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 OA = 1, OB = 1, OC = 1. Show that the side length of the cube is 1/3.
+941 To solve this problem, we can use similar triangles. Let the side length of the cube be s, and let the midpoints of AB, AC, and BC be P, Q, and R, respectively. Then OP, OQ, and OR are the diagonals of the faces of the cube, and so they are equal to \($\sqrt{2}s$\).
Consider triangle AOB. By the Pythagorean theorem, we have \($OA^2 + OB^2 = 1 + 1 = 2$\), so \($AO^2 = 2$\). Since angle AOB is 90 degrees, we can use similarity to see that triangle OAP is similar to triangle OAB, so \($\frac{OA}{OP} = \frac{OB}{OP}$\). Solving, we find that \($OP = \frac{\sqrt{2}}{2}$\).
Using similar triangles, we can find that \($OQ = \frac{\sqrt{2}}{2}$ and $OR = \frac{\sqrt{2}}{2}$\). Since the sum of the squares of the lengths of the diagonals of a cube is equal to three times the square of the side length, we have
\($$(OP^2 + OQ^2 + OR^2) = 3s^2$$\)
Substituting the values we found for $OP$, $OQ$, and $OR$, we have
\($$\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = 3s^2$$\)
Simplifying and solving for \($s$\), we find that \($s = \frac{1}{3}$\). Thus, the side length of the cube inscribed in the tetrahedron is \($\frac{1}{3}$\).
