IMO 2004 SL G3

Origin: KOR | Category: Geometry

Problem

Let O be the circumcenter of an acute-angled triangle ABC with \angleB < \angleC. The line AO meets the side BC at D. The circumcenters of the triangles ABD and ACD are E and F, respectively. Extend the sides BA and CA beyond A, and choose on the respective extension points G and H such that AG = AC and AH = AB. Prove that the quadrilateral EFGH is a rectangle if and only if \angleACB −\angleABC = 60◦.

Solution

It is important to note that since \beta < \gamma, \angleADC = 90◦−\gamma + \beta is acute. It is elementary that \angleCAO = 90◦−\beta. Let X and Y respectively be the intersections of FE and GH with AD. We trivially get X \inEF \perpAD and \triangleAGH ∼= \triangleACB. Consequently, \angleGAY = \angleOAB = 90◦−\gamma = 90◦−\angleAGY . Hence, GH \perpAD and thus GH \parallelFE. That EFGH is a rectangle is now equivalent to FX = GY and EX = HY . We have that GY = AG sin \gamma = AC sin \gamma and FX = AF sin \gamma (since \angleAFX = \gamma). Thus, FX = GY ⇔CF = AF = AC ⇔\angleAFC = 60◦⇔\angleADC = 30◦. Since \angleADC = 180◦−\angleDCA −\angleDAC = 180◦−\gamma −(90◦−\beta), it immediately follows that FX = GY ⇔\gamma −\beta = 60◦. We similarly obtain EX = HY ⇔\gamma −\beta = 60◦, proving the statement of the problem.