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Bulgaria competition problems

Geometry Difficulty 4.9 AIME Prove it Bulgaria

Let $O$ be an interior point of an acute-angled $\triangle ABC$ . Denote by $A_1$ , $B_1$ and $C_1$ its projection on the sides $BC$ , $AC$ and $AB$ . Let $P$ be the intersecting point of the lines through $A$ and $B$ , orthogonal to $B_1C_1$ and $A_1C_1$ , respectively. If $H$ is the projection of $P$ on $AB$ , prove that the points $A_1$ , $B_1$ , $C_1$ and $H$ are con-cyclic.

Solution

Since $\not\asymp PBC_1 = \not\asymp OC_1A_1$ and $\not\asymp OC_1A_1 = \not\asymp OBA_1$ (why?), then $\not\asymp PBA = \not\asymp OBC$ . Analogously $\not\asymp PAB = \not\asymp OAC$ . Denote by $M$ and $N$ the projections of $P$ on $BC$ and $AC$ . Since
$BC_1 \cdot BH = BO \cdot BP \cos \angle OBA \cos \angle PBA = BA_1 \cdot BM,$
then the points $H$ , $C_1$ , $A_1$ and $M$ lie on a circle with center at the midpoint $Q$ of $OP$ (since the bisectors of $HC_1$ and $MA_1$ pass through $Q$ ). We get in the same way that the points $H$ , $C_1$ , $N$ and $B_1$ lie on a circle with center $Q$ . Hence the points $A_1$ , $B_1$ , $C_1$ , $H$ , $M$ and $N$ are con-cyclic.

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