Triangles with Equal Area IV

Here is Problem 4 from the IMO 2007:

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

• R_a, R_b be the orthogonal projections of R on CA, CB, respectively,
• Q_a, Q_b be the orthogonal projections of Q on RR_a, RR_b, respectively,
• P_a, P_b be the orthogonal projections of P on RR_a, RR_b, respectively,
• O' be the orthogonal projections of O on CR.

ΔPOQ is isosceles because ∠OQP = ∠OPQ = 90° - ∠ACB /2 so PO' = QO', moreover O' is midpoint of chord CR, therefore QC = PR. It makes two right triangles CLQ and PP_aR equal implying QL = RP_a. As constructed QL||RP_a therefore QLP_aR is parallelogram. Similarly PKQ_bR is also parallelogram.

The rest is a proof without words:

of which the only one that may require an explanation is Area(ΔQP_aR) = Area(ΔPQ_aR) which is true because trianles PQP_a and PQ_aP_a have the same base (PP_a) 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