SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Question

In triangle ABC, such that ACB = 45^, let O and H be the circumcenter and orthocenter, respectively. The line passing through O and perpendicular to CO intersects AC and BC at K and L, respectively. Prove that the perimeter of KLH is equal to the diameter of the circumcircle of triangle ABC.

Accepted Answer

Suppose that AH, BH cut (O) at the second points E, F. By angle chasing, we can see that H, E are symmetric with respect to BC. Note that HBC = 90^ - C = 45^ so CBE = CBH = 45^. Thus CFE = 45^, which implies that COE = 90^. Similarly, COF = 90^, then four points E, L, K, F are collinear. From this, we conclude that LH = LE and KH = KF which mean aligned perimeter(HKL) & = HL + LK + KH & = EL + LK + KF = EF = diameter(O). aligned