IMO 2009 Problem 2

Problem 2 of International Mathematical Olympiad 2009

Let ABC be a triangle with circumcentre O. The points P and Q are interior points of the sides CA and AB, respectively. Let K, L and M be the midpoints of the segments BP, CQ and PQ, respectively, and let Г be the circle passing through K, L and M. Suppose that the line PQ is tangent to the circle Г. Prove that OP = OQ.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to former manager Rick Frier) QP tangents with the small circle at M, we have ∠QMK = ∠MLK.

M, K and L are midpoints of PQ, BP and QC, respectively; therefore,

KM || QB, KM = ½ QB (*), ML || PC, ML = ½ PC (**)

and ∠QMK = ∠MQA, or ∠MLK = ∠MQA.

ML || PC and KM || QB therefore ∠QAP = ∠KML

The two triangles QAP and KML are similar since their respective angles are equal. Therefore, ML/QA = KM/AP

From (*) and (**) AP × PC = QA × QB (***)

Extend PQ and QP to meet the larger circle at U and V, respectively.

In the larger circle UV intercepts AB at Q, we have

QU × QV = QA × QB or QU × (QP + PV) = QA × QB (i)

UV intercepts AC at P, we have UP × PV = AP × PC or (QU + QP) × PV = AP x PC (ii)

From (i) and (***) QU × (QP + PV) = AP × PC

Therefore, from (ii) QU × (QP + PV) = (QU + QP) × PV

Or PV = QU and M is also the midpoint of UV and OM ⊥UV

Therefore OP = OQ