HMMT February

Geometry Difficulty 5.1 AIME, harder Find the answer

Let $P A B C$ be a tetrahedron such that $\angle A P B=\angle A P C=\angle B P C=90^{\circ}, \angle A B C=30^{\circ}$ , and $A P^{2}$ equals the area of triangle $A B C$ . Compute $\tan \angle A C B$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Observe that $\begin{aligned} \frac{1}{2} \cdot A B \cdot A C \cdot \sin \angle B A C & =[A B C]=A P^{2} \\ & =\frac{1}{2}\left(A B^{2}+A C^{2}-B C^{2}\right) \\ & =A B \cdot A C \cdot \cos \angle B A C \end{aligned}$ so $\tan \angle B A C=2$ . Also, we have $\tan \angle A B C=\frac{1}{\sqrt{3}}$ . 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 $\tan \angle A C B+2+\frac{1}{\sqrt{3}}=$ $\tan \angle A C B \cdot 2 \cdot \frac{1}{\sqrt{3}}$ , so $\tan \angle A C B=8+5 \sqrt{3}$ .

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