Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Prove it Belarus

In an acute-angled triangle $ABC$ the orthocenter is $H$ . $I_H$ is the incenter of $\triangle BHC$ . The bisector of $\angle BAC$ intersects the perpendicular from $I_H$ to the side $BC$ at point $K$ . Let $F$ be the foot of the perpendicular from $K$ to $AB$ .

Prove that $2KF + BC = BH + HC$ .

(A. Voidelevich)

Solution

First, easy counting of angles shows that $AK \parallel HI_H$ . By condition, $KI_H \parallel AH$ , so we conclude that $AKIH_H$ is a parallelogram. Hence $AK = HI_H$ . Let $T$ be the tangency point of the incircle of $\triangle BHC$ with the side $BH$ , then $I_HT \perp HT$ ; thus from the above $\triangle EKA = \triangle THI_H$ . So $EK = TH$ . By the well-known formula $TH = 0.5(BH + HC - BC)$ whence the required equality follows.

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