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。
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