帕普斯定理

设U，V，W，X，Y和Z为平面上6条直线。如果： （1）U与V的交点，X与W的交点，Y与Z的交点共线，且 （2）U与Z的交点，X与V的交点，Y与W的交点共线， 则一定有（3）U与W的交点，X与Z的交点，Y与V的交点共线。这个定理叫做帕普斯六角形定理（英語：Pappus's hexagon theorem）。 也就是说， 如果 ⟨ U × V , X × W , Y × Z ⟩ = 0 {\displaystyle \langle U\times V,X\times W,Y\times Z\rangle =0} 且 ⟨ U × Z , X × V , Y × W ⟩ = 0 {\displaystyle \langle U\times Z,X\times V,Y\times W\rangle =0} 则 ⟨ U × W , X × Z , Y × V ⟩ = 0. {\displaystyle \langle U\times W,X\times Z,Y\times V\rangle =0.} 這個定理是帕斯卡定理的一個特例，當這個圓錐曲線退化成兩條直線的時候。 设 α = ⟨ U × V , X × W , Y × Z ⟩ {\displaystyle \alpha =\langle U\times V,X\times W,Y\times Z\rangle } β = ⟨ U × Z , X × V , Y × W ⟩ {\displaystyle \beta =\langle U\times Z,X\times V,Y\times W\rangle } γ = ⟨ U × W , X × Z , Y × V ⟩ {\displaystyle \gamma =\langle U\times W,X\times Z,Y\times V\rangle } 我们需要证明如果 α {\displaystyle \alpha } = 0且 β {\displaystyle \beta } = 0，则 γ {\displaystyle \gamma } = 0。 利用恒等式 ⟨ A , B , C ⟩ = ⟨ C , A , B ⟩ = ⟨ B , C , A ⟩ {\displaystyle \langle A,B,C\rangle =\langle C,A,B\rangle =\langle B,C,A\rangle } 可将 α {\displaystyle \alpha } 、 β {\displaystyle \beta } 及 γ {\displaystyle \gamma } 表述为以下形式： α = ⟨ U × V , X × W , Y × Z ⟩ {\displaystyle \alpha =\langle U\times V,X\times W,Y\times Z\rangle } β = ⟨ Y × W , U × Z , X × V ⟩ {\displaystyle \beta =\langle Y\times W,U\times Z,X\times V\rangle } γ = ⟨ X × Z , Y × V , U × W ⟩ {\displaystyle \gamma =\langle X\times Z,Y\times V,U\times W\rangle } 利用恒等式 ⟨ A , B , C ⟩ = A ⋅ ( B × C ) {\displaystyle \langle A,B,C\rangle =A\cdot (B\times C)} A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C {\displaystyle A\times (B\times C)=(A\cdot C)B-(A\cdot B)C} 可得 α = ( U × V ) ⋅ ( ( X × W ) × ( Y × Z ) ) {\displaystyle \alpha =(U\times V)\cdot ((X\times W)\times (Y\times Z))} β = ( Y × W ) ⋅ ( ( U × Z ) × ( X × V ) ) {\displaystyle \beta =(Y\times W)\cdot ((U\times Z)\times (X\times V))} γ = ( X × Z ) ⋅ ( ( Y × V ) × ( U × W ) ) {\displaystyle \gamma =(X\times Z)\cdot ((Y\times V)\times (U\times W))} 以及 α = ( U × V ) ⋅ ( ⟨ X , W , Z ⟩ Y − ⟨ X , W , Y ⟩ Z ) {\displaystyle \alpha =(U\times V)\cdot (\langle X,W,Z\rangle Y-\langle X,W,Y\rangle Z)} β = ( Y × W ) ⋅ ( ⟨ U , Z , V ⟩ X − ⟨ U , Z , X ⟩ V ) {\displaystyle \beta =(Y\times W)\cdot (\langle U,Z,V\rangle X-\langle U,Z,X\rangle V)} γ = ( X × Z ) ⋅ ( ⟨ Y , V , W ⟩ U − ⟨ Y , V , U ⟩ W ) {\displaystyle \gamma =(X\times Z)\cdot (\langle Y,V,W\rangle U-\langle Y,V,U\rangle W)} 利用数量积的分配律，可得： α = ⟨ X , W , Z ⟩ ⟨ U , V , Y ⟩ − ⟨ X , W , Y ⟩ ⟨ U , V , Z ⟩ {\displaystyle \alpha =\langle X,W,Z\rangle \langle U,V,Y\rangle -\langle X,W,Y\rangle \langle U,V,Z\rangle } β = ⟨ U , Z , V ⟩ ⟨ Y , W , X ⟩ − ⟨ U , Z , X ⟩ ⟨ Y , W , V ⟩ {\displaystyle \beta =\langle U,Z,V\rangle \langle Y,W,X\rangle -\langle U,Z,X\rangle \langle Y,W,V\rangle } γ = ⟨ Y , V , W ⟩ ⟨ X , Z , U ⟩ − ⟨ Y , V , U ⟩ ⟨ X , Z , W ⟩ {\displaystyle \gamma =\langle Y,V,W\rangle \langle X,Z,U\rangle -\langle Y,V,U\rangle \langle X,Z,W\rangle } 利用恒等式 ⟨ A , B , C ⟩ = ⟨ C , A , B ⟩ = ⟨ B , C , A ⟩ {\displaystyle \langle A,B,C\rangle =\langle C,A,B\rangle =\langle B,C,A\rangle } ⟨ A , B , C ⟩ = − ⟨ A , C , B ⟩ = − ⟨ C , B , A ⟩ = − ⟨ B , A , C ⟩ {\displaystyle \langle A,B,C\rangle =-\langle A,C,B\rangle =-\langle C,B,A\rangle =-\langle B,A,C\rangle } 可得： α = ⟨ X , W , Z ⟩ ⟨ U , V , Y ⟩ − ⟨ X , W , Y ⟩ ⟨ U , V , Z ⟩ {\displaystyle \alpha =\langle X,W,Z\rangle \langle U,V,Y\rangle -\langle X,W,Y\rangle \langle U,V,Z\rangle } β = − ⟨ U , Z , X ⟩ ⟨ Y , W , V ⟩ + ⟨ X , W , Y ⟩ ⟨ U , V , Z ⟩ {\displaystyle \beta =-\langle U,Z,X\rangle \langle Y,W,V\rangle +\langle X,W,Y\rangle \langle U,V,Z\rangle } γ = ⟨ U , Z , X ⟩ ⟨ Y , W , V ⟩ − ⟨ X , W , Z ⟩ ⟨ U , V , Y ⟩ {\displaystyle \gamma =\langle U,Z,X\rangle \langle Y,W,V\rangle -\langle X,W,Z\rangle \langle U,V,Y\rangle } 把这些等式相加，得： α + β + γ = 0 {\displaystyle \alpha +\beta +\gamma =0} γ = − ( α + β ) {\displaystyle \gamma =-(\alpha +\beta )} 因此，如果 α {\displaystyle \alpha } = 0且 β {\displaystyle \beta } = 0，则 γ {\displaystyle \gamma } = 0。 证毕。 布列安桑定理 帕斯卡定理 笛沙格定理