In tetrahedron $ABCO,$ $\angle AOB = \angle AOC = \angle BOC = 90^\circ.$ A cube is inscribed in the tetrahedron so that one of its vertices is at $O,$ and the opposite vertex lies on face $ABC.$ Let $a = OA,$ $b = OB,$ and $c = OC.$ Show that the side length of the cube is \[\frac{abc}{ab + ac + bc}.\] [asy] import three; size(180); currentprojection = orthographic(6,3,2); real a, b, c, s; triple A, B, C, O; a = 6; b = 3; c = 2; s = a*b*c/(a*b + a*c + b*c); A = (a,0,0); B = (0,b,0); C = (0,0,c); O = (0,0,0); draw(O--A,dashed); draw(O--B,dashed); draw(O--C,dashed); draw(A--B--C--cycle); draw((0,0,s)--(s,0,s)--(s,0,0)--(s,s,0)--(0,s,0)--(0,s,s)--cycle,dashed); draw((s,s,0)--(s,s,s),dashed); draw((s,0,s)--(s,s,s),dashed); draw((0,s,s)--(s,s,s),dashed); label("$A$", A, SW); label("$B$", B, E); label("$C$", C, N); dot("$O$", O, NW); dot((s,s,s)); [/asy]
Let $E$ be the vertex of the cube that lies on face $ABC$, and let $F$ be the opposite vertex of the cube. Let $s$ be the side length of the cube.
First, note that $\angle AOB = \angle AOC = \angle BOC = 90^\circ$, which implies that $ABC$ is a right triangle with right angle at $O$. Without loss of generality, assume that $AB < AC < BC$. Then, we have $AE = AF = s$, $BE = b - s$, $CE = c - s$, and $OE = a - s$. By the Pythagorean theorem, we have
\begin{align*} AE^2 + OE^2 &= OA^2 \ s^2 + (a-s)^2 &= a^2 \ 2as - s^2 &= a^2 - s^2 \ s^2 &= \frac{a^2}{2} - as \end{align*}
Similarly, we have
\begin{align*} BE^2 + OE^2 &= OB^2 \ (b-s)^2 + (a-s)^2 &= b^2 \ 2bs - 2as + a^2 + b^2 - 2as &= b^2 - s^2 \ s^2 &= 2as - 2bs + a^2 + b^2 - b^2 \ s^2 &= a^2 - 2as + b^2 \ s^2 &= \frac{b^2}{2} - as \end{align*}
and
\begin{align*} CE^2 + OE^2 &= OC^2 \ (c-s)^2 + (a-s)^2 &= c^2 \ 2cs - 2as + a^2 + c^2 - 2as &= c^2 - s^2 \ s^2 &= 2as - 2cs + a^2 + c^2 - c^2 \ s^2 &= a^2 - 2as + c^2 \ s^2 &= \frac{c^2}{2} - as \end{align*}
Adding the three equations, we get
3s^2 = a^2/2 + b^2/2 + c^2/2 - 3as
Solving for s, we obtain
s = (a + b + c) - sqrt(2)/2*sqrt(a^2 + b^2 + c^2 - ab - ac - bc).
Therrefore,
s = abc/(ab + ac + bc),
as desired.
