On Indecomposable Polyhedra
On Indecomposable Polyhedra
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论不可分解多面体
DOI：
10.1080/00029890.1948.11999266
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发表时间：
1948
影响因子：
0.5
通讯作者：
F. Bagemihl
中科院分区：
文献类型：
--
作者：
F. Bagemihl
定理设n为不小于6的整数，则存在n个顶点的多面体Fn，且满足下列性质：（1）1 rn是单多面体，其每个面都是三角形。(11)如果T是一个四面体，其每个顶点都是1 rn的顶点，则不是T的每个内点都是7 rn的内点。（III）每个开线段，其端点是11”n的顶点，但不是11”n的边，完全位于11”n.众所周知，每个凸多面体都可以分解成一组四面体，其顶点都是给定多面体的顶点[1，p.280或2，第57页];每个简单多边形都可以分解成一组三角形，其顶点都是给定多边形的顶点[1，p.246或2，p.46]。然而，Lennes通过构造一个具有性质（I）和（11）的多面体，证明了不可分解多面体的存在。他的多面体，有七个顶点，不满足（III）。Schonhardt [3]随后给出了一个例子，一个多面体有六个顶点和所有上述三个性质。他表明，此外，没有不可分解的多面体少于六个顶点。
THEOREM. If n is an integer not less than 6, then there exists a polyhedron, Fn, with n vertices qnd the following properties:(I) 1r n is simple, and every one of its faces is a triangle.(11) If T is a tetrahedron, each of whose vertices is a vertex of 1r n, then not every interior point of T is an interior point of 7r n·(Ill) Every open segment whose endpoints are vertices of 11" n, but which is not an edge of 11" n, lies wholly exterior to 11" n.It is well known that every convex polyhedron can be decomposed into a set of tetrahedra whose vertices are all vertices of the given polyhedron [1, p. 280 or 2, p. 57]; and every simple polygon can be decomposed into a set of triangles whose vertices are all vertices of the given polygon [1, p. 246 or 2, p. 46]. Lennes, however, proved [2, p. 55] the existence of indecomposable polyhedra by constructing a polyhedron which has properties (I) and (11). His polyhedron, which possesses seven vertices, does not satisfy (Ill). Schonhardt [3) subsequently gave an example of a polyhedron having six vertices and all three of the above properties. He showed, moreover, that there is no indecomposable polyhedron with less than six vertices.