能否把这个公式推广到四面体

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$A_1A_{12}:A_1A_2=k_{12}$等等，
则$A_{12}A_{13},A_{23}A_{21},A_{31}A_{32}$围成的三角形的面积与$S_{A_1A_2
A_3}$之比为
\[\frac{(k_{12}k_{23}k_{31}(\sum k_{21}-1)+k_{21}k_{32}k_{13}(\sum k_{12}-1)-\sum k_{12}k_{13}k_{21}k_{31})^2}{\left|\prod(k_{12}k_{31}+k_{32}k_{13}-k_{12}k_{32})\right|}\]
分母为0当且仅当$A_{12}A_{13},A_{23}A_{21},A_{31}A_{32}$中有两者平行或重合
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平行四边形$A_1A_2A_3A_4$，$A_1A_{12}:A_1A_2=k_{12}$等等，
则$A_{12}A_{13},A_{23}A_{21},A_{34}A_{32},A_{41}A_{43}$围成的四边形的面积与$S_{A_1A_2
A_3A_4}$之比为
\[-\frac{k_{12}^2 k_{41} \left(k_{14}+k_{41}-1\right){}^2}{2 k_{14} \left(k_{12} k_{41}+k_{14} k_{43}\right)}+\frac{-k_{32}^2 k_{21}^2-k_{23} k_{32} k_{21}^2+k_{32} k_{21}^2+k_{32}^2 k_{21}-k_{23} k_{34} k_{21}+k_{23} k_{32} k_{34} k_{21}-k_{32} k_{34} k_{21}+k_{34} k_{21}+k_{23} k_{34}^2-k_{23} k_{32} k_{34}^2+k_{23} k_{32} k_{34}}{2 \left(k_{21} k_{32}+k_{23} k_{34}\right)}+\frac{\left(k_{32} \left(k_{12}+2 k_{34}-2\right)-2 k_{34}\right) k_{41}}{2 k_{32}}+\frac{k_{14} k_{23} k_{12}^2-k_{23} k_{12}^2+k_{23} k_{41} k_{12}^2-k_{14} k_{21} k_{12}-2 k_{14} k_{23} k_{12}+2 k_{14} k_{21} k_{23} k_{12}-k_{21} k_{23} k_{12}+k_{23} k_{12}+k_{21} k_{23} k_{32} k_{12}-k_{23} k_{32} k_{12}+k_{14} k_{21} k_{41} k_{12}-k_{14} k_{21}^2+k_{14} k_{21}+k_{14} k_{21}^2 k_{23}+k_{14} k_{23}-2 k_{14} k_{21} k_{23}+k_{14} k_{21}^2 k_{32}-k_{14} k_{21} k_{32}}{2 \left(k_{14} k_{21}+k_{12} k_{23}\right)}-\frac{k_{14} k_{12}-k_{12}-k_{14}+k_{14} k_{34}}{2 k_{14}}-\frac{\left(k_{14} k_{34}-k_{12} k_{32}\right) \left(k_{41}^2-2 k_{41}\right)}{2 k_{14} k_{32}}+\frac{k_{41} \left(k_{34} k_{32}-k_{32}-k_{34} k_{41}\right){}^2}{2 k_{32} \left(k_{34} k_{41}+k_{32} k_{43}\right)}\]
分母为0当且仅当$A_{12}A_{13},A_{23}A_{21},A_{34}A_{32},A_{41}A_{43}$中有两者平行或重合

hbghlyj

平行四边形的那个公式能不能化简呢

hbghlyj

设$P_1,P_2,P_3$分别在$A_2A_3,A_3A_1,A_1A_2$上,$A_2P_1=Q_1A_3=m_1,A_3P_2=Q_2A_1=m_2,A_1P_3=Q_3A_2=m_3,$由共角定理,$S_{A_1P_2P_3}=\frac{(a_2-m_2)m_3}{a_2a_3}S_{ABC}$,但$S_{P_1P_2P_3}=S_{ABC}-\sum S_{A_1P_2P_3}=S_{ABC}\left(1-\sum\frac{(a_2-m_2)m_3}{a_2a_3}\right)=S_{ABC}\left(1-\sum k_1+\sum k_2k_3\right)$
作代换$k_i\to 1-k_i$后不变,所以
两个三角形的顶点 ,都在一个已知三角形的边上,并且到边的中点距离相等,则这两个三角形面积相等.
作代换$k_i\to 1-k_{i+1}$后不变,所以
设三角形$P_1P_2P_3$内接于三角形$A_1A_2A_3$,又设$P_1Q_2$平行于$A_1A_2$,交$A_1A_3$于$Q_2$,等,则$S_{P_1P_2P_3}=S_{Q_1Q_2Q_3}$,特别地,设$P_1P_2P_3$共线,则$Q_1Q_2Q_3$共线.
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对于一楼的三角形，设
$M=k_{12}k_{23}k_{31}(\sum k_{21}-1)+k_{21}k_{32}k_{13}(\sum k_{12}-1)-\sum k_{12}k_{13}k_{21}k_{31}$
$N=k_{12} k_{31}-k_{12} k_{32}+k_{13} k_{32}$
作代换$k_i\to 1-k_i$后
$M=M'+\left(k_{12}+k_{21}-1\right) \left(k_{13}+k_{31}-1\right) \left(k_{23}+k_{32}-1\right)$
$N=N'+k_{13}+k_{31}-1$