1985 IMO Problems/Problem 5

Problem

A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$ ). Prove that $\angle OMB = 90^{\circ}$ .

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. $M$ is the Miquel Point of quadrilateral $ACNK$ , so there is a spiral similarity centered at $M$ that takes $KA$ to $NC$ . Let $M_1$ be the midpoint of $KA$ and $M_2$ be the midpoint of $NC$ . Thus the spiral similarity must also send $M_1$ to $M_2$ and must also be the center of another spiral similarity that sends $KN$ to $M_1 M_2$ , so $BMM_1 M_2$ is cyclic. $OM_1 B M_2$ is also cyclic with diameter $BO$ and thus $M$ must lie on the same circumcircle as $B$ , $M_1$ , and $M_2$ so $\angle OMB = 90^{\circ}$ .

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1985 IMO Problems/Problem 5

Problem

Solution