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Saudi Arabian Mathematical Competitions

Geometry Difficulty 6.4 National Olympiad Prove it Saudi Arabia

Given two circles $O_{1}$ and $O_{2}$ intersect at $A$ and $B$ . Let $d_{1}$ and $d_{2}$ be two lines through $A$ and be symmetric with respect to $AB$ . The line $d_{1}$ cuts $O_{1}$ and $O_{2}$ at $G$ , $E$ ( $E \neq A$ ), respectively; the line $d_{2}$ cuts $O_{1}$ and $O_{2}$ at $F$ , $H$ ( $F \neq A$ ), respectively; such that $E$ is between $A$ , $G$ and $F$ is between $A$ , $H$ .
Let $J$ be the intersection of $EH$ and $FG$ . The line $BJ$ cuts $O_{1}$ , $O_{2}$ at $K$ , $L$ ( $K, L \neq B$ ), respectively. Let $N$ be the intersection of $O_{1}K$ and $O_{2}L$ . Prove that the circle $(NLK)$ is tangent to $AB$ .

Solution

We have
$\begin{aligned} \angle AKO_{1} & = 90^{\circ} - \frac{1}{2} \angle AO_{1}K = 90^{\circ} - \angle ABK = 90^{\circ} - \angle ABL \\ & = 90^{\circ} - \frac{1}{2} \angle AO_{2}L = \angle ALO_{2} \end{aligned}$
Therefore $ALNK$ is a cyclic quadrilateral.

Let $O$ be the circumcenter of quadrilateral $ALNK$ . Since $B$ is the intersection of $(AGF)$ and $(AEH)$ then $B$ is the Miquel point of complete quadrilateral $AEJF.GH$ or $B \in (JFH)$ .

We obtain
$\text{BAO = BAK + KAO = BAF + FAK + 90 - ALK = 1 2 ( BF + FK ) + 90 - AHB = 1 2 ( BG + FK ) - GJB + 90 = 90 .}$
From this, $AB$ is a tangent of $(O)$ . We are done.

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