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[几何] 凸五面体仅有两种可能——四棱锥或三棱柱

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hbghlyj posted 2025-7-17 13:16 |Read mode
任一面最多与其它 \(F-1=4\) 个面相邻,故最大边数 \(k\le4\)。因此  
\[
f_k = 0\quad(\forall k\ge5),
\quad
f_3 + f_4 = 5.
\]
设 \(f_k\) 为 \(k\) 边形面数,则  
\[
\sum_{k\ge3} f_k = F = 5
\]每条棱邻接两面,故  
\[
2E = \sum_{k\ge3} k\,f_k
\]欧拉公式 \(V - E + F = 2\) 写为  
\[
V = E - 3
\]每顶点至少三条棱,得
\[
2E = \sum_v \deg(v) \ge 3V = 3(E-3)
\implies
E \le 9
\]
计数与同余约束  
由  
\[
2E = 3f_3 + 4f_4 \le 18,
\quad
f_3 + f_4 = 5,
\quad
3f_3 + 4f_4 \equiv 0 \pmod2,
\]
整数解仅为  
\[
(f_3,f_4) = (4,1)\quad\text{或}\quad(2,3).
\]

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original poster hbghlyj posted 2025-7-17 13:07
Determine all convex polyhedra with 6 faces
seven hexahedra and their duals

oeis.org/A000944 Number of polyhedra (or 3‑connected simple planar graphs) with n nodes
$$
f(4)=1,\;f(5)=2,\;f(6)=7,\;f(7)=34,\;f(8)=257,\;\dots
$$

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