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One can form a polygon of $4 n$ sides by intersecting $n$ congruent squares (treated as closed sets, i.e., filled squares):
         Three Squares

Q1. For which of the $k=3,4,\ldots,4n$ can the intersection of $n$ congruent squares result in a $k$-gon? Perhaps not all can be achieved?

Q2. Can the intersection of $n$ congruent cubes result in a polyhedron of $6 n$ faces? I believe so, but an explicit construction would be useful.

Q3. For which $k$ can $k$-face polyhedra result from the intersection of $n$ congruent cubes?

Q4. The questions extend to $\mathbb{R}^d$.

Question Q2 in particular occurred to me as a possible exercise to build 3D intuition.

Added. Here are two cubes intersected to produce a polyhedron of 12 faces (although one can hardly verify that from this single, not well-lighted image!):
         Two Cubes

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Correct me if I am wrong, but my count would be $8n$ sides. From that count, one obvious necessary condition would be that $k$ be divisible by 8 (if you rule out the possibility that two squares coincide). – Manolito Pérez Jan 30 '12 at 17:03
@Manolito: For the $n{=}3$ squares illustrated, the intersection consists of $4n=12$ edges. Note one can intersect two squares and get a triangle, so $k{=}3$ is achievable. – Joseph O'Rourke Jan 30 '12 at 17:19
OK, now I see. I misunderstood when you said 'filled' squares. – Manolito Pérez Jan 30 '12 at 17:34
@Manolito: I tried to revise to block that misintepretation: sorry! – Joseph O'Rourke Jan 30 '12 at 17:59

Any configuration of regular polytopes, where no two polytopes have a pair of parallell planes, and every plane is symmetric wrt. to the configuration, and the intersection is nonempty, should give the max faces.

For Q2 and n odd, we can rotate each cube around an internal diagonal like thisenter image description here

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