The problem was to fit a cube of side larger than one through a unit cube. This is known as "Prince Rupert's Cube"; a keyword search yields a lot of information about it. In the notes below, I will not prove every claim. In particular, I do not have the details of a rigorous proof of the largest square that can be inscribed in a cube. A reference I have not checked is L. Bankoff, Regular octahedron inscribed in a cube, Math. Mag. 25 (1951) 48-49. For extensions of this problem to higher dimensions, start with Unsolved Problems In Geometry, H. T. Croft, K. J. Falconer, and R. K. Guy, Springer, Problem B4.
It is not hard to see that a cube of side-length 1.03 can fit through a unit cube. Hold the unit cube by one vertex and let it hang freely below that vertex. The projection onto a horizontal plane is a regular hexagon and the largest square that fits inside that hexagon has side-length √2(√3 − 1). A square hole extruded through this square (more precisely, a tiny amount smaller than the square so that the cube does snot fall apart) will allow a cube of side-length 1.035 to pass through.
Let s = √3. To be precise, rotate the unit cube by arccos(s) around the line defined by (0, 0, 0) and (1, 1, 0). From this, one learns that the hexagon is formed from the points
(0, 0)
(1, 1)
(1/s, -1/s)
(1 + 1/s, 1 − 1/s)
((1 + s)/2, (1 − s)/2)
((3 − s)/6, (3 + s)/6)
The side-length of this hexagon is √[2]/s. Now, one can show that the largest square that fits into a hexagon has side-length 3 − s times the hexagon length and starts at a point 2 − s of the way from one vertex to the next, as in
This yields the assertion about the square-hole size. These figures show the hole in the square:
But now comes a surprise: One can do better.
Consider the four points on the surface of a unit cube: (0, 1/4, 1), (3/4, 1, 1),(1, 3/4, 0), (1/4, 0, 0). They form a square of side length 3√[2]/4 ~ 1.0607. The square is oriented in a close-to-vertical but not true vertical position. It is the largest square that one can inscribe in a cube. For any positive epsilon, a square hole perpendicular to this square and epsilon smaller leaves the cube in one piece and allows a square to pass through. Thus the cube of side-length 3√[2]/4 is Prince Rupert's Cube:
Here is video of a model created by Leigh Jerrard:
I received two solutions, from Michael Elgersma and John Snyder, both with the correct larger answer of 1.06.
