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Saudi Arabian Mathematical Competitions

Geometry Difficulty 6.1 National Olympiad Prove it Saudi Arabia

Let $ABC$ be a triangle inscribed in circle $(O)$ , with its altitudes $BE$ , $CF$ intersecting at orthocenter $H$ ( $E \in AC$ , $F \in AB$ ). Let $M$ be the midpoint of $BC$ , $K$ be the orthogonal projection of $H$ on $AM$ . $EF$ intersects $BC$ at $P$ . Let $Q$ be the intersection of the tangent to $(O)$ which passes through $A$ with $BC$ , $T$ be the reflection of $Q$ through $P$ . Prove that $\angle OKT = 90^{\circ}$ .

Solution

Let $AD$ be the altitude of triangle $ABC$ , $AA'$ be the diameter of $(O)$ . $AO$ meets $EF$ at $N$ . Let $L$ be the reflection of $A$ through $N$ .

We have $AO \perp EF$ then quadrilateral $NECA'$ is cyclic. Then
$AO \cdot AL = \frac{1}{2} AA' \cdot 2AN = AA' \cdot AN = AE \cdot AC = AH \cdot AD = AK \cdot AM.$
From this we obtain $KOLM$ is cyclic.

On the other side, $AQ \parallel EF$ and $P$ is the midpoint of $QT$ then $AL TQ$ is a trapezoid with its midline $PN$ . We get
$\angle LTC = \angle EPC = \angle HAO = \angle MOL.$
Hence $OMTL$ is cyclic, which follows that $O, K, M, T, L$ are concyclic or $K$ lies on circle with diameter $OT$ .

In other words, $\angle OKT = 90^{\circ}$ . $\square$

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