Solution IV by Bui Quang Tuan

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

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:

Area( ΔQLR) = Area( ΔQP_aR) = Area( ΔPQ_aR) = Area( ΔPQ_bR) = Area( ΔPKR),

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

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