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Geometry Difficulty 4.8 AIME Prove it United States

Problem:

In triangle $ABC$ , $AB = 32$ , $AC = 35$ , and $BC = x$ . What is the smallest positive integer $x$ such that $1 + \cos^2 A$ , $\cos^2 B$ , and $\cos^2 C$ form the sides of a non-degenerate triangle?

Solution

Solution:

By the triangle inequality, we wish $\cos^2 B + \cos^2 C > 1 + \cos^2 A$ . The other two inequalities are always satisfied, since $1 + \cos^2 A \geq 1 \geq \cos^2 B, \cos^2 C$ . Rewrite the above as
$2 - \sin^2 B - \sin^2 C > 2 - \sin^2 A$
so it is equivalent to $\sin^2 B + \sin^2 C < \sin^2 A$ . By the law of sines, $\sin A : \sin B : \sin C = BC : AC : AB$ . Therefore,
$\sin^2 B + \sin^2 C < \sin^2 A \Longleftrightarrow CA^2 + AB^2 < x^2$
Since $CA^2 + AB^2 = 35^2 + 32^2 = 1225 + 1024 = 2249$ , the smallest possible value of $x$ such that $x^2 > 2249$ is $48$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.