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
![{\displaystyle \langle U\times V,X\times W,Y\times Z\rangle =0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b57b303b6dabe8abdac12570f4f11b5b15b3042e)
且
![{\displaystyle \langle U\times Z,X\times V,Y\times W\rangle =0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d8f07fe9e77deff2b0e93466f588a8f16a915ef)
则
![{\displaystyle \langle U\times W,X\times Z,Y\times V\rangle =0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/701c11ea775794daaee89d4b5d1177ec803fd96a)
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
![{\displaystyle \alpha =\langle U\times V,X\times W,Y\times Z\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f4df20ed753fd0dcdd8a237421410a89ba617bf)
![{\displaystyle \beta =\langle U\times Z,X\times V,Y\times W\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb557ba30c45172d88ebf8b023f80d5d4e11aa2c)
![{\displaystyle \gamma =\langle U\times W,X\times Z,Y\times V\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e9b42b5c595669f30d48fee5eef4661519ef3f0)
我们需要证明如果
= 0且
= 0,则
= 0。
第一步[编辑]
利用恒等式
![{\displaystyle \langle A,B,C\rangle =\langle C,A,B\rangle =\langle B,C,A\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/93d376cc47905d7184d2c8ce18aa8b889ce0f53e)
可将
、
及
表述为以下形式:
![{\displaystyle \alpha =\langle U\times V,X\times W,Y\times Z\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f4df20ed753fd0dcdd8a237421410a89ba617bf)
![{\displaystyle \beta =\langle Y\times W,U\times Z,X\times V\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/277a6549eb686f57b816c68fe102e6bdee4cc15b)
![{\displaystyle \gamma =\langle X\times Z,Y\times V,U\times W\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/71bbcef3c2655092bc1b0bebe27a14c3209e3bc3)
第二步[编辑]
利用恒等式
![{\displaystyle \langle A,B,C\rangle =A\cdot (B\times C)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76baaf3fe59406692c6385c469abf729667195f3)
![{\displaystyle A\times (B\times C)=(A\cdot C)B-(A\cdot B)C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e316a888fb86916f7f552cd2c5a49d2b9589a99)
可得
![{\displaystyle \alpha =(U\times V)\cdot ((X\times W)\times (Y\times Z))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/916c525bf3f80f95064dd9ee596a879bceea2cf8)
![{\displaystyle \beta =(Y\times W)\cdot ((U\times Z)\times (X\times V))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d01d67efec7a4a62649eb8b49a158833dc594b2)
![{\displaystyle \gamma =(X\times Z)\cdot ((Y\times V)\times (U\times W))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edba066e85cd0379d58b82ae509e468af08cea8d)
以及
![{\displaystyle \alpha =(U\times V)\cdot (\langle X,W,Z\rangle Y-\langle X,W,Y\rangle Z)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4aebf26e7f33dc32afc886195d2b73ddcb3f1846)
![{\displaystyle \beta =(Y\times W)\cdot (\langle U,Z,V\rangle X-\langle U,Z,X\rangle V)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/611b79e41e370fbb3677cc120dc13d23459c0bb0)
![{\displaystyle \gamma =(X\times Z)\cdot (\langle Y,V,W\rangle U-\langle Y,V,U\rangle W)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b3bdc648e17259518bd543c560e529ab78f7fec)
第三步[编辑]
利用数量积的分配律,可得:
![{\displaystyle \alpha =\langle X,W,Z\rangle \langle U,V,Y\rangle -\langle X,W,Y\rangle \langle U,V,Z\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1c1fc0081e97d4297c28ec1abc4b0424432f224)
![{\displaystyle \beta =\langle U,Z,V\rangle \langle Y,W,X\rangle -\langle U,Z,X\rangle \langle Y,W,V\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f90917d68532ac8091be8e41eb8e3f141318de3)
![{\displaystyle \gamma =\langle Y,V,W\rangle \langle X,Z,U\rangle -\langle Y,V,U\rangle \langle X,Z,W\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/66fecac1b69724da5549833c08fe539b6b190963)
第四步[编辑]
利用恒等式
![{\displaystyle \langle A,B,C\rangle =\langle C,A,B\rangle =\langle B,C,A\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/93d376cc47905d7184d2c8ce18aa8b889ce0f53e)
![{\displaystyle \langle A,B,C\rangle =-\langle A,C,B\rangle =-\langle C,B,A\rangle =-\langle B,A,C\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cee2d01f0a6411439722157651c2044a0645422)
可得:
![{\displaystyle \alpha =\langle X,W,Z\rangle \langle U,V,Y\rangle -\langle X,W,Y\rangle \langle U,V,Z\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1c1fc0081e97d4297c28ec1abc4b0424432f224)
![{\displaystyle \beta =-\langle U,Z,X\rangle \langle Y,W,V\rangle +\langle X,W,Y\rangle \langle U,V,Z\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a83d776b2c77ff2bc0c22c4d26e1a40c242831c9)
![{\displaystyle \gamma =\langle U,Z,X\rangle \langle Y,W,V\rangle -\langle X,W,Z\rangle \langle U,V,Y\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b49c29f8036ca002eadb6ff96b3cfdb159d57995)
第五步[编辑]
把这些等式相加,得:
![{\displaystyle \alpha +\beta +\gamma =0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46d08d3c4638952323fc98557fa51d003553f8f4)
![{\displaystyle \gamma =-(\alpha +\beta )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/324fc7f313e5307185e219b0358ddaa8647878b3)
因此,如果
= 0且
= 0,则
= 0。
证毕。