1815年的蝴蝶定理（5楼高观下）7楼经典解析法 OK 37楼变式 - 初等数学讨论 - 悠闲数学娱乐论坛(第2版)

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发表于 2017-4-16 00:31 | 只看该作者

本帖最后由 abababa 于 2017-4-16 00:42 编辑

定理二：四边形$A_1A_2A_3A_4$一组对边$A_1A_2,A_3A_4$交于$P$，过$P$作割线交另两组对边$A_2A_3, A_4A_1, A_1A_3, A_2A_4$于$M_1,M_2,N_1,N_2$，则$\frac{1}{PM_1}+\frac{1}{PM_2} = \frac{1}{PN_1}+\frac{1}{PN_2}$

证明：设$A_1A_4 \cap A_2A_3 = T, A_1A_3 \cap A_2A_4 = B, TB \cap PM_2 = C, TB \cap A_1A_2 = D, PB \cap A_1A_4 = B_1, PB \cap A_2A_3 = B_2$。

于是$(PB,B_1B_2) \barwedge T(PB,B_1B_2) \barwedge (PC,M_1M_2) \barwedge T(PD,A_1A_2) \barwedge B(PD,A_1A_2) \barwedge (PC,N_1N_2)$，因此$(PC,M_1M_2) = (PC,N_1N_2) = (PB,B_1B_2) = -1$，所以$(PM_1,CM_2) = (PN_1,CN_2)$。

由$(PC,M_1M_2) = -1$知$\frac{CM_1}{PM_1} = -\frac{CM_2}{PM_2}$，所以$\frac{1}{PM_1} = -\frac{CM_2}{CM_1} \cdot \frac{1}{PM_2}$，所以$\frac{1}{PM_1}+\frac{1}{PM_2} = \frac{1}{PM_2} \cdot \frac{CM_1-CM_2}{CM_1} = \frac{1}{PM_2} \cdot \frac{M_2M_1}{CM_1}$。

由$(PC,N_1N_2) = -1$知$\frac{CN_1}{PN_1} = -\frac{CN_2}{PN_2}$，所以$\frac{1}{PN_1} = -\frac{CN_2}{CN_1} \cdot \frac{1}{PN_2}$，所以$\frac{1}{PN_1}+\frac{1}{PN_2} = \frac{1}{PN_2} \cdot \frac{CN_1-CN_2}{CN_1} = \frac{1}{PN_2} \cdot \frac{N_2N_1}{CN_1}$。

由$(PM_1,CM_2) = (PN_1,CN_2)$知$\frac{PC}{M_1C}:\frac{PM_2}{M_1M_2} = \frac{PC}{N_1C}:\frac{PN_2}{N_1N_2}$，所以$\frac{1}{PM_2} \cdot \frac{M_2M_1}{CM_1} = \frac{1}{PN_2} \cdot \frac{N_2N_1}{CN_1}$。

由以上三式知$\frac{1}{PM_1}+\frac{1}{PM_2} = \frac{1}{PN_1}+\frac{1}{PN_2}$

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abababa

42^#

发表于 2017-4-16 00:33 | 只看该作者

本帖最后由 abababa 于 2017-4-16 00:37 编辑

定理三：椭圆$\Gamma_1$含于椭圆$\Gamma_2$，过$\Gamma_1$内一点$T$作$\Gamma_2$的三条弦$A_{21}A_{22},B_{21}B_{22},P_{21}P_{22}$，分别在$\Gamma_1$上截出弦$A_{11}A_{12},B_{11}B_{12},P_{11}P_{12}$，直线$A_{12}B_{21} \cap P_{21}P_{22} = M_1, A_{22}B_{11} \cap P_{21}P_{22} = N_1,A_{11}B_{22} \cap P_{11}P_{12} = M_2, A_{21}B_{12} \cap P_{11}P_{12} = N_2$，则$(\frac{1}{TM_1}+\frac{1}{TN_1})-(\frac{1}{TP_{11}}+\frac{1}{TP_{21}}) = (\frac{1}{TM_2}+\frac{1}{TN_2})-(\frac{1}{TP_{12}}+\frac{1}{TP_{22}})$

证明：设$A_{12}B_{11}, A_{22}B_{21}, A_{11}B_{12}, A_{21}B_{22}$分别交$P_{11}P_{12}$于$E_1,F_1,E_2,F_2$。

由定理二知
\[\frac{1}{TE_1}+\frac{1}{TF_1} = \frac{1}{TM_1}+\frac{1}{TN_1}(*) \qquad\qquad \frac{1}{TE_2}+\frac{1}{TF_2} = \frac{1}{TM_2}+\frac{1}{TN_2}(**)\]

由定理一知
\[\text{在~} \Gamma_1 \text{~中有} \frac{1}{TP_{11}}-\frac{1}{TP_{12}} = \frac{1}{TE_1}-\frac{1}{TE_2} \qquad\qquad \text{在~} \Gamma_2 \text{~中有}\frac{1}{TP_{21}}-\frac{1}{TP_{22}} = \frac{1}{TF_1}-\frac{1}{TF_2} (***)\]

$(*)-(**)$并注意$(***)$得$(\frac{1}{TM_1}+\frac{1}{TN_1})-(\frac{1}{TM_2}+\frac{1}{TN_2}) = (\frac{1}{TP_{11}}+\frac{1}{TP_{21}})-(\frac{1}{TP_{12}}+\frac{1}{TP_{22}})$，即$(\frac{1}{TM_1}+\frac{1}{TN_1})-(\frac{1}{TP_{11}}+\frac{1}{TP_{21}}) = (\frac{1}{TM_2}+\frac{1}{TN_2})-(\frac{1}{TP_{12}}+\frac{1}{TP_{22}})$。

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abababa

43^#

发表于 2017-4-16 00:36 | 只看该作者

本帖最后由 abababa 于 2017-4-16 00:39 编辑

定理四：设椭圆$\Gamma_1$含于$\Gamma_2$含于$\Gamma_3$，过$\Gamma_1$内一点$K$作三条直线，分别截三椭圆$\Gamma_3,\Gamma_2,\Gamma_1$得弦$P_3T_3,P_2T_2,P_1T_1,A_3C_3,A_2C_2,A_1C_1,B_3D_3,B_2D_2,B_1D_1$，直线$A_1B_3,A_2B_2,A_3B_1,C_1D_3,C_2D_2,C_3D_1$分别交$P_3T_3$于$M_1,M_2,M_3,N_1,N_2N_3$，则
\[(\frac{1}{KM_1}+\frac{1}{KM_2}+\frac{1}{KM_3})-(\frac{1}{KP_1}+\frac{1}{KP_2}+\frac{1}{KP_3}) = (\frac{1}{KN_1}+\frac{1}{KN_2}+\frac{1}{KN_3})-(\frac{1}{KT_1}+\frac{1}{KT_2}+\frac{1}{KT_3})\]

证明：将定理一应用于$\Gamma_2$中的四边形$A_2B_2C_2D_2$得
\[\frac{1}{KM_2}-\frac{1}{KP_2} = \frac{1}{KN_2}-\frac{1}{KT_2}\]

再由定理三得
\[(\frac{1}{KM_1}+\frac{1}{KM_3})-(\frac{1}{KP_1}-\frac{1}{KP_3}) = (\frac{1}{KN_1}+\frac{1}{KN_3})-(\frac{1}{KT_1}+\frac{1}{KT_3})\]

两式相加即有
\[(\frac{1}{KM_1}+\frac{1}{KM_2}+\frac{1}{KM_3})-(\frac{1}{KP_1}+\frac{1}{KP_2}+\frac{1}{KP_3}) = (\frac{1}{KN_1}+\frac{1}{KN_2}+\frac{1}{KN_3})-(\frac{1}{KT_1}+\frac{1}{KT_2}+\frac{1}{KT_3})\]

一般地，当$n$个椭圆层层内含时，此定理仍成立，即为
\[\sum_{i=1}^{n}\frac{1}{KM_i}-\sum_{i=1}^{n}\frac{1}{KP_i} = \sum_{i=1}^{n}\frac{1}{KN_i}-\sum_{i=1}^{n}\frac{1}{KT_i}\]

当$n$为奇数时，对椭圆$\Gamma_{\frac{n+1}{2}}$应用定理一，对椭圆$\Gamma_i, \Gamma_{n+1-i} (i=1,2,\cdots,\frac{n-1}{2})$应用定理三。当$n$为偶数时只要有限次应用定理三。

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abababa

44^#

发表于 2017-4-16 00:40 | 只看该作者

本帖最后由 abababa 于 2017-4-16 00:41 编辑

定理五：完全四边形$ABCD$的一组对边$AC \cap BD = O$，过$O$作两直线$A'C', B'D'$分别交另两组对边于$A',C'$和$B',D'$，直线$A'D' \cap AC = E, B'C' \cap AC = F$，则$\frac{1}{OA}-\frac{1}{OC} = \frac{1}{OE}-\frac{1}{OF}$

证明：考察六边形$D'AA'C'CB'$中的三组对边的交点$D'A \cap C'C = D, AA' \cap CB' = B, A'C' \cap B'D' = O$，由于$DBO$共线，由 Pascal 定理知$D'AA'C'CB'$内接于圆锥曲线$\Gamma$，由定理一知$\frac{1}{OA}-\frac{1}{OC} = \frac{1}{OE}-\frac{1}{OF}$。

此即四边形中的 Candy 定理。当$OA = OC$时$OE = OF$，即为四边形中的蝴蝶定理。

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