Cube Inscribed In Cube

The length of the side of the smallest such cube is $\dfrac{3}{5}$ .

The red rectangles in this figure below are all identical, so that vertex-to-vertex distances are similarly identical (see Note).

Let $a,b$ be the dimensions of the red rectangles. Then the squares of the edge and the diagonal of a face of the small cube must satisfy

$\sqrt{2} \sqrt{ a^2 + \left(1-b \right)^2+ \left(b-a \right)^2 } = \sqrt { { b }^{ 2 }+{ \left( 1-a \right) }^{ 2 }+{ \left( 1-a-b \right) }^{ 2 } }$

so that

$b=\dfrac{1}{2}\left(1+3a-\sqrt{ 1-2a+5a^2 } \right)$

so that the edge of the small cube has length

$\sqrt{ 1-2a+6a^2-2a \sqrt {1-2a+5a^2}}$

the minimum of which is $\dfrac{3}{5}$ , with $a=\dfrac{2}{5}$ and $b=\dfrac{3}{5}$

Note: A loose proof why: To consider any other way a cube can be inscribed other than having all the red rectangles being identical, consider the first two rectangles having dimensions $a, b$ and $c, d$ . Then in succession the rest of the tangent points are determined by the first four variables $a, b, c, d$ , and for it to come back to the point of beginning, two degrees of freedom have to be given up. The resulting box formed in this manner is not necessarily a cube, but could be rhombic in cross section, so in order for it to be a true cube, one more degree of freedom has to be given up. That leaves just one degree of freedom which determines the size of the cube. Ergo, all the red rectangles need to be identical, as that is a solution and therefore the only one.

The cube vertices coordinates lying on the face are assumed and the volume of the inscribed cube is minimized to get the desired answer.