Problem 4 of the Austrian Mathematical Olympiad 2009

Let D, E and F be the midpoints of the sides of the triangle ABC (D on BC, E on CA and F on AB). Further let HaHbHc be the triangle formed by the base points of the altitudes of the triangle ABC. Let P, Q and R be the midpoints of the sides of the triangle HaHbHc (P on HbHc, Q on HcHa and R on HaHb).

Show: The lines PD, QE and RF share a common point.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to poet Thu Hong of Ft. Collins, Colorado)

Since BHcC and BHbC are right triangles, BHcHbC is cyclic, and with D being the midpoint of diameter BC, DHc = DHb. Combining with P being the midpoint of HbHc, we have DP ┴ HbHc.

Similarly, AHcHaC is cyclic, EHc = EHa and EQ ┴ HaHc. Also FR ┴ HaHb.

Let I be the intersection of DP and EQ. Since I is on DP and EQ and DP ┴ HbHc, EQ ┴ HaHc, we have IHb = IHc and IHa = IHc or IHa = IHb or IR ┴ HaHb since R is also the midpoint of HaHb.

Combining with FR ┴ HaHb, the three points F, I and R are collinear, or the lines PD, QE and RF share a common point I.