A regular tetrahedron is a triangular pyramid in which each face is an equilateral triangle.  If the height of a regular tetrahedron is 20 inches then what is the length of each edge of the tetrahedron? Express your answer in simplest radical form.
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 $\left(\frac{s\sqrt{3}}{3}\right)^2+DM^2=s^2$.  Substituting $20$ inches for $DM$, we solve to find $s=\boxed{10\sqrt{6}}$ inches.

[asy]

import three;

size(2.5inch);

currentprojection = orthographic(1/3,-1,1/4);

triple A = (0,0,0);

triple B = (1,0,0);

triple C = (0.5,sqrt(3)/2,0);

triple D = (0.5,sqrt(3)/4,sqrt(6)/3);

triple E = (1/2,0,0);

triple M = (0.5,sqrt(3)/6,0);

dot(A); dot(B); dot(C); dot(D); dot(M); dot(E);

label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,S); label("$D$",D,N);

label("$M$",M,SE); label("$E$",E,S);

draw(A--B--C--D--A--C);

draw(B--D);

draw(D--M);

draw(M--A);

draw(C--E,dashed);

[/asy]