四面体内切圆切点分所在面的面积比

hejoseph

假设四面体 $ABCD$ 的各棱、各面面积已知，内切球切面 $BCD$ 于点 $P$，求三角形 $PBC$、$PCD$、$PDB$ 的面积比。
对于三角形，内切圆与对边切点连线是共点的，到四面体似乎并不一定共点了。

hbghlyj

hejoseph 发表于 2024-4-28 09:24
对于三角形，内切圆与对边切点连线是共点的，到四面体似乎并不一定共点了。

满足这个条件的四面体称为inspherical tetrahedraheldermann-verlag.de/jgg/jgg08/j8h1haj2.pdf

hbghlyj

The Gergonne and Nagel centers of a tetrahedron证明了存在一个点P，AP,BP,CP,DP将相对的面分成3个三角形，共12个三角形，使公共边是一条棱的两个三角形的面积相等。

The Gergonne center [4] of a triangle $A B C$ is defined to be the point of intersection of the cevians $A A^{\prime}, B B^{\prime}, C C^{\prime}$ where $A^{\prime}, B^{\prime}, C^{\prime}$ are the points where the incircle of $A B C$ touches the sides $B C, C A, A B$ (Figure 1). Such cevians are concurrent by Ceva's Theorem since $A^{\prime}, B^{\prime}, C^{\prime}$ can be equivalently defined by the algebraic relations
\begin{align}
A B^{\prime}=A C^{\prime}, \quad B C^{\prime}=B A^{\prime}, \quad C A^{\prime}=C B^{\prime}
\end{align}
The Nagel center is defined similarly by taking $A^{\prime}, B^{\prime}, C^{\prime}$ to be the points where the excircles touch the sides (Figure 1). It can be easily seen that the points $A^{\prime}, B^{\prime}, C^{\prime}$ can also be defined by the relations
\begin{align}
B C^{\prime}=C B^{\prime}, C A^{\prime}=A C^{\prime}, \quad A B^{\prime}=B A^{\prime}
\end{align}
This note is a result of the question whether Gergonne and Nagel centers can be defined for a general tetrahedron in a similar way.

The generalized Ceva's Theorem [6] states that the cevians $A A^{\prime}, B B^{\prime}, C C^{\prime}, D D^{\prime}$ are concurrent if and only if one can assign masses $M_A, M_B, M_C, M_D$ to the vertices in such a way that the center of mass of each three of them is one of the points $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$. From Figure 2, this is equivalent to the conditions
\begin{align*}
& M_A: M_B: M_C=x: y: z, \quad M_B: M_C: M_D=Z: Y: x, \\
& M_C: M_D: M_A=X: y: Z, \quad M_D: M_A: M_B=z: Y: X .
\end{align*}
These in turn can be seen to be equivalent to the conditions
\begin{align}
X x=Y y=Z z \quad(=t, \text { say })
\end{align}
Denoting the areas of the faces by $\alpha, \beta, \gamma$, we see that
\begin{align}
x+y+z=\delta \\
x+Y+Z=\alpha \\
X+y+Z=\beta \\
X+Y+z=\gamma
\end{align}
Adding (5), (6) and (7) and subtracting (4), we get
\begin{align}
X+Y+Z=(\alpha+\beta+\gamma-\delta) / 2
\end{align}
Subtracting (8) from (5) and using the fact that $X=t / x$, we obtain
\[
x-t / x+(-\alpha+\beta+\gamma-\delta) / 2=0
\]
and hence
\[
x^2+x(-\alpha+\beta+\gamma-\delta) / 2-t=0 .
\]

hejoseph

谢谢，我的方法是确定了切点的面积比后发现一般是不共点的。再寻求共点的那些跟文档里的差不多了。

hejoseph

面积比的结果：设点 $A$ 所对的面的面积为 $S_A$，棱 $AB$ 的二面角为 $\theta_{AB}$，以此类推，那么 $\triangle PBC$、$\triangle PBD$、$\triangle PCD$ 的面积比为 $S_D(1+\cos\theta_{BC}):S_C(1+\cos\theta_{BD}):S_B(1+\cos\theta_{CD})$。