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Mongolian National Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Mongolia

Two circles $\omega_1$ and $\omega_2$ intersected at points $A$ and $B$ . Line through $B$ is intersect the $\omega_1$ at point $C$ and intersect the $\omega_2$ at point $D$ . The line $AD$ intersect the $\omega_1$ at point $E$ different from $A$ and the line $AC$ intersect the $\omega_2$ at point $F$ different from $A$ . If $O$ is circumcenter of triangle $AEF$ then prove that $OB \perp CD$ .

(Proposed by B. Bat-Od)

Solution

We draw the circumcircle of $AEF$ , and denote $\angle EAC = \alpha$ . So $\angle EAF = 180^\circ - \alpha$ and from here we get $\angle EOF = 2\alpha$ . Other hand $\alpha = \angle EBC = \angle EAC = \angle FAD = \angle FBD$ , so $\angle EBF = 180^\circ - 2\alpha$ . From here we have $\angle EOF + \angle EBF = 180^\circ$ , hence $EOFB$ is cyclic. Therefore, from $EO = FO$ we have $\angle EBO = \angle OBF$ . From here we have $\angle OBC = \angle CBE + \angle EBO = \angle OBF + \angle FBD = \angle OBD$ . We know $\angle OBC + \angle OBD = 180^\circ$ so $OB \perp CD$ .

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