IMO 1991 SL 2

Origin: JAP

Problem

For an acute triangle ABC, M is the midpoint of the segment BC, P is a point on the segment AM such that PM = BM, H is the foot of the perpendicular line from P to BC, Q is the point of intersection of segment AB and the line passing through H that is perpendicular to PB, and finally, R is the point of intersection of the segment AC and the line passing through H that is perpendicular to PC. Show that the circumcircle of \triangleQHR is tangent to the side BC at point H.

Solution

Let HQ meet PB at Q′ and HR meet PC at R′. From MP = MB = MC we have \angleBPC = 90o. So PR′HQ′ is a rectangle. Since PH is perpen- dicular to BC, it follows that the circle with diameter PH, through P, R′, H, Q′, is tangent to BC. It is now sufficient to show that QR is parallel to Q′R′. Let CP meet AB at X, and BP meet AC at Y . Since P is on the median, it follows (for example, by Ceva’s theorem) that A B C P M X Q Y R Q′ R′ H AX/XB = AY/Y C, i.e. that XY is parallel to BC. Consequently, PY/BP = PX/CP. Since HQ is parallel to CX, we have QQ′/HQ′ = PX/CP and similarly RR′/HR′ = PY/BP. It follows that QQ′/HQ′ = RR′/HR′, hence QR is parallel to Q′R′ as required. Second solution. It suffices to show that \angleRHC = \angleRQH, or equivalently RH : QH = PC : PB. We assume PC : PB = 1 : x. Let X \inAB and Y \inAC be points such that MX \perpPB and MY \perpPC. Since MX bisects \angleAMB and MY bisects AMC, we deduce AX : XB = AM : MB = AY : Y C ⇒XY \parallelBC ⇒ ⇒\triangleXY M ∼\triangleCBP ⇒XM : MY = 1 : x. Now from CH : HB = 1 : x2 we obtain RH : MY = CH : CM = 1 : 1+x2 and QH : MX = BH : BM = x2 : 1+x2 . Therefore RH : QH = 1 + x2 MY : 2x2 1 + x2 MX = 1 : x.