Problem (Geometry):
Six solid regular tetrahedra are placed on a flat surface so that their bases form a regular hexagon $\mathcal{H}$ with side length 1 , and so that the vertices not lying in the plane of $\mathcal{H}$ (the "top" vertices) are themselves coplanar. A spherical ball of radius $r$ is placed so that its center is directly above the center of the hexagon. The sphere rests on the tetrahedra so that it is tangent to one edge from each tetrahedron. If the ball's center is coplanar with the top vertices of the tetrahedra, compute $r$.

Solution:
Let $O$ be the center of the sphere, $A$ be the top vertex of one tetrahedron, and $B$ be the center of the hexagon.



Then $B O$ equals the height of the tetrahedron, which is $\frac{\sqrt{6}}{3}$. Because $A$ is directly above the centroid of the bottom face, $A O$ is two-thirds the length of the median of one triangular face, so $A O=\frac{2}{3}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{3}$. The radius of the sphere is the altitude to hypotenuse $\overline{A B}$ of $\triangle A B O$, so the area of $\triangle A B O$ can be represented in two ways: $[A B O]=\frac{1}{2} A O \cdot B O=\frac{1}{2} A B \cdot r$. Substitute given and computed values to obtain $\frac{1}{2}\left(\frac{\sqrt{3}}{3}\right)\left(\frac{\sqrt{6}}{3}\right)=\frac{1}{2}(1)(r)$, from which $r=\frac{\sqrt{18}}{9}=\frac{\sqrt{2}}{3}$.

Final answer: $\frac{\sqrt{2}}{3}$