Geometry

Let's denote the side length of the cube by s.

Since the cube is inscribed in the tetrahedron, the opposite vertex of the cube lies on the face ABC. Let's call this vertex D. Then, since AOD is a right triangle, we can use the Pythagorean theorem to find the length of AD:

AD^2 = AO^2 + OD^2 = 1^2 + s^2

Similarly, we can find the lengths of BD and CD:

BD^2 = BO^2 + OD^2 = 1^2 + s^2 CD^2 = CO^2 + OD^2 = 1^2 + s^2

Now, consider the triangle ABC. We know that angles AOB, AOC, and BOC are all right angles, so this is a right triangle. Using the Pythagorean theorem, we can find the length of BC:

BC^2 = AB^2 + AC^2 = (OA^2 + OB^2) + (OA^2 + OC^2) = 2 + 2 = 4

Therefore, BC = 2.

Now, consider the triangle ABD. We know that angles AOB and AOD are right angles, so this is also a right triangle. Using the Pythagorean theorem, we can find the length of AB:

AB^2 = AD^2 + BD^2 = (1^2 + s^2) + (1^2 + s^2) = 2s^2 + 2

Therefore, AB = sqrt(2s^2 + 2).

Similarly, we can find the length of AC:

AC^2 = AD^2 + CD^2 = (1^2 + s^2) + (1^2 + s^2) = 2s^2 + 2

Therefore, AC = sqrt(2s^2 + 2).

Now, consider the triangle BCD. We know that BC = 2, BD = sqrt(2s^2 + 2), and CD = sqrt(2s^2 + 2). This is a triangle with sides of known length, so we can use the Law of Cosines to find the angle at vertex C:

cos(C) = (BC^2 + BD^2 - CD^2) / (2 * BC * BD) = (4 + 2s^2 - 2s^2 - 2) / (4 * sqrt(2s^2 + 2)) = (2 - 1) / (2 * sqrt(2s^2 + 2)) = 1 / (2 * sqrt(2s^2 + 2))

Since C is the angle between the faces AOC and BOC of the tetrahedron, we know that cos(C) = 1/3 (using the fact that the angle between two faces of a regular tetrahedron is arccos(-1/3)). Therefore:

1 / (2 * sqrt(2s^2 + 2)) = 1 / 3

Multiplying both sides by 6 * sqrt(2s^2 + 2), we get:

6 = 2 * sqrt(2s^2 + 2)

Squaring both sides, we get:

36 = 8s^2 + 8

Solving for s, we get:

s^2 = 1/3

Therefore, s = 1/sqrt(3) = sqrt(3)/3, so the side length of the cube is 1/3, as required.