Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National olympiad Prove it Saudi Arabia

Let $(O)$ be a circle, and $BC$ be a chord of $(O)$ such that $BC$ is not a diameter. Let $A$ be a point on the larger arc $BC$ of $(O)$ , $E$ and $F$ be two feet of the perpendiculars from $B$ and $C$ to $AC$ , $AB$ respectively.

1. Prove that two tangents of $(AEF)$ at $E$ and $F$ intersect to each other at a fixed point $M$ when $A$ moves on the larger arc $BC$ of $(O)$ .

2. Let $T$ be the intersection of $EF$ and $BC$ , $H$ be the orthocenter of $ABC$ . Prove that $TH$ is perpendicular to $AM$ .

Solution

Denote $M$ as the midpoint of $BC$ . Since $\angle AEH = \angle AFH = 90^\circ$ then $A, H, E, F$ belong to the same circle and the center of this circle is midpoint $N$ of $AH$ . It is easy to check that $NE = NH$ and $ME = MB$ , so
$\angle MEN = \angle MEB + \angle NEB = \angle MBE + \angle NHE = \angle EAH + \angle AHE = 90^\circ.$
Then $ME$ is the tangent line of $(AEF)$ at $E$ .

By same way, we have $FM$ is the tangent line of $(AEF)$ at $F$ . These imply that the tangent lines of $(AEF)$ at $E, F$ meet at the fixed point $M$ .

2) Consider the diameter $AA'$ of $(O)$ then it is easy to see that $BHC A'$ is a parallelogram and $H, M, A'$ are collinear.

Suppose that $MH \cap (O) = D \neq A'$ then $\angle ADH = \angle ADA' = 90^\circ$ which means $D \in (AEF)$ .

By consider three radical axis of three circles $(O)$ , $(BFEC)$ , $(AEHF)$ , we have $AD, EF, BC$ are concurrent at $T$ , which is the radical center. In triangle $ATM$ , we have $AH \perp TM$ and $MH \perp AT$ , then $H$ is the orthocenter. From this, we can conclude that $TH \perp AM$ . $\square$

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