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Problem 989

AMC 12 late, AIME early

Geometry Difficulty 4.5 Prove it Problems of Ukrainian Authors · Ukraine

Angular bisectors $BD$ and $CE$ are drawn in the triangle $ABC$ with angle $\angle BAC \ge 60^\circ$ . Prove that $AD + AE \le BC$ .

Fig.30

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Denote the sides of triangle by $a$ , $b$ , $c$ . By the property of angular bisectors,

$AD = \frac{ch}{a+c}$ , $AE = \frac{ch}{a+b}$ (pic.30), thus it is enough to prove: $\frac{ch}{a+c} + \frac{ch}{a+b} \le a$ .

After simple transformations, we have: $(a^2 - bc)(a+b+c) \ge 0$ , and so it is enough to prove: $a^2 - bc \ge 0$ .

By the cosine theorem, $a^2 = b^2 + c^2 - 2bc \cos A \ge b^2 + c^2 - 2bc \cdot \frac{1}{2} \ge 2bc - bc = bc$ , which required.

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