Triangles with Equal Area VII

Here is Problem 4 from the IMO 2007:

The solution below is by Vo Duc Dien.

From R draw two lines perpendicular to BC and AC meeting them at T and S, respectively. Also, let the perpendicular from P to AC meet AC at F.

To prove that the two triangles RPK and RQL have the same area suffice it to show that

Since CR is the bisector of ∠ACB, ΔKPC = ΔFPC and, similarly, ΔTRC = ΔSRC. Thus TK = SF and PK = PF, implying that (1) is equivalent to PF×SF = SL×QL, or

But SL / SF = RQ / RP (since the three lines RS, QL, and PF are parallel), so that (2) is equivalent to PF / QL = RQ / RP, or

Also note that QL / PF = QC / PC, implying that (3) is equivalent to

Now, QC = QP + PC and RP = RQ + QP which reduces (4) to 1 + QP / PC = 1 + QP / RQ, or simply to

Thus (5) is equivalent to (1) such that proving (5) solves the problem.

Note that O is the circumcenter of ΔABC. Let M be the foot of the perpendicular from O to RC.

Angles MOP and PCK that have perpendicular sides are equal, and for the same reason ∠MOQ = ∠PCF. Since ∠PCK = ∠PCF, angles MOP and MOQ are equal, making triangles MOP and MOQ congruent, wherefrom

(7)

OR = OC

(since R is on the circumcircle of ΔABC with the circumcenter O). This makes ΔCOR isosceles. In particular, ∠ROM = ∠COM and then also

From (6)-(8), ΔROQ = ΔCOP and, finally, PC = RQ which is exactly (5), the condition we set out to prove.

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

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