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Imagine a polyhedron of 7 triangles and 7 quadrilaterals, where each quadrilateral is parallel to one of the triangles. Is a relaxed version of that possible -- any number of sides is allowable, where each side is parallel to a polygon with a different number of sides.

If that's possible or easy, can the faces have roughly the same area? Can an inner sphere be tangent to all faces?

Ed Pegg
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    There is no such polyhedron even without the parallelness condition. It would have $\frac{3\cdot 7+4\cdot 7}2$ edges. – Hagen von Eitzen Jun 01 '17 at 20:59
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    If faces can be paired into parallels and are tangent to a sphere, then reflaction at the center permutes the touching points, which is all we need to reconstruct the polyhedron. Consequently, oppisng faces are congruent. – Hagen von Eitzen Jun 01 '17 at 21:01
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    A truncated tetrahedron is an example of such polyhedron (in a relaxed version, that is): each triangle is parallel to certain hexagon. However, the faces are of different area, and the inner sphere doesn't touch them all, and if you try do match any of these requirements, you achieve that at a cost: in effect, your polyhedron ceases to be what it used to be, and becomes an octahedron. – Ivan Neretin Jun 14 '17 at 14:13

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