jmc

Recall that a median of a triangle is a line segment drawn from a vertex of the triangle to the midpoint of the opposite side. The three medians of a triangle intersect in a common point called the centroid of the triangle. The centroid divides each median into two segments whose lengths have ratio 2:1.

Call the four vertices of the tetrahedron $A$, $B$, $C$, and $D$. Also, define $E$ to be the midpoint of $AB$ and $M$ to be the centroid of triangle $ABC$. Let $s$ be the side length of the tetrahedron. From the Pythagorean theorem applied to right triangle $AEC$, we find that $CE=\sqrt{s^2-(s/2{)}^2}=s\sqrt{3}/2$. Since $M$ is the centroid of triangle $ABC$, $AM=\frac{2}{3}(CE)=\frac{2}{3}\left(\frac{s\sqrt{3}}{2}\right)=\frac{s\sqrt{3}}{3}$. Finally, applying the Pythagorean theorem to $AMD$, we find ${\left(\frac{s\sqrt{3}}{3}\right)}^2+D{M}^2=s^2$. Substituting $20$ inches for $DM$, we solve to find $s=\boxed{10\sqrt{6}}$ inches.