Altitudes $AA_1, BB_1, CC_1$ of acute triangle $ABC$ intersect at point $H$. On the tangent drawn from point $C$ to the circle $(AB_1C_1)$, the perpendicular $HQ$ is drawn (the point $Q$ lies inside the triangle $ABC$). Prove that the circle passing through the point $B_1$ and touching the line $AB$ at point $A$ is also tangent to line $A_1Q$.
Problem
Source: St Petersburg 2022 11.5
Tags: geometry
crazyeyemoody907
30.09.2022 08:47
What's $\overline{HQ}$ perpendicular to again? Also, (this is a statement) it should work for either tangent.
a_507_bc
30.09.2022 09:25
$HQ$ is perpendicular to the tangent from $C$ to $(AH)$ and $Q$ is on this tangent. Otherwise probably the condition $Q$ to be inside the triangle (and hence the tangent being inside the triangle) is superfluous.
NguyenHieu
03.10.2022 16:54
Let $CF$ be the tangent from $C$ to $(AH)$. Denote $(D)$ be the circle passing through $B_1,A$ and touching $AB$, $E=AF\cap (D)$. Notice that $Q,B_1,A_1,C,H$ all lie on $(CH)$, so $\angle FQB_1=180^\circ-\angle CQB_1=180^\circ-\angle B_1A_1C=180^\circ-\angle BAC$. Also, $\angle FEB_1=180^\circ-\angle AEB_1=\angle BAC$ (because $(D)$ touches $AB$). Thus, $(E,F,B_1,Q)$ are concyclic. By angle changing, we have $\angle QEB_1=\angle QFB_1=\angle FAB_1=\angle EAB_1$ or $EQ$ is the tangent of $(D)$. Now, we only need to prove that $A_1,Q,E$ are collinear. Let $AF\cap BC=G$. By angle chaging, $\angle ABC=\angle AB_1C_1=\angle AFC_1\implies (B,G,F,C_1)$ are concyclic. We have $\angle EQB_1=\angle EFB_1=\angle EFQ-\angle B_1FQ=\angle EFQ-\angle FAC=\angle AC_1F-\angle FAC=\angle AGB-\angle GAC=\angle ACB\implies \angle EQB_1+\angle A_1QB_1=180^\circ$ or $A_1,Q,E$ are collinear.
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