Triangles with Equal Area IV
Here is Problem 4 from the IMO 2007:
|In triangle ABC the bisector of angle BCA intersects the circumcircle again at R, the perpendicular bisector of BC at P, and the perpendicular bisector of AC at Q. The midpoint of BC is K and the midpoint of AC is L. Prove that the triangles RPK and RQL have the same area.|
The solution by Bui Quang Tuan is the nearest to the manner which is, to my taste, the most natural to treat this problem. Still, it requires additional construction. Let
- Ra, Rb be the orthogonal projections of R on CA, CB, respectively,
- Qa, Qb be the orthogonal projections of Q on RRa, RRb, respectively,
- Pa, Pb be the orthogonal projections of P on RRa, RRb, respectively,
- O' be the orthogonal projections of O on CR.
ΔPOQ is isosceles because ∠OQP = ∠OPQ = 90° - ∠ACB /2 so
The rest is a proof without words:
|Area( ΔQLR) = Area( ΔQPaR) = Area( ΔPQaR) = Area( ΔPQbR) = Area( ΔPKR),|
of which the only one that may require an explanation is Area( ΔQPaR) = Area( ΔPQaR) which is true because trianles PQPa and PQaPa have the same base (PPa) and lie between the same parallel lines and thus have the same area.
Note: also observes that the proof goes through for any P and Q on CR that are reflections of each other in the midpoint of CR.
2007 IMO, Problem 4
- Triangles with Equal Area
- Solution II by A. Batominovski
- Solution III by A. Bogomolny
- Solution IV by Bui Quang Tuan
- Solution V by Bui Quang Tuan
- Solution VI by Tom Verhoeff
- Solution VII by Vo Duc Dien