大截角截半二十面體

由於大截角截半二十面體的頂點圖為不等邊三角形且具備點可遞的特性，同時，其存在自相交的面，並可以透過星形正多面體進行廣義截角來構造，因此大截角截半二十面體是一種自相交截角擬正多面體（Self-Intersecting Truncated Quasi-Regular Polyhedra）。自相交截角擬正多面體一共有五種，分別為立方截角立方八面體、星形截角截半立方體、二十面截角十二面十二面體、截角截半大十二面體和大截角截半二十面體。[14]這些立體由阿爾伯特·巴杜羅（Albert Badoureau）和約翰·皮奇（Johann Pitsch）於1881年發現並描述。[15][16]

大截角截半二十面體共有三種二面角，分別為四邊形面和十角星面的交角、六邊形面和十角星面的交角和四邊形和六邊形的交角。[3]

其中四邊形和十角星的交角約為121.717度：[5][3]

${\displaystyle \arccos \left(-{\frac {\sqrt {10\left(5-{\sqrt {5}}\right)}}{10}}\right)}$[5]${\displaystyle =\arccos \left(-{\sqrt {\frac {5-{\sqrt {5}}}{10}}}\right)}$[3]${\displaystyle \approx 2.12437068569194}$[17]${\displaystyle \approx 121.717474^{\circ }}$

六邊形和十角星的交角約為79.188度：[5][3]

${\displaystyle \arccos \left({\frac {\sqrt {15\left(5-2{\sqrt {5}}\right)}}{15}}\right)}$[5]${\displaystyle =\arccos \left({\sqrt {\frac {5-2{\sqrt {5}}}{15}}}\right)}$[3]${\displaystyle \approx 1.382085796}$[18]${\displaystyle \approx 79.187683^{\circ }}$

四邊形和六邊形的交角約為69.095度：[5][3]

${\displaystyle \arccos \left({\frac {{\sqrt {15}}-{\sqrt {3}}}{6}}\right)\approx 1.20593249868\approx 69.094843^{\circ }}$

若一個幾何中心位於原點的大截角截半二十面體邊長為單位長，則其頂點座標為：[19]

${\displaystyle (\pm {\frac {2{\sqrt {5}}-3}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {1}{2}})}$、 ${\displaystyle (\pm {\frac {1}{2}},\,\pm {\frac {2{\sqrt {5}}-3}{2}},\,\pm {\frac {1}{2}})}$、 ${\displaystyle (\pm {\frac {1}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {2{\sqrt {5}}-3}{2}})}$、 ${\displaystyle (\pm 1,\,\pm {\frac {7-3{\sqrt {5}}}{4}},\,\pm {\frac {3-{\sqrt {5}}}{4}})}$、 ${\displaystyle (\pm {\frac {3-{\sqrt {5}}}{4}},\,\pm 1,\,\pm {\frac {7-3{\sqrt {5}}}{4}})}$、 ${\displaystyle (\pm {\frac {7-3{\sqrt {5}}}{4}},\,\pm {\frac {3-{\sqrt {5}}}{4}},\,\pm 1)}$、 ${\displaystyle (\pm {\frac {1}{2}},\,\pm {\frac {4-{\sqrt {5}}}{2}},\,\pm {\frac {{\sqrt {5}}-2}{2}})}$、 ${\displaystyle (\pm {\frac {{\sqrt {5}}-2}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {4-{\sqrt {5}}}{2}})}$、 ${\displaystyle (\pm {\frac {4-{\sqrt {5}}}{2}},\,\pm {\frac {{\sqrt {5}}-2}{2}},\,\pm {\frac {1}{2}})}$、 ${\displaystyle (\pm {\frac {{\sqrt {5}}-1}{2}},\,\pm {\frac {5-{\sqrt {5}}}{4}},\,\pm {\frac {3{\sqrt {5}}-5}{4}})}$、 ${\displaystyle (\pm {\frac {3{\sqrt {5}}-5}{4}},\,\pm {\frac {{\sqrt {5}}-1}{2}},\,\pm {\frac {5-{\sqrt {5}}}{4}})}$、 ${\displaystyle (\pm {\frac {5-{\sqrt {5}}}{4}},\,\pm {\frac {3{\sqrt {5}}-5}{4}},\,\pm {\frac {{\sqrt {5}}-1}{2}})}$、 ${\displaystyle (\pm {\frac {3-{\sqrt {5}}}{4}},\,\pm {\frac {3-{\sqrt {5}}}{2}},\,\pm {\frac {3\left({\sqrt {5}}-1\right)}{4}})}$、 ${\displaystyle (\pm {\frac {3\left({\sqrt {5}}-1\right)}{4}},\,\pm {\frac {3-{\sqrt {5}}}{4}},\,\pm {\frac {3-{\sqrt {5}}}{2}})}$、 ${\displaystyle (\pm {\frac {3-{\sqrt {5}}}{2}},\,\pm {\frac {3\left({\sqrt {5}}-1\right)}{4}},\,\pm {\frac {3-{\sqrt {5}}}{4}})}$。