(1) 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.
(2) Simplfy abc(1/(ab) + 1/(ac) + 1/(bc)).
(1) Since the angles of the tetrahedron are all 90°, the sides of the tetrahedron are all equal. Since the cube is inscribed in the tetrahedron, the side length of the cube must also be equal to the side length of the tetrahedron. Since the side length of the tetrahedron is 1, the side length of the cube must also be 1. Therefore, the side length of the cube is 1/3.
(2) abc(1/(ab) + 1/(ac) + 1/(bc)) = abc(ab + ac + bc)/(abacbc) = abc/abc = 1.