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Geometry Difficulty 4.8 AIME Prove it Brazil

Find all n>5n > 5 for which it is possible to find a convex polyhedron with all nn faces congruent such that each face has another face parallel to it.

Solution

Since for each face there is another parallel to it, nn must be even. For n=4kn = 4k, attach two congruent regular pyramids with 2k2k lateral faces by its bases. For n=4k+2n = 4k + 2, attach two congruent regular pyramids with 2k+12k + 1 lateral faces and twist one of them. Possible coordinates for the vertices of this solid are (cosπj2k+1,sinπj2k+1,(1)j(1cosπ2k+1))(\cos \frac{\pi j}{2k+1}, \sin \frac{\pi j}{2k+1}, (-1)^j(1 - \cos \frac{\pi}{2k+1})), j=1,2,,4k+1j = 1, 2, \dots, 4k + 1 and (0,0,±(1+cosπ2k+1))(0, 0, \pm (1 + \cos \frac{\pi}{2k+1})).

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