[Geometry]
Problem: Regular tetrahedra $J A N E, J O H N$, and $J O A N$ have non-overlapping interiors. Compute $\tan \angle H A E$.

Solution: First note that $\overline{J N}$ is a shared edge of all three pyramids, and that the viewpoint for the figure below is from along the line that is the extension of edge $\overline{J N}$.

<img_3460>

Let $h$ denote the height of each pyramid. Let $X$ be the center of pyramid JOAN, and consider the plane passing through $H, A$, and $E$. By symmetry, the altitude in pyramid $J O H N$ through $H$ and the altitude in pyramid $J A N E$ through $E$ pass through $X$. Thus points $H, X$, and $A$ are collinear, as are points $E, X$, and $O$. Hence $A H=O E=2 h$. Using the result that the four medians in a tetrahedron are concurrent and divide each other in a $3: 1$ ratio, it follows that $A X=O X=\frac{3 h}{4}$ and $X E=O E-O X=\frac{5 h}{4}$. Applying the Law of Cosines to triangle $A X E$ yields $\cos \angle X A E=\cos \angle H A E=\frac{2-2 h^{2}}{3 h}$. Suppose, without loss of generality, that the common side length of the pyramids is 1 . Then $h=\sqrt{\frac{2}{3}}$ and $\cos \angle H A E=\frac{\sqrt{6}}{9}$. Hence $\sin \angle H A E=\frac{\sqrt{75}}{9}$ and therefore $\tan \angle H A E=\frac{5 \sqrt{2}}{\mathbf{2}}$.

Final answer: $\frac{5 \sqrt{2}}{2}$