均分田地，田埂最短问题

数学星空

对于N=5,根据楼上109# mathe提示重新修正了方程组：

我们记\(BC=2a,CD=CD'=2b,DJ=2c,DF=2d,\widehat{CD},\widehat{DI},\widehat{DF}\)对应的圆半径，圆弧度,圆周角分别\(R_0,2t_0,\theta_0,R_1,2t_1,\theta_1,R_2,2t_2,\theta_2\)

\(A\)到\(DD'\)的垂足记为\(H\)并令\(AH=x\),\(\angle ADH=t\)

我们有曲边三边形IFD的面积：\(S=2cd\sin(\frac{2\pi}{3}-t_1-t_2)+\frac{1}{2}(\frac{c}{\sin(t_1)})^2(2t_1-\sin(2t_1))+\frac{1}{2}(\frac{d}{\sin(t_2)})^2(2t_2-\sin(2t_2))+

\frac{1}{2}(\frac{2\pi}{3}-2(t_1+t_2)-\sin(\frac{2\pi}{3}-2(t_1+t_2))=\frac{\pi}{5}\)

曲边四边形BIDC的面积\(S=\frac{1}{2}(\frac{\pi}{3}+2(t_1-t_0))+c\sin(t_1)+\frac{x^2}{2}\frac{\cos(t)}{\sin(t)}-b(1-2a+x)\sin(\frac{\pi}{3}+t_0)-\frac{1}{2}(\frac{c}{\sin(t_1)})^2(2t_1-\sin(2t_1))+\frac{1}{2}(\frac{b}{\sin(t_0)})^2(2t_0-\sin(2t_0))=\frac{\pi}{5}\)

曲边五边形D'CDFF'的面积：\(S=2b^2\sin(\frac{2\pi}{3}+2t_0)+4[2b\sin(\frac{\pi}{3}+t_0)+d\cos(\frac{\pi}{2}-t_2-2t_0)]d\sin(\frac{\pi}{2}-t_2-2t_0)+\frac{1}{2}(4(t_2+t_0)-\sin(4t_2+4t_0))-(\frac{d}{\sin(t_2)})^2(2t_2-\sin(2t_2))-(\frac{b}{\sin(t_0)})^2(2t_0-\sin(2t_0))=\frac{\pi}{5}\)

在三角形CD'I中有：\(4b^2\sin(\frac{\pi}{3}+t_0)^2+x^2=1+4c^2-4c\cos(t_1)=1+4d^2-4d\cos(t_2)\)

对于D'点三条弧长半径倒数和为0有：\(\frac{\sin(t_0)}{b}+\frac{\sin(t_1)}{c}=\frac{\sin(t_2)}{d}\)

在三角形CDH中得到

\(x\cos(t)=2b\sin(\frac{\pi}{3}+t_0)\)

\(1-2a+x=2b\cos(\frac{\pi}{3}+t_0)\)

求解上面各个方程得到：

\(a =0 .445366231773976752082217753283, b =0 .125703193486028338536903388397, c =0 .400810373580294262765867719395, d = 0.455035788271461515582880131740\)

\( t = 0.0505122043270516188486948079902, t_0 = 0.0240799997132231123427553708159e, t_1 = 0.108762599328710501102437634466, t_2 =0 .211976978684874490113628985041\)

\(x = 0.0111569196689654896519789625679,\theta_0= 1.37968235416969^{\circ}, \theta_1= 6.23163791007230^{\circ}, \theta_2= 12.1453862319245^{\circ}\)

\( R_0 = 5.22073687273590, R_1 = 3.69246124938494, R_2 = 2.16278919549312\)

\(L=2a+4R_0t_0+4R_1t_1+4R_2t_2=4.83384664352739253889673598464\)

数学星空

\(N=8\)的构型猜测是如下图形

zeroieme

数学星空 发表于 2019-6-17 22:03
对于N=6,借用57# 给出的图形

N=6似乎99#结果更好。
N=8，觉得上下两细长的“6边形”可以成为不与圆周相交的内部5边形，靠近圆周部分合并成为左右部分的公共边。

mathe

正方形方案46#或36#模式打败47#模式

其中左图(46#)总长2.93994625160691，右图(47#)总长2.95599055531302
左图坐标:
        G(0.41416332015583,0.77339985066382)
        H(0.61979599979948,0.65851304019061)
        J(0.41416332015588,0.22660014933615)
        K(0.61979599979951,0.34148695980942)
        N(0.16634685271279,0.49999999999997)
        E(0.40304087778984,1.00000000000000)
        F(0.00000000000000,0.49999999999996)
        I(1.00000000000000,0.74627531842534)
        L(1.00000000000000,0.25372468157474)
        M(0.40304087778992,0.00000000000000)
        Theta(G=>H)=-0.08398801242370
        Theta(N=>G)=-0.21275474647953
        Theta(J=>N)=-0.21275474647960
        Theta(K=>J)=-0.08398801242368
        Theta(H=>K)=0.06988674220821
        Theta(K=>L)=0.22685601669499
        Theta(M=>J)=-0.04904464131952
        Theta(F=>N)=0.00000000000002
        Theta(E=>G)=0.04904464131964
        Theta(I=>H)=0.22685601669510
右图坐标
        G(0.52243964608848,0.70705864450124)
        H(0.47756035391152,0.29294135549876)
        J(0.45507281580503,0.63411451662830)
        K(0.54492718419496,0.36588548337170)
        E(0.48118555746766,1.00000000000000)
        F(0.00000000000000,0.67985611939168)
        I(1.00000000000000,0.64855768966193)
        L(1.00000000000000,0.32014388060832)
        M(0.00000000000000,0.35144231033807)
        N(0.51881444253235,0.00000000000000)
        Theta(J=>G)=-0.02171387696832
        Theta(K=>J)=-0.00000000000001
        Theta(H=>K)=0.02171387696832
        Theta(G=>I)=0.12189231452232
        Theta(N=>H)=0.13990707327684
        Theta(M=>H)=-0.12189231452231
        Theta(F=>J)=-0.10017843755400
        Theta(E=>G)=0.13990707327683
        Theta(L=>K)=-0.10017843755399

mathe

112#结果为 7.14747448265212, 所以远远不如104#的解
各坐标值为: (-0.28014501064343,0.23980771562651),(-0.14036749198807,0.00334374348045),(-0.29124585915831,-0.22619641240636)
(-0.16016077751332,0.98709094076804),(-0.94018148259430,0.34067400808806)
各圆弧弧度的一半为0.03407923799949，0.19598766199318，0.32005533926486

mathe

正方形分成7个区域先猜测一个对称的图

总长度3.28072646427294
各点坐标有: (0.50000000000000,0.77256804096864) (0.66644546360513,0.66400317016946) (1.00000000000000,0.73420538399341)
各条圆弧的圆周角弧度分别有0.05435998567898，0.10871997135796，0.20743940212017

正方形分成8个区域猜想如下左右对称的图:

总长度3.64544405543832
各点坐标有G(0.50000000000005,0.78650448163742)  H(0.65181072849233,0.69521642810606) K(0.69020529431282,0.40632922424041) J(0.63730164122633,0.34725087921656)
I(1.00000000000000,0.78190277811128)  L(1.00000000000000,0.37152655751760)  M(0.69582210643012,0.00000000000000)
各条圆弧的圆周角弧度分别有(G=>H)=-0.01779668021960  (H=>K)=0.16772405840176 (K=>J)=0.01702889085755 (M=>J)=0.16695626903967
(K=>L)=0.11187200961702 (I=>H)=0.24400270757952  (J=>V)=0.18968623751879 (V和J左右对称）

mathe

另外大家觉得单位圆和单位正方形均分为n份，分界线总长度最短，那么这两个这短长度在n趋向无穷时比例会是多少？

zeroieme

mathe 发表于 2019-6-20 11:58
112#结果为 7.14747448265212, 所以远远不如104#的解
各坐标值为: (-0.28014501064343,0.23980771562651), ...

我猜想圆8等份最优解应当是中心双5边两份，外围6份。

zeroieme

mathe 发表于 2019-6-20 14:37
另外大家觉得单位圆和单位正方形均分为n份，分界线总长度最短，那么这两个这短长度在n趋向无穷时比例会是多 ...

n趋向无穷就当正6边形拼接计算。
\(\frac{3 l}{\frac{3 \sqrt{3} l^2}{2}}\) 代入\(\frac{3 \sqrt{3} l^2}{2}=\frac{1}{n}\)

结果\(\sqrt{2} \sqrt[4]{3} \sqrt{n}\)
比例趋向无穷

zeroieme

mathe 发表于 2019-6-20 14:09
正方形分成7个区域先猜测一个对称的图

总长度3.28072646427294

8等份中心双5边形方案，就是7等份中央横切一刀。