe other ways to do this but projective is fast Let $R = AN \cap PQ$. Then $-1=(N, D; C, B) \stackrel{A}{=} (R, H; Q, P)$, but since $H$ is the midpoint of $PQ$ it follows that $R$ is the point at infinity, so $PQ \parallel AN \perp MH$. $\Box$
Note that $X$ is the spiral center that takes $PHQ \to BMC$. In particular, we have $\angle XCQ = \angle XMH = \angle XBP$. Extend $XM$ to meet $(ABC)$ again at $Y'$. But $\angle XY'A = \angle XCA = \angle XMH$, so $AY' \parallel MH \perp PQ$. It follows that $Y, Y'$ are isogonal wrt $\angle BAC$, so $\angle YCB = \angle CXY'$. But by homothety $\angle CXY'=\angle CC'B$, so the tangency follows. $\blacksquare$
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