$ABCD$ is a regular tetrahedron (right triangular pyramid).  If $M$ is the midpoint of $\overline{CD}$, then what is $\tan\angle AMB$?
The tetrahedron is shown below.  In order to find $\tan\angle AMB$, we build a right triangle with $\angle AMB$ among its angles.  The foot of the altitude from $A$ to face $BCD$ is the centroid, $G$, of triangle $BCD$.

[asy]

import three;

currentprojection = orthographic(1.5,1.1,-1);

triple A = (1,1,1);

triple B = (1,0,0);

triple C = (0,1,0);

triple D = (0,0,1);

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

draw(A--D,dashed);

draw(C--D--B,dashed);

label("$A$",A,NW);

label("$B$",B,W);

label("$C$",C,S);

label("$D$",D,NW);

triple M = (0,0.5,0.5);

draw(A--M--B,dashed);

label("$M$",M,NE);

triple G = B/3 + 2*M/3;

draw(A--G,dashed);

label("$G$",G,S);

[/asy]

Since $\overline{BM}$ is a median of $\triangle BCD$, point $G$ is on $\overline{BM}$ such that $GM = \frac13BM$.  Furthermore, we have $AM = BM$, so $GM = \frac{AM}{3}$.  The Pythagorean Theorem gives us  \[AG = \sqrt{AM^2 - GM^2} = \sqrt{AM^2 - \frac{AM^2}{9}} = AM\cdot \sqrt{\frac89} = \frac{2\sqrt{2}}{3}\cdot AM.\] Finally, we have  \[\tan \angle AMB = \tan\angle AMG =\frac{AG}{GM} = \frac{(2\sqrt{2}/3)AM}{AM/3} = \boxed{2\sqrt{2}}.\]