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Geometry Difficulty 5.2 AIME, harder Prove it Croatia

Let $\triangle ABC$ be a right triangle with the right angle at vertex $C$ and let $M$ be the midpoint of the leg $\overline{BC}$ . Prove that $\sin(\angle MAB) \le \frac{1}{3}$ . When is the equality achieved?

Solution

Since $\angle MAB = \angle CAB - \angle CAM$ , according to the sine addition theorem we have
$\sin(\angle MAB) = \sin(\angle CAB) \cos(\angle CAM) - \cos(\angle CAB) \sin(\angle CAM).$

Let $|BC| = a$ , $|CA| = b$ , $|AB| = c$ and $|AM| = t$ . From right triangles $ABC$ and $AMC$ we get
$\sin(\angle MAB) = \frac{a}{c} \cdot \frac{b}{t} - \frac{b}{c} \cdot \frac{a/2}{t} = \frac{ab}{2ct}.$

Since $c^2 = a^2 + b^2$ and $t^2 = \frac{a^2}{4} + b^2$ , that is equivalent to
$9a^2b^2 \le (a^2 + b^2)(a^2 + 4b^2)$
i.e. $(a^2 - 2b^2)^2 \ge 0$ . Now it is clear that the given inequality holds in every right triangle. The equality is achieved when $a = b\sqrt{2}$ .

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