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Geometry Difficulty 5.1 AIME, harder Find the answer

Let PABCP A B C be a tetrahedron such that ∠APB=∠APC=∠BPC=90∘,∠ABC=30∘\angle A P B=\angle A P C=\angle B P C=90^{\circ}, \angle A B C=30^{\circ}, and AP2A P^{2} equals the area of triangle ABCA B C. Compute tan⁡∠ACB\tan \angle A C B.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Observe that 12⋅AB⋅AC⋅sin⁡∠BAC=[ABC]=AP2=12(AB2+AC2−BC2)=AB⋅AC⋅cos⁡∠BAC\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⁡∠BAC=2\tan \angle B A C=2. Also, we have tan⁡∠ABC=13\tan \angle A B C=\frac{1}{\sqrt{3}}. Also, for any angles α,β,γ\alpha, \beta, \gamma summing to 180∘180^{\circ}, one can see that tan⁡α+tan⁡β+tan⁡γ=tan⁡α⋅tan⁡β⋅tan⁡γ\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \cdot \tan \beta \cdot \tan \gamma. Thus we have tan⁡∠ACB+2+13=\tan \angle A C B+2+\frac{1}{\sqrt{3}}= tan⁡∠ACB⋅2⋅13\tan \angle A C B \cdot 2 \cdot \frac{1}{\sqrt{3}}, so tan⁡∠ACB=8+53\tan \angle A C B=8+5 \sqrt{3}.

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