截角超立方體：修订间差异

2022年12月28日 (三) 11:29的最新版本

截角超立方体可以通过在每条棱距离顶点 $1 / (\sqrt{2} + 2)$ 处截断超立方体的每一个角来得到。每个截断的角会产生一个正四面体。

一个棱长为2的截角超立方体的每个顶点的笛卡儿坐标系坐标为:

(\pm 1, \pm (1 + \sqrt{2}), \pm (1 + \sqrt{2}), \pm (1 + \sqrt{2}))

正交投影
考克斯特平面	B₄	B₃ / D₄ / A₂	B₂ / D₃
Graph
二面体群	[8]	[6]	[4]
考克斯特平面	F₄	A₃
Graph
二面体群	[12/3]	[4]

展开图	三维正交投影

T. Gosset: On the Regular and Semi-Regular Figures in Space of n Dimensions, Messenger of Mathematics, Macmillan, 1900
H.S.M. Coxeter:
- Coxeter, Regular Polytopes, (3rd edition, 1973), Dover edition, ISBN 0-486-61480-8, p. 296, Table I (iii): Regular Polytopes, three regular polytopes in n-dimensions (n≥5)
- H.S.M. Coxeter, Regular Polytopes, 3rd Edition, Dover New York, 1973, p. 296, Table I (iii): Regular Polytopes, three regular polytopes in n-dimensions (n≥5)
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John H. Conway, Heidi Burgiel, Chaim Goodman-Strass, The Symmetries of Things 2008, ISBN 978-1-56881-220-5 (Chapter 26. pp. 409: Hemicubes: 1_n1)
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