HMMT February, 2023

Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Let $PABC$ be a tetrahedron such that $\angle APB = \angle APC = \angle BPC = 90^{\circ}$ , $\angle ABC = 30^{\circ}$ , and $AP^{2}$ equals the area of triangle $ABC$ . Compute $\tan \angle ACB$ .

Solution

Solution:

Observe that
$\begin{aligned} \frac{1}{2} \cdot AB \cdot AC \cdot \sin \angle BAC & = [ABC] = AP^{2} \\ & = \frac{1}{2}\left(AB^{2} + AC^{2} - BC^{2}\right) \\ & = AB \cdot AC \cdot \cos \angle BAC \end{aligned}$
so $\tan \angle BAC = 2$ . Also, we have $\tan \angle ABC = \frac{1}{\sqrt{3}}$ . Also, for any angles $\alpha, \beta, \gamma$ summing to $180^{\circ}$ , one can see that $\tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \cdot \tan \beta \cdot \tan \gamma$ . Thus we have $\tan \angle ACB + 2 + \frac{1}{\sqrt{3}} = \tan \angle ACB \cdot 2 \cdot \frac{1}{\sqrt{3}}$ , so $\tan \angle ACB = 8 + 5\sqrt{3}$ .

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