等边六边形三线共点猜想

hujunhua

已知：六边形ABCDEF的六边等长，
试证明：三角形ACE和DFB的三对对应边中点连线共点。

mathe

hujunhua

二楼的反例说明猜想不成立。
的确不应该成立。
等边六边形容易仿射变换为一个一般六边形，但中点和三线共点是仿射不变的，
所以，如果对等边六边形成立，那么对一般六边形也应成立。
但对于一般六边形，构图三线不共点很显著。

aimisiyou

hujunhua 发表于 2024-3-7 21:47
二楼的反例说明猜想不成立。
的确不应该成立。
等边六边形容易仿射变换为一个一般六边形，但中点和三线共点 ...

差点我就信了。还好随意画个图，发现结论不对。

mathe

应该猜测是正确的，随机产生了两组高精度结果

1. ? gencase()
   
2. A=[0.E-115, 0.E-115]
   
3. B=[1.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000, 0.E-115]
   
4. C=[1.381850275666201968036446434110270592790773666315704857773927146527976835777665916036476336173946291, 0.9242241973534590190083094621090604933181603037872083917290885105574539139704279583565213690674598776]
   
5. D=[0.7236454410352553861671102299959019514193842353209913570854820850755299826813612096659545632745311826, 1.677063088264687291505776851439934517517317798213891183288753805639749246786399033298639228483101750]
   
6. E=[-0.2763545589647446138328897700040980485806157646790086429145179149244700173186387903340454367254688174, 1.677063088264687291505776851439934517517317798213891183288753805639749246786399033298639228483101750]
   
7. F=[-0.6582048346309465818693362041143686413713894309947135006884450614524468530963047063705217728994151079, 0.7528388909112282724974673893308740241991574944266827915596652950822953328159710749421178594156418720]
   
8. N=[0.3618227205176276930835551149979509757096921176604956785427410425377649913406806048329772816372655913, 0.8385315441323436457528884257199672587586588991069455916443769028198746233931995166493196142415508748, 1]
   
9. O=[0.3618227205176276930835551149979509757096921176604956785427410425377649913406806048329772816372655913, 0.8385315441323436457528884257199672587586588991069455916443769028198746233931995166493196142415508748, 1]
   
10. ? gencase()
    
11. A=[0.E-115, 0.E-115]
    
12. B=[1.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000, 0.E-115]
    
13. C=[1.420717849240105919075815073879202884039792248173960512887904662690624892908706611236488029180884277, 0.9071915405970116889397117940967804345185283123433039514347066674576359106831368367496276676432150199]
    
14. D=[1.037465194827528186138376558848505456538953376012649215260561239872068715680071799430651659244128707, 1.830835088122632235658233718365070922747442448343009830134246769420472223760322669798202234498960602]
    
15. E=[0.03746519482752818613837655884850545653895337601264921526056123987206871568007179943065165924412870675, 1.830835088122632235658233718365070922747442448343009830134246769420472223760322669798202234498960602]
    
16. F=[-0.3832526544125777329374385150306974275008388721613112976273434228185561772286348118058363699367555704, 0.9236435475256205467185219242682904882289141359997058786995401019628363130771858330485745668557455822]
    
17. N=[0.5187325974137640930691882794242527282694766880063246076302806199360343578400358997153258296220643534, 0.9154175440613161178291168591825354613737212241715049150671233847102361118801613348991011172494803011, 1]
    
18. O=[0.5187325974137640930691882794242527282694766880063246076302806199360343578400358997153258296220643534, 0.9154175440613161178291168591825354613737212241715049150671233847102361118801613348991011172494803011, 1]
    

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nyy

hujunhua猜想！
我感觉这个猜想很不错，确实有可能是成立的！

mathe

不知道换成空间六边形是否还成立，至少我选择了正方体六条封闭的边，竟然在空间中也相交到一点，就是正方体中心。

mathe

空间六边形不行

mathe

以下用大写字母表示位置矢量，黑体小写字母表示自由向量。
  
如图，设等边六边形`ABCDEF`的顶点`A`在原点，边长为1. 即`A=\boldsymbol {0}`,
记`B=\boldsymbol{a}`, 六边形其余 5 边对应向量依次为\(\boldsymbol{b},\boldsymbol{c},\boldsymbol{d},\boldsymbol{e},\boldsymbol{f}\), 它们长度都为1，而且\(\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}+\boldsymbol{d}+\boldsymbol{e}+\boldsymbol{f}=\boldsymbol{0}\)
于是六边形`ABCDEF`的其余4个顶点依次为\(C=\boldsymbol{a}+\boldsymbol{b},D=\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}, E=-\boldsymbol{e}-\boldsymbol{f},F=-\boldsymbol{f}\)
记六角星的各边中点为`GHIJKL`, 即有\(\begin{cases}
2G=F+B=\boldsymbol{a}-\boldsymbol{f}\\
2H=A+C=\boldsymbol{a}+\boldsymbol{b}\\
2I=B+D = 2\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}=(\boldsymbol{a}-\boldsymbol{d}-\boldsymbol{e}-\boldsymbol{f})\\
2J=C+E=\boldsymbol{a}+\boldsymbol{b}-\boldsymbol{e}-\boldsymbol{f}\\
2K=D+F=\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}-\boldsymbol{f}\\
2L=E+A= -\boldsymbol{e}-\boldsymbol{f}\\

\end{cases}\)
容易看到六角星对边中点连线的向量`2(J-G)=\boldsymbol{b}-\boldsymbol{e}, 2(L-I)=\boldsymbol{d}-\boldsymbol{a}, 2(H-K)=\boldsymbol{f}-\boldsymbol{c}`
设GJ和KH的交点为
\(M=G+u(J-G)=K+v(H-K)\),`u,v`为实数。
立得       \((\boldsymbol{b}-\boldsymbol{e})u=(\boldsymbol{f}-\boldsymbol{c})v+(\boldsymbol{b}+\boldsymbol{c})\)
两边点乘`(\boldsymbol{c}+\boldsymbol{f})`（消去`v`）可得
\((\boldsymbol{c}+\boldsymbol{f})·(\boldsymbol{b}-\boldsymbol{e})u= (\boldsymbol{c}+\boldsymbol{f})·(\boldsymbol{b}+\boldsymbol{c})\)
若\((\boldsymbol{c}+\boldsymbol{f})·(\boldsymbol{b}-\boldsymbol{e})≠0\)    则可得    \(\D u = \frac{(\boldsymbol{c}+\boldsymbol{f})·(\boldsymbol{b}+\boldsymbol{c})}{(\boldsymbol{c}+\boldsymbol{f})·(\boldsymbol{b}-\boldsymbol{e})}\)
同样设GJ和IL的交点为
\(N=G+u'(J-G)=I+w(L-I)\), `u',w`为实数，
仿上可得       \(\D u'=\frac{(\boldsymbol{d}+\boldsymbol{e})·(\boldsymbol{a}+\boldsymbol{d})}{(\boldsymbol{e}-\boldsymbol{b})·(\boldsymbol{a}+\boldsymbol{d})}\)

只要能够证明$u=u'$，那么就可以得到M和N重叠，从而证明本题。

注意到    \(\boldsymbol{b}+\boldsymbol{c}+\boldsymbol{f} = -(\boldsymbol{a}+\boldsymbol{d}+\boldsymbol{e})\to(\boldsymbol{b}+\boldsymbol{c}+\boldsymbol{f})^2=(\boldsymbol{a}+\boldsymbol{d}+\boldsymbol{e})^2\)(*平方指内积自乘，下同*）
由于\(\boldsymbol{b}^2=\boldsymbol{c}^2=\boldsymbol{f}^2=\boldsymbol{a}^2=\boldsymbol{d}^2=\boldsymbol{e}^2\)
两边消去后即得      \(\boldsymbol{b}·\boldsymbol{c}+\boldsymbol{c}·\boldsymbol{f}+\boldsymbol{b}·\boldsymbol{f}=\boldsymbol{a}·\boldsymbol{d}+\boldsymbol{d}·\boldsymbol{e}+\boldsymbol{a}·\boldsymbol{e}\)
所以    \((\boldsymbol{c}+\boldsymbol{f})·(\boldsymbol{b}+\boldsymbol{c})=1+\boldsymbol{b}·\boldsymbol{c}+\boldsymbol{c}·\boldsymbol{f}+\boldsymbol{b}·\boldsymbol{f}=1+\boldsymbol{a}·\boldsymbol{d}+\boldsymbol{d}·\boldsymbol{e}+\boldsymbol{a}·\boldsymbol{e}=(\boldsymbol{d}+\boldsymbol{e})·(\boldsymbol{a}+\boldsymbol{d})\)
另外一方面
\((\boldsymbol{a}+\boldsymbol{d}+\boldsymbol{c}+\boldsymbol{f})·(\boldsymbol{b}-\boldsymbol e)=-(\boldsymbol{b}+\boldsymbol e)·(\boldsymbol{b}-\boldsymbol e)=0\)
\(\to(\boldsymbol{c}+\boldsymbol{f})·(\boldsymbol{b}-\boldsymbol{e})=(\boldsymbol{e}-\boldsymbol{b})·(\boldsymbol{a}+\boldsymbol{d})\)
由此我们得到  \(\D u =u'\)

即M和N重叠。

数学星空

三条对边中点连接的长度均为两个内接三角形A1B1C1和A2B2C2的外接圆半径R,等六边形的边长也为R