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[几何] 四面体 $T$ 和 $T'$ 的体积相同

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hbghlyj Posted at 2024-11-11 04:31:51 |Read mode
Rational Elliptic Surfaces and the Trigonometry of Tetrahedra   Theorem 1.3.

设 $T$ 为一个四面体,其棱长为 $l_{ij}$,二面角为 $\alpha_{ij}$。
假设在同一空间中存在一个四面体 $T'$,其棱长为 $l_{ij}'$,其中 $1\leq i<j\leq4$,使得
\begin{align*}
&l_{12}'=l_{12}, \ \  \
l_{13}'=\frac{l_{14}+l_{23}+l_{24}-l_{13}}{2}, \ \ \
l_{14}'=\frac{l_{13}+l_{23}+l_{24}-l_{14}}{2},\\
&l_{34}'=l_{34}, \ \ \
l_{24}'=\frac{l_{13}+l_{14}+l_{23}-l_{24}}{2}, \ \ \
l_{23}'=\frac{l_{13}+l_{14}+l_{24}-l_{23}}{2}.
\end{align*}
则四面体 $T'$ 的相应二面角 $\alpha_{ij}'$ 满足
\begin{align*}
& \alpha_{12}'=\alpha_{12},\ \  
  \alpha_{13}'=\frac{\alpha_{14}+\alpha_{23}+\alpha_{24}-\alpha_{13}}{2},\ \
  \alpha_{14}'=\frac{\alpha_{13}+\alpha_{23}+\alpha_{24}-\alpha_{14}}{2},\\
& \alpha_{34}'=\alpha_{34},\ \
  \alpha_{24}'=\frac{\alpha_{13}+\alpha_{14}+\alpha_{23}-\alpha_{24}}{2},\ \
  \alpha_{23}'=\frac{\alpha_{13}+\alpha_{14}+\alpha_{24}-\alpha_{23}}{2}.
\end{align*}
此外,四面体 $T$ 和 $T'$ 的体积相同。

如何证明呢?

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 Author| hbghlyj Posted at 2024-11-11 04:38:45
Combinatorial and Geometrical Origins of Regge Symmetries有一幅插图: Screenshot 2024-11-10 203800.png
如何证明呢?

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 Author| hbghlyj Posted at 2024-11-11 04:50:16
The Regge symmetry, confocal conics, and the Schläfli formula也有一幅插图: Screenshot 2024-11-10 204934.png
如何证明呢?

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 Author| hbghlyj Posted at 2024-11-15 16:29:14
上述“对偶”的棱长的四面体的二面角、体积的关系该如何证明?

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