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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.1 AIME, harder Prove it Saudi Arabia

Let $\triangle ABC$ be an acute triangle, with $\widehat{A} > \widehat{B} \geq \widehat{C}$ . Let $D$ , $E$ and $F$ be the tangency points between the incircle of the triangle and sides $BC$ , $CA$ , $AB$ , respectively. Let $J$ be a point on $(BD)$ , $K$ a point on $(DC)$ , $L$ a point on $(EC)$ and $M$ a point on $(FB)$ , such that
$AF = FM = JD = DK = LE = EA$
Let $P$ be the intersection point between $AJ$ and $KM$ and let $Q$ be the intersection point between $AK$ and $JL$ . Prove that $PJKQ$ is cyclic.

Solution

Let $I$ be the incenter of triangle $ABC$ . Because $ID = IE = IF$ , $AF = FM = JD = DK = LE = EA$ , and the angles at $D$ , $E$ , and $F$ are right, we have by Pythagoras $IA = IM = IJ = IK = IL$ . Therefore, the pentagon $AMJKL$ is cyclic.

On the other hand, $AM = 2AF = 2EA = LA$ . We deduce for the circumcircle of the pentagon $AMJKL$ that $\text{AM = LA}$ . This implies that $\measuredangle QKP = \measuredangle QJP$ , and therefore the quadrilateral $PJKQ$ is cyclic.

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