Place two points on two adjacent edges of a unit cube, each at a distance of 3/4 from the point where the two edges meet, and two more points symmetrically on the opposite face of the cube. Then these four points form a square with side length

The parts of the unit cube that remain, after emptying this hole, form two triangular prisms and two irregular tetrahedra, connected by thin bridges at the four vertices of the square. Each prism has as its six vertices two adjacent vertices of the cube, and four points along the edges of the cube at distance 1/4 from these cube vertices. Each tetrahedron has as its four vertices one vertex of the cube, two points at distance 3/4 from it on two of the adjacent edges, and one point at distance 3/16 from the cube vertex along the third adjacent edge.^[2]

Prince Rupert's cube is named after Prince Rupert of the Rhine. According to a story recounted in 1693 by English mathematician John Wallis, Prince Rupert wagered that a hole could be cut through a cube, large enough to let another cube of the same size pass through it. Wallis showed that in fact such a hole was possible (with some errors that were not corrected until much later), and Prince Rupert won his wager.^[3][4]

Wallis assumed that the hole would be parallel to a space diagonal of the cube. The projection of the cube onto a plane perpendicular to this diagonal is a regular hexagon, and the best hole parallel to the diagonal can be found by drawing the largest possible square that can be inscribed into this hexagon. Calculating the size of this square shows that a cube with side length

${\displaystyle {\sqrt {6}}-{\sqrt {2}}\approx 1.03527}$ ,

slightly larger than one, is capable of passing through the hole.^[3]

Approximately 100 years later, Dutch mathematician Pieter Nieuwland found that a better solution may be achieved by using a hole with a different angle than the space diagonal. In fact, Nieuwland's solution is optimal. Nieuwland died in 1794, a year after taking a position as a professor at the University of Leiden, and his solution was published posthumously in 1816 by Nieuwland's mentor, Jean Henri van Swinden.^[3][4][5]

Since then, the problem has been repeated in many books on recreational mathematics, in some cases with Wallis' suboptimal solution instead of the optimal solution.^{[1][2][6][7][8][9][10][11][12]}

The construction of a physical model of Prince Rupert's cube is made challenging by the accuracy with which such a model needs to be measured, and the thinness of the connections between the remaining parts of the unit cube after the hole is cut through it. For the maximally sized inner cube with length ≈1.06 relative to the length 1 outer cube, constructing a model is "mathematically possible but practically impossible".^[13] On the other hand, using the orientation of the maximal cube but making a smaller hole, big enough only for a unit cube, leaves additional thickness that allows for structural integrity.^[14]

For the example using two cubes of the same size, as originally proposed by Prince Rupert, model construction is possible. In a 1950 survey of the problem, D. J. E. Schrek published photographs of a model of a cube passing through a hole in another cube.^[15] Martin Raynsford has designed a template for constructing paper models of a cube with another cube passing through it; however, to account for the tolerances of paper construction and not tear the paper at the narrow joints between parts of the punctured cube, the hole in Raynsford's model only lets cubes through that are slightly smaller than the outer cube.^[16]

Since the advent of 3D printing, construction of a Prince Rupert cube of the full 1:1 ratio has become easy.^[17]

A polyhedron ${\displaystyle P}$  is said to have the Rupert property if a polyhedron of the same or larger size and the same shape as ${\displaystyle P}$  can pass through a hole in ${\displaystyle P}$ .^[18] All five Platonic solids—the cube, regular tetrahedron, regular octahedron,^[19] regular dodecahedron, and regular icosahedron—have the Rupert property. Of the 13 Archimedean solids, it is known that at least these ten have the Rupert property: the cuboctahedron, truncated octahedron, truncated cube, rhombicuboctahedron, icosidodecahedron, truncated cuboctahedron, truncated icosahedron, truncated dodecahedron,^[20] and the truncated tetrahedron^[21][22], as well as the truncated icosidodecahedron^[23][24]. It has been conjectured that all 3-dimensional convex polyhedra have this property^[18], but also, to the contrary, that the rhombicosidodecahedron does not have Rupert's property^[23][24].

Do all convex polyhedra have the Rupert property?