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42nd International Mathematical Olympiad

IMO 2001 Problems and Solutions

Washinton, DC, USA
July 1-14, 2001

Problem 5

In a triangle ABC, let AP bisect BAC, with P on BC, and let BQ bisect ABC, with Q on CA.

It is known that BAC = 60° and that AB + BP = AQ + QB.

What are the possible angles of triangle ABC?


Denote the angles of ABC by = 60°, , and . Extend AB to P' so that BP' = BP, and construct P'' on AQ so that AP'' = AP'. Then BP'P is an isosceles triangle with base angle /2. Since AQ + QP'' = AB + BP' = AB + BP = AQ + QB, it follows that QP'' = QB. Since AP'P'' is equilateral and AP bisects the angle at A, we have PP' = PP''.

Claim. Points B, P, P'' are collinear, so P'' coincides with C.

Proof. Suppose to the contrary that BPP'' is a nondegenerate triangle. We have that PBQ = PP'B = P P''Q = /2. Thus the diagram appears as below, or else with P is on the other side of BP''. In either case, the assumption that BPP'' is nondegenerate leads to BP = PP'' = PP', thus to the conclusion that BPP' is equilateral, and finally to the absurdity /2 = 60° so + = 60° + 120° = 180°.

Thus points B, P, P'' are collinear, and P'' = C as claimed.

Since triangle BCQ is isosceles, we have 120° - = = /2, so = 80 and = 40°. Thus ABC is a 60-80-40 degree triangle.