IMO 1967 LL POL41

Origin: POL

Problem

A line l is drawn through the intersection point H of the altitudes of an acute-angled triangle. Prove that the symmetric images la, lb, lc of l with respect to sides BC, CA, AB have one point in common, which lies on the circumcircle of ABC.

Solution

It is well known that the points K, L, M, symmetric to H with respect to BC, CA, AB respectively, lie on the circumcircle k of the triangle ABC. For K, this follows from an elementary calculation of angles of triangles HBC and noting that ∡KBC = ∡HBC = ∡KAC. For other points the proof is analogous. Since the lines la, lb pass through K and L and lb is obtained from la by rotation about C for an angle 2\gamma = \angleLCK, it fol- lows that the intersection point P of la and lb is at the circumcircle of KLC, that is, k. Similarly, lb and lc meet at a point on k; hence they must pass through the same point P. A B C K M L H lc la lb P l