IMO 2000 SL G3

Origin: IND | Category: Geometry

Problem

Let O be the circumcenter and H the orthocenter of an acute triangle ABC. Show that there exist points D, E, and F on sides BC, CA, and AB respectively such that OD + DH = OE + EH = OF + FH and the lines AD, BE, and CF are concurrent.

Solution

Let L be the point symmetric to H with respect to BC. It is well known that L lies on the circumcircle k of \triangleABC. Let D be the intersection point of OL and BC. We similarly define E and F. Then OD + DH = OD + DL = OL = OE + EH = OF + FH.

We shall prove that AD, BE, and CF are concurrent. Let line AO meet BC at D′. It is easy to see that \angleOD′D = \angleODD′; hence the per- pendicular bisector of BC bisects DD′ as well. Hence BD = CD′. If we define E′ and F ′ analogously, we have CE = AE′ and AF = BF ′. Since the lines AD′, BE′, CF ′ meet at O, it follows that BD DC \cdot CE EA \cdot AF F B = B C A O H L D D′ E E′ F F ′ BD′ D′C \cdot CE′ E′A \cdot AF ′ F ′B = 1. This proves our claim by Ceva’s theorem.