Take a projector gener­ating parallel rays. It's obvious that a cube may have a square shadow. But what is the maximum number of vertices that may have the polygon forming the shadow of a cube? If the diag­onal of the cube is parallel to the rays, the shadow will be a regular hexagon!

Let's turn both the projector and the screen. The shadow on the screen is a square. Does our object have to be a cube?

Add another screen and projector in direc­tion perpen­dic­ular to the first one. Now we have two orthog­onal projec­tions forming squares. Is it defi­nitely a cube now?

And what if three orthog­onal projec­tions? are squares? Is there a body except for cubes forming three orthog­onal square shadows?

It's easy to imagine non-convex bodies, e.g. a cube with holes, giving such projec­tions. And what if we consider only convex bodies, or even regular poly­topes?

It turns out that even a regular poly­tope different from a cube may form square shadows in three orthog­onal direc­tions.

Indeed, we can inscribe a regular tetra­he­dron in a cube! Four vertices of the tetra­he­dron will coin­cide with the vertices of the cube. All the edges of the tetra­he­dron will be diag­o­nals of the cube's faces, and as follows, equal.

If we look through one of the cube's faces, such a tetra­he­dron will «fill» the whole projec­tion along the direc­tion perpen­dic­ular to the face.

It means that if a cube is placed so that it's three orthog­onal projec­tions are squares, i.e. the screens are parallel to the faces, the inscribed tetra­he­dron will give the same shadows, three squares.

Other etudes in “Polyhedra’s outer geometry”