IMO 1975 SL 8

Origin: NET

Problem

On the sides of an arbitrary triangle ABC, triangles BPC, CQA, and ARB are externally erected such that ∡PBC = ∡CAQ = 45◦, ∡BCP = ∡QCA = 30◦, ∡ABR = ∡BAR = 15◦. Prove that ∡QRP = 90◦and QR = RP.

Solution

Let K and L be the feet of perpendiculars from P and Q to BC and AC respectively.

Let M, N be points on AB (or- dered A −N −M −B) such that RMN is a right isosceles triangle with \angleR = 90◦. By sine theorem we have BM BA = BM BR \cdot BR BA = sin 15◦ sin 45◦. Since BK BC = sin 45◦sin 30◦ cos 15◦ = sin 15◦ sin 45◦, we deduce that MK \parallelAC and MK = AL. Similarly, NL \parallelBC A B C K L M N P Q R and NL = BK. It follows that the vectors −−\to RN, −−\to NL, −\to LQ are the images of −−\to RM, −−\to KP, −−\to MK respectively under a rotation of 90◦, and consequently the same holds for their sums −−\to RQ and −\to RP. Therefore, QR = RP and \angleQRP = 90◦. Second solution. Let ABS be the equilateral triangle constructed in the exterior of \triangleABC. Obviously, the triangles BPC, BRS, ARS, AQC are similar. Let f be the rotational homothety centered at B that maps P onto C, and let g be the rotational homothety about A that maps C onto Q. The composition h = g ◦f is also a rotational homothety; its angle is \anglePBC +\angleCAQ = 90◦, and the coefficient is BC BP \cdot AQ AC = 1. Moreover, R is a fixed point of h because f(R) = S and g(S) = R. Hence R is the center of h, and the statement follows from h(P) = Q. Remark. There are two more possible approaches: One includes using complex numbers and the other one is mere calculating of RP, RQ, PQ by the cosine theorem. Second remark. The problem allows a generalization: Given that \angleCBP = \angleCAQ = \alpha, \angleBCP = \angleACQ = \beta, and \angleRAB = \angleRBA = 90◦−\alpha −\beta, show that RP = RQ and \anglePRQ = 2\alpha.