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Indian National Mathematical Olympiad, 2011

Geometry Difficulty 4.5 AIME Prove it India

Let $D$ , $E$ , $F$ be points on the sides $BC$ , $CA$ , $AB$ respectively of a triangle $ABC$ such that $BD = CE = AF$ and $\angle BDF = \angle CED = \angle AFE$ . Prove that $ABC$ is equilateral.

Solution

Consider the triangle $BDF$ , $CED$ and $AFE$ with $BD$ , $CE$ and $AF$ as bases. The sides $DF$ , $ED$ and $FE$ make equal angles $\theta$ with the bases of respective triangles. If $B \geq C \geq A$ , then it is easy to see that $FD \geq DE \geq EF$ . Now using the triangle $FDE$ , we see that $B \geq C \geq A$ gives $DE \geq EF \geq FD$ . Combining, you get $FD = DE = EF$ and hence $A = B = C = 60^\circ$ .

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