IMO 1993 SL 22

Origin: GBR

Problem

A, B, C, D are four points in the plane, with C, D on the same side of the line AB, such that AC \cdot BD = AD \cdot BC and ∡ADB = 90◦+ ∡ACB. Find the ratio AB \cdot CD AC \cdot BD, and prove that circles ACD, BCD are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicular.)

Solution

Let M be the point inside \angleADB that satisfies DM = DB and DM \perp

DB. Then \angleADM = \angleACB and AD/DM

AC/CB. It follows that the triangles ADM, ACB are similar; hence \angleCAD = \angleBAM (because \angleCAB = \angleDAM) and AB/AM = AC/AD. Consequently the triangles CAD, BAM are sim- ilar and therefore AC AB

CD BM

CD \sqrt 2BD. Hence AB\cdotCD AC\cdotBD = \sqrt 2. A B C D M T U Let CT, CU be the tangents at C to the circles ACD, BCD respectively. Then (in oriented angles) \angleTCU = \angleTCD+\angleDCU = \angleCAD+\angleCBD = 90◦, as required. Second solution to the first part. Denote by E, F, G the feet of the per- pendiculars from D to BC, CA, AB. Consider the pedal triangle EFG. Since FG = AD sin \angleA, from the sine theorem we have FG : GE : EF = (CD \cdot AB) : (BD \cdot AC) : (AD \cdot BC). Thus EG = FG. On the other hand, \angleEGF = \angleEGD + \angleDGF = \angleCBD + \angleCAD = 90◦implies that EF : EG = \sqrt 2 : 1; hence the required ratio is \sqrt 2. Third solution to the first part. Under inversion centered at C and with power r2 = CA\cdotCB, the triangle DAB maps into a right-angled isosceles triangle D∗A∗B∗, where D∗A∗= AD \cdot BC CD , D∗B∗= AC \cdot BD CD , A∗B∗= AB \cdot CD CD . Thus D∗B∗: A∗B∗= \sqrt 2, and this is the required ratio.