In tetrahedron ABCO, angle AOB = AOC = BOC = 90. 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 ABC/(ab + ac + bc).
We are given that tetrahedron ABCO has angle AOB = AOC = BOC = 90. This means that triangle OAB, triangle OAC, and triangle OBC are all right triangles.
We are also given that a = OA, b = OB, and c = OC. This means that the sides of triangle OAB are a, b, and c.
A cube is inscribed in the tetrahedron so that one of its vertices is at O, and the opposite vertex lies on face ABC. This means that the cube is tangent to face ABC.
The side length of the cube is the distance from O to the opposite vertex. This distance is also the length of the altitude from O to face ABC.
The altitude from O to face ABC is the perpendicular bisector of edge BC. This means that the altitude from O to face ABC is also the median of triangle ABC.
The median of triangle ABC is half the length of side BC. This means that the altitude from O to face ABC is half the length of side BC.
The length of side BC is ab + ac + bc. This is because the length of side BC is the sum of the lengths of sides AB, AC, and BC.
Therefore, the side length of the cube is abc/(ab + ac + bc).
I greatly appreciate your effort to help, however, I belive I found an error in your reasoning.
When you say "The side length of the cube is the distance from O to the opposite vertex." that does not really make sense since the diagonal of a cube is x\(\sqrt{3} \)
Also I think the whole solutions feels rather strange, from what I tried using similar triangles gets me farther
