A convex polyhedron has at least three faces which are pentagons. What is the minimum number of faces the polyhedron might have?
I have a polyhedron with seven faces but I don't know whether it is possible with six:
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6 faces ? A cube ! – Jean Marie May 01 '16 at 06:47
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@JeanMarie at least three faces which are pentagons – Salomo May 01 '16 at 06:47
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All right: the text and, mainly, the title are misleading; I advise the asker to write the title as : "Polyhedra with at least 3 pentagonal faces" – Jean Marie May 01 '16 at 06:56
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Does it require convex faces? – Riccardo Orlando May 01 '16 at 07:46
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1I have edited the title...I think it does require convex faces, but I'm not sure – May 01 '16 at 09:28
2 Answers
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At least $3$ edges meet at each vertex. Thus (denoting the numbers of vertices, edges and faces by $V$, $E$ and $F$, respectively), we have $3V\le 2E$. With $V-E+F=2$, this yields $F\ge2+\frac E3$. For $F=6$, we'd need $E\le12$. The $3$ pentagons have $15$ edges, and each pair of them can share at most one edge, so the pentagons themselves already have $12$ distinct edges, so $E=12$. But the $3$ pentagons have $15$ vertices, and each pair of them can share at most two vertices, so $V\ge15-3\cdot2=9$, contradicting $V=2+E-F=2+12-6=8$.
Thus your polyhedron has the minimal number of faces.
joriki
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@joriki has said everything about seven faces being minimal, but it turns out that the cube with one vertex chopped off is not the only solution with seven total faces. Chopping three vertices of a tetrahedron works too.
Oscar Lanzi
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