IMO 1970 SL 6

Origin: FRA

Problem

In the triangle ABC let B′ and C′ be the midpoints of the sides AC and AB respectively and H the foot of the altitude passing through the vertex A. Prove that the circumcircles of the triangles AB′C′, BC′H, and B′CH have a common point I and that the line HI passes through the midpoint of the segment B′C′.

Solution

Let F be the midpoint of B′C′, A′ the midpoint of BC, and I the inter- section point of the line HF and the circle circumscribed about \triangleBHC′. Denote by M the intersection point of the line AA′ with the circum- scribed circle about the triangle ABC. Triangles HB′C′ and ABC are similar. Since \angleC′IF = \angleABC = \angleA′MC, \angleC′FI = \angleAA′B = \angleMA′C,

2C′F = C′B′, and 2A′C = CB, it follows that \triangleC′IB′ ∼\triangleCMB, hence \angleFIB′ = \angleA′MB = \angleACB. Now one concludes that I belongs to the circumscribed circles of \triangleAB′C′ (since \angleC′IB′ = 180◦−\angleC′AB′) and \triangleHCB′. Second Solution. We denote the angles of \triangleABC by \alpha, \beta, \gamma. Evidently \triangleABC ∼\triangleHC′B′. Within \triangleHC′B′ there exists a unique point I such that \angleHIB′ = 180◦−\gamma, \angleHIC′ = 180◦−\beta, and \angleC′IB′ = 180◦−\alpha, and all three circles must contain this point. Let HI and B′C′ intersect in F. It remains to show that FB′ = FC′. From \angleHIB′ + \angleHB′F = 180◦ we obtain \angleIHB′ = \angleIB′F. Similarly, \angleIHC′ = \angleIC′F. Thus circles around \triangleIHC′ and \triangleIHB′ are both tangent to B′C′, giving us FB′2 = FI \cdot FH = FC′2.