IMO 1995 SL G8

Origin: COL | Category: Geometry

Problem

Let ABC be a triangle. A circle passing through B and C in- tersects the sides AB and AC again at C′ and B′, respectively. Prove that BB′, CC′, and HH′ are concurrent, where H and H′ are the orthocenters of triangles ABC and AB′C′ respectively.

Solution

Let BB′ cut CC′ at P. Since \angleB′BC′ = \angleB′CC′, it follows that \anglePBH = \anglePCH. Let D and E be points such that BPCD and HPCE are parallelograms (consequently, so is BHED). Triangles BAC and C′AB′ are similar, from which we deduce that \triangleB′H′C′ and \triangleBHC are similar, as well as \triangleB′PC′ and \triangleBDC. Hence B′PC′H′ and BDCH are similar, from which we obtain \angleH′PB′ = \angleHDB. Now \angleCDE = \anglePBH = \anglePCH = \angleCHE implies that HCED is a cyclic quadrilateral. Therefore \angleBPH = \angleDCE = \angleDHE = \angleHDB = \angleH′PB′; hence HH′ also passes through P. B A C C′ B′ H H′ P D E Second solution. Observe that \triangleHBC ∼\triangleH′B′C′, \anglePBH = \anglePCH and \anglePB′H′ = \anglePC′H′. By Ceva’s theorem in trigonometric form applied to \triangleBPC and the point H, we have sin \angleBPH sin \angleHPC = sin \angleHBP sin \angleHBC \cdot sin \angleHCB sin \angleHCP = sin \angleHCB sin \angleHBC . Similarly, Ceva’s theorem for \triangleB′PC′ and point H′ yields sin \angleB′PH′ sin \angleH′PC′ = sin \angleH′C′B′ sin \angleH′B′C′ . Thus it follows that sin \angleB′PH′ sin \angleH′PC′ = sin \angleBPH sin \angleHPC , which finally implies that \angleBPH = \angleB′PH′.