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福田网站建设罗湖网站建设,深圳手机商城网站设计制作,json做网站的数据库,英文网站建设用途符号约定 齐次坐标 \(a,b\) 等价#xff08;\(\exists \lambda, a \lambda b\)#xff09;记作 \(a\sim b\)所有的齐次坐标都记录为用圆括号包裹的三元组。#xff08;有的资料会把直线的齐次坐标记录为方括号包裹的三元组#xff09;#xff08;使用本文的记录方法可以更…符号约定齐次坐标\(a,b\)等价\(\exists \lambda, a \lambda b\)记作\(a\sim b\)所有的齐次坐标都记录为用圆括号包裹的三元组。有的资料会把直线的齐次坐标记录为方括号包裹的三元组使用本文的记录方法可以更突显点、线的代数共性而非几何区别Disargues 定理题目描述有不重合的\(6\)点\(A_1,A_2,B_1,B_2,C_1,C_2\)满足两两不三点共线则下面两个命题等价\(p_1:\)线\(A_1\times A_2, B_1\times B_2, C_1\times C_2\)三线共点。\(p_2:\)设线\(a_i B_i\times C_i, b_i C_i\times A_i, c_i A_i\times B_i\ (i\in\{1,2\})\)则点\(a_1\times a_2, b_1\times b_2, c_1\times c_2\)三点共线。证明观察发现\(p_1\Rightarrow p_2\)和\(p_1\Leftarrow p_2\)对偶所以只需证\(p_1\Rightarrow p_2\).现证\(p_1\Rightarrow p_2\).设\(A_1\times A_2,B_1\times B_2,C_1\times C_2\)的交点为\(P\)则\(A_1,B_1,C_1,P\)两两不三点共线若\(A_1,B_1,P\)三点共线由\(A_1,A_2,P\)三点共线则\(A_1,A_2,B_1\)三点共线与题目条件矛盾所以可设\[\begin{gathered} A_1 (1, 0, 0)\\ B_1 (0, 1, 0)\\ C_1 (0, 0, 1)\\ P (1, 1, 1) \end{gathered} \]由于\(P\)不与\(A_2,B_2,C_2\)重合否则将会出现三点共线所以设\(\lambda_1,\lambda_2,\lambda_3\)使得\[\begin{gathered} A_2 P \lambda_1A_1 (1 \lambda_1,\ 1,\ 1)\\ B_2 P \lambda_2B_1 (1,\ 1 \lambda_2,\ 1)\\ C_2 P \lambda_3C_1 (1,\ 1,\ 1 \lambda_3)\\ \end{gathered} \]则\[\begin{gathered} a_1 B_1 \times C_1 (1, 0, 0)\\ b_1 C_1 \times A_1 (0, 1, 0)\\ c_1 A_1 \times B_1 (0, 0, 1)\\ a_2 B_2 \times C_2 (\lambda_2\lambda_3 \lambda_2 \lambda_3,\ -\lambda_3,\ -\lambda_2)\\ b_2 C_2\times A_2 (-\lambda_3,\ \lambda_1\lambda_3 \lambda_1 \lambda_3,\ -\lambda_1)\\ c_2 A_2\times B_2 (-\lambda_2,\ -\lambda_1,\ \lambda_1\lambda_2 \lambda_1 \lambda_2) \end{gathered} \]于是\[\begin{gathered} a_1\times a_2 (0, \lambda_2, -\lambda_3)\\ b_1\times b_2 (-\lambda_1,0,\lambda_3)\\ c_1\times c_2 (\lambda_1, -\lambda_2, 0) \end{gathered} \]观察发现\(a_1\times a_2 b_1\times b_2 c_1\times c_2 0\)它们三个线性相关\(p_2\)得证。额外可以将\(\triangle A_2B_2C_2\in\)平面\(\beta\)看做\(\triangle A_1B_1C_1\in\)平面\(\alpha\)以点\(P\)为中心在\(\beta\)上的投影即可直观证明。Menelaus 定理和 Ceva 定理符号约定设点\(P_1 a, P_2 b, Q_1 a \lambda_1b, Q_2 a \lambda_2 b\)记\[(P_1P_2, Q_1Q_2) \frac{\lambda_1}{\lambda_2} \]实际上就是\(P_1,P_2,Q_1,Q_2\)的交比。问题描述有点\(P_1,P_2,P_3,Q_1,Q_2,Q_3,Q_1,Q_2,Q_3\)两两不重合点\(P_1,P_2,P_3\)不三点共线点\(Q_1,Q_2,Q_3\)三点共线点\(Q_1,Q_1\)在线\(P_2\times P_3\)上点\(Q_2,Q_2\)在线\(P_3\times P_1\)上点\(Q_3,Q_3\)在线\(P_1\times P_2\)上。设\[\begin{gathered} k_1 (P_2P_3,Q_1Q_1)\\ k_2 (P_3P_1,Q_2Q_2)\\ k_3 (P_1P_2,Q_3Q_3) \end{gathered} \]则有Menelaus 定理\(Q_1,Q_2,Q_3\)三点共线\(\Leftrightarrow k_1k_2k_3 1\).Ceva 定理\(P_1\times Q_1, P_2\times Q_2, P_3\times Q_3\)三线共点\(\Leftrightarrow\)\(k_1k_2k_3 -1\).证明设点\(P_1 (1, 0, 0), P_2 (0, 1, 0), P_3 (0, 0, 1)\)线\(Q_1\times Q_2 (u, v, w)\).则\[\begin{gathered} P_1\times P_2 (0, 0, 1)\\ P_2\times P_3 (1, 0, 0)\\ P_3\times P_1 (0, 1, 0) \end{gathered} \]求交点得\[\begin{gathered} Q_1 (0,-w,v) \sim P_2 - \frac{v}{w}P_3\\ Q_2 (-w,0,u) \sim P_3 - \frac{w}{u}P_1\\ Q_3 (-v,u,0) \sim P_1 - \frac{u}{v}P_2 \end{gathered} \]需要注意的是\(u,v,w\)均非\(0\)否则\(Q_1,Q_2,Q_3,P_1,P_2,P_3\)将会出现重合。Menelaus 定理正定理证明可设\(Q_1\times Q_2 (u,v,w)\)同理可得\(u,v,w\)均非\(0\)且\[\begin{gathered} Q_1 \sim P_2 - \frac{v}{w}P_3\\ Q_2 \sim P_3 - \frac{w}{u}P_1\\ Q_3 \sim P_1 - \frac{u}{v}P_2 \end{gathered} \]于是\[k_1k_2k_3 \frac{(-\frac{v}{w})(-\frac{w}{u})(-\frac{u}{v})}{(-\frac{v}{w})(-\frac{w}{u})(-\frac{u}{v})} 1 \]得证。Menelaus 逆定理证明设\(t_1,t_2,t_3\)使得\[\begin{gathered} Q_1 (0,1,t_1) \sim P_2 t_1P_3\\ Q_2 (t_2,0,1) \sim P_3 t_2P_1\\ Q_3 (1,t_3,0) \sim P_1 t_3P_2 \end{gathered} \]则\[k_1k_2k_3 \frac{(-\frac{v}{w})(-\frac{w}{u})(-\frac{u}{v})}{t_1t_2t_3} 1 \]即\[t_1t_2t_3 -1 \]计算\[\det(Q_1,Q_2,Q_3) \left|\begin{matrix} 0 t_2 1\\ 1 0 t_3\\ t_1 1 0 \end{matrix}\right| t_1t_2t_3 1 0 \]所以\(Q_1,Q_2,Q_3\)三点共线。得证。Ceva 定理正定理证明设\(P_1\times Q_1,P_2\times Q_2,P_3\times Q_3\)交点为\(E (1, 1, 1)\)因为\(P_1,P_2,P_3,E\)不三点共线否则将出现\(Q_1,Q_2,Q_3,P_1,P_2,P_3\)之间的重合所以存在射影变换将\(P_1,P_2,P_3,E\)映射为\((1,0,0),(0,1,0),(0,0,1),(1,1,1)\)则\[\begin{gathered} P_1\times Q_1 \sim P_1\times E (0, -1, 1)\\ P_2\times Q_2 \sim P_2\times E (1, 0, -1)\\ P_3\times Q_3 \sim P_3\times E (-1, 1, 0)\\ \end{gathered} \]求直线交点例如\(Q_1 \sim (P_2\times P_3)\times (P_1\times E)\)可得\[\begin{gathered} Q_1 (0, 1, 1) P_2 P_3\\ Q_2 (1, 0, 1) P_3 P_1\\ Q_3 (1, 1, 0) P_1 P_2\\ \end{gathered} \]于是\[k_1k_2k_3 \frac{(-\frac{v}{w})(-\frac{w}{u})(-\frac{u}{v})}{1\times 1\times 1} -1 \]得证。Ceva 定理逆定理证明设\(t_1,t_2,t_3\)使得\[\begin{gathered} Q_1 (0,1,t_1) \sim P_2 t_1P_3\\ Q_2 (t_2,0,1) \sim P_3 t_2P_1\\ Q_3 (1,t_3,0) \sim P_1 t_3P_2 \end{gathered} \]则\[k_1k_2k_3 \frac{(-\frac{v}{w})(-\frac{w}{u})(-\frac{u}{v})}{t_1t_2t_3} -1 \]即\[t_1t_2t_3 1 \]于是设\[\begin{gathered} l_1 P_1 \times Q_1 (0, -t_1, 1)\\ l_2 P_2 \times Q_2 (1, 0, -t_2)\\ l_3 P_3 \times Q_3 (-t_3, 1, 0)\\ \end{gathered} \]于是\(\det(l_1,l_2,l_3) -t_1t_2t_3 1 0\)故\(l_1,l_2,l_3\)交于一点。得证。额外Melelaus 定理可以利用相似轻松证明Ceva 定理可以利用面积法轻松证明。