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Given a 3D convex polyhedron $P$ with $n$ vertices, when $n$ is large, is there a tight upper bound (or at least one that holds "most of the time") on the number of (triangular faces) $P$ has?

Here triangular faces means if a face isn't triangle then we break it into triangles. For example a face that's a square or rectangle breaks down into two triangular faces.

Of course $C_n^3$ is a bound but it's too loose. It'd be nice if there's a bound of lower order.

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A convex polyhedron with triangular faces and $n$ vertices has exactly $2n-4$ faces.

This follows from Euler's formula $V-A+F=2$ because $2A=3F$. Indeed, every face has three edges which are counted twice since each edge belongs to exactly two faces.

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  • $\begingroup$ Thanks! Yes I'm talking about polyhedrons. $\endgroup$
    – Vim
    Commented Nov 14, 2017 at 23:34
  • $\begingroup$ Could you give some further explanation on the claim $2A=3F$? I don't feel it's very trivial (although intuitive), and maybe it's based on an algebraic topology theorem I don't know. $\endgroup$
    – Vim
    Commented Nov 14, 2017 at 23:56
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    $\begingroup$ @Vim, it's because if you count the edges by faces, you count each edge twice (it's an edge of exactly two faces). And the faces, being triangles, have 3 edges each. Hence the coefficients 2 and 3 in the equality. $\endgroup$
    – ByteEater
    Commented Nov 18, 2022 at 14:47

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