Question

An angle $x$ is chosen at random from the interval $0^\circ \lt x \lt 90^\circ.$ Let $p$ be the probability that the numbers $\sin^2 x,$ $\cos^2 x,$ and $\sin x \cos x$ are not the lengths of the sides of a triangle. Given that $p = \frac{d}{n},$ where $d$ is the number of degrees in $\arctan m$ and $m$ and $n$ are positive integers with $m + n \lt 1000,$ find $m + n.$

Answer

Solution:

Replacing $x$ by $90^\circ - x$ swaps $\sin x$ and $\cos x,$ so the failure probability on $(45^\circ, 90^\circ)$ matches that on $(0^\circ, 45^\circ),$ and it suffices to consider $0^\circ \lt x \le 45^\circ.$ There $\cos^2 x \ge \sin x \cos x \ge \sin^2 x,$ so the three numbers fail to form a triangle exactly when $\cos^2 x \ge \sin^2 x + \sin x \cos x.$

Since $\cos^2 x - \sin^2 x = \cos 2x$ and $\sin x \cos x = \frac{1}{2}\sin 2x,$ this says $\cos 2x \ge \frac{1}{2} \sin 2x,$ i.e. $\tan 2x \le 2.$ Because tangent increases on this range, that happens exactly for $x \le \frac{1}{2}\arctan 2.$

Therefore $p = \frac{\frac{1}{2}\arctan 2}{45^\circ} = \frac{\arctan 2}{90^\circ},$ so $m = 2$ and $n = 90,$ with $m + n = 92 \lt 1000,$ and the answer is $92.$

2003 AIME I Problem 11

Solution:

Problem 11 in Other Years