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Problem 940

AIME late

Geometry Difficulty 5.9 Prove it

Show that if for a certain triangle the following relationships hold simultaneously:

$\begin{aligned} \frac{a^{3}+b^{3}+c^{3}}{a+b+c} & =c^{2} \ldots \quad \text { and } \\ \sin \alpha \sin \beta & =\sin ^{2} \gamma \ldots \end{aligned}$

then the triangle is equilateral.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

In a triangle, the sines of the angles are proportional to the opposite sides; thus, instead of 2), we can write: $a b=c^{2}$ . Already from 1),

$a^{3}+b^{3}+c^{3}=(a+b) c^{2}+c^{3} \quad \text{or rather} \quad a^{3}+b^{3}=a b(a+b) \ldots$

However,

$a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$

and thus, since

$\begin{aligned} & \quad a+b \neq 0 \\ & a^{2}-a b+b^{2}=a b \quad \text{or rather} \quad a^{2}-2 a b+b^{2}=(a-b)^{2}=0 \end{aligned}$

It follows that: $a=b$ , and since $c^{2}=a b$ ,

$c^{2}=a^{2}=b^{2} \quad \text{that is} \quad c=b=a$

Francis Say (Cistercian St. Stephen's Grammar School VI., Székesfehérvár.)

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.