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Saudi Arabian Mathematical Competitions

Geometry Difficulty 4.8 AIME Prove it Saudi Arabia

In triangle $ABC$ , such that $\angle ACB = 45^{\circ}$ , let $O$ and $H$ be the circumcenter and orthocenter, respectively. The line passing through $O$ and perpendicular to $CO$ intersects $AC$ and $BC$ at $K$ and $L$ , respectively. Prove that the perimeter of $KLH$ is equal to the diameter of the circumcircle of triangle $ABC$ .

Solution

Suppose that $AH$ , $BH$ cut $(O)$ at the second points $E$ , $F$ . By angle chasing, we can see that $H$ , $E$ are symmetric with respect to $BC$ . Note that
$\angle HBC = 90^{\circ} - \angle C = 45^{\circ}$
so $\angle CBE = \angle CBH = 45^{\circ}$ .
Thus $\angle CFE = 45^{\circ}$ , which implies that $\angle COE = 90^{\circ}$ . Similarly, $\angle COF = 90^{\circ}$ , then four points $E$ , $L$ , $K$ , $F$ are collinear. From this, we conclude that $LH = LE$ and $KH = KF$ which mean
$\begin{aligned} \text{perimeter}(HKL) & = HL + LK + KH \\ & = EL + LK + KF = EF = \operatorname{diameter}(O). \end{aligned}$

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