IMO 2003 SL G1

Origin: FIN | Category: Geometry

Problem

Let ABCD be a cyclic quadrilateral. Let P, Q, R be the feet of the perpendiculars from D to the lines BC, CA, AB, respectively. Show that PQ = QR if and only if the bisectors of \angleABC and \angleADC are concurrent with AC.

Solution

Denote by K and L the intersec- tions of the bisectors of \angleABC and \angleADC with the line AC, respec- tively. Since AB : BC = AK : KC and AD : DC = AL : LC, we have to prove that PQ = QR ⇔AB BC = AD DC . (1) Since the quadrilaterals AQDR and QPCD are cyclic, we see that A B C D P Q R \angleRDQ = \angleBAC and \angleQDP = \angleACB. By the law of sines it fol- lows that AB BC = sin(\angleACB) sin(\angleBAC) and that QR = AD sin(\angleRDQ), QP = CD sin(\angleQDP). Now we have

AB BC = sin(\angleACB) sin(\angleBAC) = sin(\angleQDP) sin(\angleRDQ) = AD \cdot QP QR \cdot CD. The statement (1) follows directly.