Let $AA_1, BB_1, CC_1$ be the heights of triangle $ABC$ and $H$ be its orthocenter. Liune $\ell$ parallel to $AC$, intersects straight lines $AA_1$ and $CC_1$ at points $A_2$ and $C_2$, respectively. Suppose that point $B_1$ lies outside the circumscribed circle of triangle $A_2 HC_2$. Let $B_1P$ and $B_1T$ be tangent to of this circle. Prove that points $A_1, C_1, P$, and $T$ are cyclic.
Problem
Source: Ukraine TST 2018 p9
Tags: geometry, circumcircle, Concyclic, orthocenter, altitudes
Limerent
11.12.2020 23:27
Let $K=(BA_1HC_1)\cap(A_2C_2H)$; $A', C' -$ reflections of $A_1, C_1$ wrt. $BH$ respectively. $$\angle A_1C_1H=\angle A_1BB_1=\angle A_1AB_1=\angle A_1A_2B_2 \implies (A_1C_1C_2A_2)$$By radical axis we get that $A_1C_1, A_2C_2, KH$ intersect at one point $S$. In order to finish by radical axis it is sufficient to prove that $S$ lies on the polar of $B_1$ or by La Hire that $B_1$ lies on the polar of $S$. Let $X=C_2H\cap A_1A_2, Z=C_2K\cap A_2H$. By Brokard's theorem we need to prove that $B_1, X, Z$ are collinear. Now note that $H$ is the center of spiral similarity sending $C'A'$ to $A_2C_2 \implies A_2, C', K$ are collinear and $C_2, A', K$ are collinear. Finish by Pascal on $A_1HC_1A'KC'$. 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