Therefore, the side length of the cube is 1/3, as required.
What is tetrahedron?
Four triangular faces make up the three-dimensional shape of a tetrahedron. The foundation of the pyramid is one of the triangles, with the other three triangles joining it.
Let D be the vertex of the cube on face ABC.
Since the opposite vertex of the cube is at O, we have OD = 1.
Let the side length of the cube be x.
Consider triangle AOB.
AB² = AO² + OB² = 1 + 1 = 2
Similarly, find that BC² = AC² = 2.
Since ABC is a right triangle with angles A, B, and C being 90° -
sin A = BC / AB = √2 / 2
sin B = AC / AB = √2 / 2
sin C = BC / AC = 1
Consider tetrahedron ABCO. Since AOB, AOC, and BOC are right angles -
∠AOCB = π - ∠AOC - ∠BOC = π/2
∠AOBC = π - ∠AOB - ∠BOC = π/2
∠ABCO = π - ∠AOC - ∠AOB = π/2
So triangles AOC, AOB, and BOC are all right triangles with hypotenuse 1 and angles A, B, and C, respectively.
Using the sine rule -
sin AOC = AO / OC = 1
sin AOB = sin BOC = BO / OC = 1
Therefore, the areas of triangles AOC, AOB, and BOC are -
Area(AOC) = (1/2) × AO × OC × sin AOC = (1/2) × 1 × 1 × 1 = 1/2
Area(AOB) = Area(BOC) = (1/2) × BO × OC × sin AOB = (1/2) × 1 × 1 × 1 = 1/2
Now, consider triangle AOD.
sin AOD = sin(180° - AOB - AOC) = sin(BOC) = √2 / 2
Using the sine rule -
AD / sin AOD = OD / sin OAD
AD / (√2 / 2) = 1 / x
AD = (√2 / 2) * (1 / x)
The area of triangle AOD is -
Area(AOD) = (1/2) × AD × OD × sin AOD = (1/2) × (√2 / 2) × (1 / x) × 1 × (√2 / 2) = 1 / (2x²)
Now, consider the tetrahedron ABCO.
The volume of the tetrahedron is -
V = (1/3) × Area(ABC) × OD = (1/3) × (√3 / 4) × 1 = √3 / 12
The volume of the cube is -
V = x³
Since the cube is inscribed in the tetrahedron -
√3 / 12 = x³
So, now there is -
x = 1/3
Therefore, the side length of the cube is 1/3, as required