**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.