Problem 1 of Asian Pacific Mathematical Olympiad 2010

Let ABC be a triangle with ∠BAC ≠ 90. Let O be the circumcenter of the triangle ABC and let Г be the circumcircle of the triangle BOC. Suppose that Г intersects the line segment AB at P different from B, and the line segment AC at Q different from C. Let ON be a diameter of the circle Г. Prove that the quadrilateral APNQ is a parallelogram.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (tặng bạn học cũ Nguyễn Hựu Trung Học Tỉnh Hạt Thuận An)

Let the circumcircle of triangle ABC be C, M be the midpoint of AC and E the intersection of C and PC. Also let r and R be the radii of C and Г, respectively.

Consider two right triangles MOC and QON with ∠ONQ = ∠OCQ (subtends arc OQ). So they are similar; therefore, MC/OC = QN/ON or MC/r = QN/2R or AC/QN = r/R or ∠ABC = ∠NPQ.

We also have ∠PQN = ∠PBN (subtends PN) = 180 - ∠OBN - ∠ABO = 90 - ∠ABO

But ∠ABO = ∠BAO, ∠OBC = ∠OCB and ∠OAC = ∠OCA or

	∠OCB + ∠OCA = ∠ACB =  90 - ∠ABO = ∠PQN

Now ∠BAC = 180 - ∠ABC - ∠ACB = 180 - ∠NPQ - ∠PQN = ∠PNQ

We also have ∠PNQ = ∠PCQ = ∠BAC or AP = PC and BE || AC and since BN is tangent to C, ∠NBE = ∠BCE = ∠BNP (subtends BP) or BE || PN

Along with BE || AC, we have PN || AC.

Now combine with ∠BAC = ∠PNQ, we conclude that APNQ is a parallelogram.