IMO 1979 SL 24

Origin: USA

Problem

A circle O with center O on base BC of an isosceles triangle ABC is tangent to the equal sides AB, AC. If point P on AB and point Q on AC are selected such that PB \times CQ = (BC/2)2, prove that line segment PQ is tangent to circle O, and prove the converse.

Solution

Clearly O is the midpoint of BC. Let M and N be the points of tangency of the circle with AB and AC, respectively, and let \angleBAC = 2ϕ. Then \angleBOM = \angleCON = ϕ. Let us assume that PQ touches the circle in X. If we set \anglePOM = \anglePOX = x and \angleQON = \angleQOX = y, then 2x + 2y = \angleMON = 180◦−2ϕ, i.e., y = 90◦−ϕ −x. It follows that \angleOQC = 180◦−\angleQOC − \angleOCQ = 180◦−(ϕ + y) −(90◦−ϕ) = 90◦−y = x + ϕ = \angleBOP. Hence the triangles BOP and CQO are similar, and consequently BP \cdot CQ = BO \cdot CO = (BC/2)2. Conversely, let BP \cdot CQ = (BC/2)2 and let Q′ be the point on (AC) such that PQ′ is tangent to the circle. Then BP \cdot CQ′ = (BC/2)2, which implies Q \equivQ′.