IMO 1994 SL G4

Origin: AUS-ARM | Category: Geometry

Problem

N is an arbitrary point on the bisector of \angleBAC. P and O are points on the lines AB and AN, respectively, such that ∡ANP = 90◦= ∡APO. Q is an arbitrary point on NP, and an arbitrary line through Q meets the lines AB and AC at E and F respectively. Prove that ∡OQE = 90◦if and only if QE = QF.

Solution

First, assume that \angleOQE = 90◦. Extend PN to meet AC at R. Then OEPQ and ORFQ are cyclic quadrilaterals; hence we have \angleOEQ = \angleOPQ = \angleORQ = \angleOFQ. It follows that \triangleOEQ ∼= \triangleOFQ and QE = QF. Now suppose QE = QF. Let S be the point symmetric to A with respect to Q, so that the quadri- lateral AESF is a parallelogram. Draw the line E′F ′ through Q so that \angleOQE′ = 90◦and E′ \inAB, F ′ \inAC. By the first part QE′ = QF ′; hence AE′SF ′ is also a paral- A B C N P R O Q E F S lelogram. It follows that E \equivE′, F \equivF ′, and \angleOQE = 90◦.