论坛 › 数学欣赏 › 反演会保持田埂最短的特性吗？

mathe 发表于 2019-4-3 15:20:32

改用Pari/gp可以计算获得:
t=1.145869754950095508247099850069722495324714873543556586284342371924883601241352833592462102584559613
d=0.2209696780891027977378513676362321110743346941807911312188610942298283317783824735498763241214062325
然后
I=
F=
B=[-0.05049072689722888952170518173930602603539630333402870218696573616533218839983143172228601348066893592, 0.9987245298366258947940528940386819558149148013150071278173336181831481433060257309566067346523348876]
C=[-0.005516997341733283071050005947320227288672836943175290939750257551052317406672856306266778809573949534, 0.1091281689694765362720605584799146840444020451405389291319828382217985837518922661795789409501447797]
D=
C1=
C2=[2.373238735071780516122518462665177432016514277997764994381435898296641075515169341887236268252632947,
-3.000334645429469396166805981966068437547593604613223408672154189470610668722859382440929592887206127]
C3=
最终弧长4.833846643527396783657715928557596371715336638885537942540338590143011601591030504537006038698250498
rotate60(x)=
{
    local(y);
    y=vector(2);
   y=x*0.5+x*sqrt(3)/2;
   y=-sqrt(3)/2*x+0.5*x;
   y
}

get_circle_line1(cc, l)=
{
    local(c1,c2,c3,delt);
    local(u,v,w,a,b,c,y2);
    u=cc;v=cc;w=cc;
    a=l;b=l;c=l;
    c1=(b*b)/(a*a)+1.0;
    c2=-b*u/a+v+2*b*c/(a*a);
    c3=-c*u/a+w+c*c/(a*a);
    delt = sqrt(c2*c2-4*c1*c3);
    y2=(-c2+delt)/(2*c1);
   [-b/a*y2-c/a,y2]
}

get_circle_line2(cc, l)=
{
    local(c1,c2,c3,delt);
    local(u,v,w,a,b,c,y2);
    u=cc;v=cc;w=cc;
    a=l;b=l;c=l;
    c1=(b*b)/(a*a)+1.0;
    c2=-b*u/a+v+2*b*c/(a*a);
    c3=-c*u/a+w+c*c/(a*a);
    delt = sqrt(c2*c2-4*c1*c3);
    y2=(-c2-delt)/(2*c1);
   [-b/a*y2-c/a,y2]
}

get_circle_circle2(c1, c2)=
{
    get_circle_line2(c1,c2-c1)
}

get_circle_circle1(c1, c2)=
{
    get_circle_line1(c1,c2-c1)
}

area(a,b)=
{
    (a*b-a*b)/2
}

atan2(s,h)=
{
    local(tv);
    tv=atan(s/h);
    if(h>0, tv,
         if(tv<0, tv+Pi, tv-Pi)
    )
}

get_arc_area(c,f,t)=
{
   local(rs,da);
   rs=(f-c)*(f-c)+(f-c)*(f-c);
   da=atan2(t-c,t-c)-atan2(f-c,f-c);
   rs*(da-sin(da))/2.0
}

get_arc_length(c, f, t)=
{
   local(rs,da);
      rs=(f-c)*(f-c)+(f-c)*(f-c);
      da=atan2(t-c,t-c)-atan2(f-c,f-c);
      sqrt(rs)*da
}

start_with_c(d,t,c)=
{
   local(u,tv,tv1,tv2,II,J);
   local(rc,b,s1,s2);
   u=(d+1/d)/2.0;tv=(1/d-d)/2.0*tan(t);
   tv2=(1/d-d)/2.0*tan(t-Pi/3);
   tv1=(1/d-d)/2.0*tan(t+Pi/3);
   J=get_circle_circle2(,[-2*u,-2*tv2,-d*d+2*d*u]);
   II=get_circle_circle1(,[-2*u,-2*tv1,-d*d+2*d*u]);
   rc=sqrt(c*c+c*c);b=c/rc;
   s1=area(b,c)+area(c,)+area(,II)+area(II,b);
   s1+=get_arc_area(,c,);
   s1+=get_arc_area(,,II);
   s1+=get_arc_area(,II,b);
   s2=area(,J)+area(J,II)+area(II,);
   s2+=get_arc_area(,,J);
   s2+=get_arc_area(,J,II);
   s2+=get_arc_area(,II,);
   
}

start_out_with_c(d,t,c)=
{
   local(u,tv,tv1,tv2,II,J);
   local(rc,b,s1,s2,alen);
   u=(d+1/d)/2.0;tv=(1/d-d)/2.0*tan(t);
   tv2=(1/d-d)/2.0*tan(t-Pi/3);
   tv1=(1/d-d)/2.0*tan(t+Pi/3);
   J=get_circle_circle2(,[-2*u,-2*tv2,-d*d+2*d*u]);
   II=get_circle_circle1(,[-2*u,-2*tv1,-d*d+2*d*u]);
   print("I=" II);
   print("F=" J);
   rc=sqrt(c*c+c*c);b=c/rc;
   print("B=" b);
   print("C="c);
   print("d="d);
   printf("u="u);
   print("tv0=" tv);
   print("tv1=" tv1);
   print("tv2=" tv2);
   alen=1.0-rc;
   alen+=2.0*abs(get_arc_length(,c,));
   alen+=2.0*abs(get_arc_length(,,II));
   alen+=2.0*abs(get_arc_length(,,J));
   printf("len="alen);
   s1=area(b,c)+area(c,)+area(,II)+area(II,b);
   s1+=get_arc_area(,c,);
   s1+=get_arc_area(,,II);
   s1+=get_arc_area(,II,b);
   s2=area(,J)+area(J,II)+area(II,);
   s2+=get_arc_area(,,J);
   s2+=get_arc_area(,J,II);
   s2+=get_arc_area(,II,);
   
}

start_with_f2(d,t, f2)=
{
    local(u,tv,c);
    u=(d+1/d)/2.0;tv=(1/d-d)/2.0*tan(t);
    c=get_circle_circle2([-2*u,-2*tv,-d*d+2*d*u], [-2*f2,-2*f2,0]);
    start_with_c(d,t,c)
}

start_out_with_f2(d,t, f2)=
{
    local(u,tv,c);
    u=(d+1/d)/2.0;tv=(1/d-d)/2.0*tan(t);
    c=get_circle_circle2([-2*u,-2*tv,-d*d+2*d*u], [-2*f2,-2*f2,0]);
    start_out_with_c(d,t,c)
}

scan_one_config(d,t)=
{
    local(u,v,tt,f2);
    u=(d+1.0/d)/2.0;v=(1/d-d)/2.0;
    tt=tan(t);
    f2=rotate60();
    start_with_f2(d,t,f2)
}

output_data(d,t)=
{
    local(u,v,tt,f2);
    u=(d+1.0/d)/2.0;v=(1/d-d)/2.0;
    tt=tan(t);
    f2=rotate60();
    start_out_with_f2(d,t,f2)
}
然后比如调用
find_d(t)=solve(d=0.20,0.24,scan_one_config(d,t)-Pi/5)
solve(t=1.14,1.15,scan_one_config(find_d(t),t)-Pi/5)

数学星空 发表于 2019-4-7 21:52:41

mathe 发表于 2019-3-31 12:39
如果不考虑面积限制，那么仅仅给出所有内点的坐标和边的连接关系，我们就应该可以确定每条边的曲率。
假设 ...
https://bbs.emath.ac.cn/forum.php?mod=attachment&aid=ODkwOHxiMWZmZTkxNHwxNTU0NjgxMzg2fDIwfDE2MTA1&noupdate=yes

由于代数计算的表达式比较复杂，我做了下列的数值计算：

\(A,B,C,O,B',C',F,F'\)

过\(AOF',BOF',COF'\)的圆分别为

\((x-x_{21})^2+(y-y_{21})^2=r_1^2\)

\((x-x_{22})^2+(y-y_{22})^2=r_2^2\)

\((x-x_{23})^2+(y-y_{23})^2=r_3^2\)


取坐标\(x_0 = 0, x_1 = 0, y_0 = 0, y_1 = 1, x_2 = -2, y_2 = -\frac{1}{2}, x_3 = \frac{1}{3}, y_3 = -\frac{2}{3}\)


可以求得：

\(r_1 = 1.563789149, r_2 = 22.01863033, r_3 = 1.460091708, x_0 = 0, x_{00} = -2.525004700, x_{01}= -.3703591928, x_1 = 0, x_{11} = 0, x_{12}= -\frac{8}{17}, x_{13}= \frac{3}{5}, x_2 = -2, x_{21} = -1.481700544, x_{22} = 4.334447332, x_{23} = -1.096022181, x_3 =\frac{1}{3}, y_0 = 0, y_{00}= -0.6648832305, y_{01}= -00.9752283492, y_1 = 1, y_{11} = 1, y_{12} = -\frac{2}{17}, y_{13} = -\frac{6}{5}, y_2 = -\frac{1}{2}, y_{21}= 0.5000000000, y_{22}= -21.58778933, y_{23}= -0.9646777573, y_3 = -\frac{2}{3}\)


即：经过AOF',BOF',COF'的三个圆：

\((x+1.481700544)^2+(y-0.5000000000)^2=2.445436503\) ,\(\theta_1=0.6509021186=37.29394426^{\circ}, \widehat{AO}= 1.017873670,AO=1,r_1=1.563789149\)

\((x-4.334447332)^2+(y+21.58778933)^2=484.8200816\),\(\theta_2 =0.09366189084= 5.366431045^{\circ},\widehat{BO}= 2.062306550,BO=2.061552813,r_2=22.01863033\)

\((x+1.096022181)^2+(y+0.9646777573)^2=2.131867796\),\(\theta_3 =0.5161977674=29.57595346^{\circ}, \widehat{CO}= 2.288803009,CO=0.7453559923,r_3=1.460091708\)


注：\(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{r_3}=\frac{1}{1.563789149}+\frac{1}{22.01863033}-\frac{1}{1.460091708}=0\)


画图：

数学星空 发表于 2019-4-8 18:45:11

对于一般三角形，若存在内点P及三条分割线圆弧\(\widehat{AP},\widehat{BP},\widehat{CP}\)将三角形分成面积相等的三等分,且使其分割线总长最小，根据mathe得到的条件：

设\(A,B,C,P\),取\(x_2=0,y_2=0,x_3=a,y_3=0\),三角形三边长及面积为\(a,b,c,s\),三条分割线半径及圆弧角\(r_1,r_2,r_3,2t_1,2t_2,2t_3\)，

线段长\(AP=x,BP=y,CP=z\)

则有下列方程：

\(\sin(t_1)yz+\sin(t_2)xz+\sin(t_3)xy=0\)

\(12ay_0\sin(t_2)^2\sin(t_3)^2-3y^2\sin(2t_2)\sin(t_3)^2+3z^2\sin(2t_3)\sin(t_2)^2-8s\sin(t_2)^2\sin(t_3)^2-6z^2t_3\sin(t_2)^2+6y^2t_2\sin(t_3)^2=0\)

\(-6\sin(t_1)^2\sin(t_2)^2a^2y_0+6\sin(t_1)^2\sin(t_2)^2b^2y_0-6\sin(t_1)^2\sin(t_2)^2c^2y_0-3x^2\sin(2t_1)\sin(t_2)^2a+3y^2\sin(2t_2)\sin(t_1)^2a-8s\sin(t_1)^2\sin(t_2)^2a+24x_0s\sin(t_1)^2\sin(t_2)^2-6y^2t_2\sin(t_1)^2a+6x^2t_1\sin(t_2)^2a=0\)

\(\sin(t_1)a^2yz-2\sin(t_1)yzx_0a-\sin(t_1)b^2yz+\sin(t_1)c^2yz-2\sin(t_2)xzx_0a+2\sin(t_3)xya^2-2\sin(t_3)xyx_0a+2\cos(t_1)yzy_0a+2\cos(t_2)xzy_0a+2\cos(t_3)xyy_0a-4\cos(t_1)yzs=0\)

\(2\sin(t_1)xyy_0a+2\sin(t_1)yzy_0a+2\sin(t_2)xzy_0a-\cos(t_1)a^2yz+2\cos(t_1)yzx_0a+\cos(t_1)b^2yz-\cos(t_1)c^2yz+2\cos(t_2)xzx_0a-2\cos(t_3)xya^2+2\cos(t_3)xyx_0a-4\sin(t_1)xys-4\sin(t_1)yzs=0\)

\(-x_0^2+y^2-y_0^2=0\)

\(-a^2+2ax_0-x_0^2-y_0^2+z^2=0\)

\(a^2x_0-ac^2+ax^2-ax_0^2-ay_0^2-b^2x_0+c^2x_0+4sy_0=0\)


取\(a=5,b=4,c=3,s=6\)得到


sin(t1)*y*z+sin(t2)*x*z+sin(t3)*x*y, 60*y0*sin(t2)^2*sin(t3)^2-3*y^2*sin(2*t2)*sin(t3)^2+3*z^2*sin(2*t3)*sin(t2)^2-48*sin(t2)^2*sin(t3)^2-6*z^2*t3*sin(t2)^2+6*y^2*t2*sin(t3)^2, -108*sin(t1)^2*sin(t2)^2*y0-15*x^2*sin(2*t1)*sin(t2)^2+15*y^2*sin(2*t2)*sin(t1)^2-240*sin(t1)^2*sin(t2)^2+144*x0*sin(t1)^2*sin(t2)^2-30*y^2*t2*sin(t1)^2+30*x^2*t1*sin(t2)^2, 18*sin(t1)*y*z-10*sin(t1)*y*z*x0-10*sin(t2)*x*z*x0+50*sin(t3)*x*y-10*sin(t3)*x*y*x0+10*cos(t1)*y*z*y0+10*cos(t2)*x*z*y0+10*cos(t3)*x*y*y0-24*cos(t1)*y*z, 10*sin(t1)*x*y*y0+10*sin(t1)*y*z*y0+10*sin(t2)*x*z*y0-18*cos(t1)*y*z+10*cos(t1)*y*z*x0+10*cos(t2)*x*z*x0-50*cos(t3)*x*y+10*cos(t3)*x*y*x0-24*sin(t1)*x*y-24*sin(t1)*y*z, -x0^2+y^2-y0^2, -x0^2-y0^2+z^2+10*x0-25, 5*x^2-5*x0^2-5*y0^2+18*x0+24*y0-45



得不到我们想要的解,谁有兴趣解一下上面的方程组看能否得到非平凡解？

我用maple只得到了下列平凡解：

{t1 = 0, t2 = t2, t3 = 0, x = 4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = 0, t2 = t2, t3 = Pi, x = 4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = 0, t2 = t2, t3 = 0, x = 4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = 0, t2 = t2, t3 = Pi, x = 4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = 0, t2 = t2, t3 = 0, x = -4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = 0, t2 = t2, t3 = Pi, x = -4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = 0, t2 = t2, t3 = 0, x = -4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = 0, t2 = t2, t3 = Pi, x = -4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = Pi, t2 = 0, t3 = 0, x = 4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = Pi, t2 = 0, t3 = Pi, x = 4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = Pi, t2 = 0, t3 = 0, x = 4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = Pi, t2 = 0, t3 = Pi, x = 4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = Pi, t2 = 0, t3 = 0, x = -4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = Pi, t2 = 0, t3 = Pi, x = -4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = Pi, t2 = 0, t3 = 0, x = -4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = Pi, t2 = 0, t3 = Pi, x = -4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = Pi, t2 = Pi, t3 = 0, x = 4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = Pi, t2 = Pi, t3 = Pi, x = 4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = Pi, t2 = Pi, t3 = 0, x = 4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = Pi, t2 = Pi, t3 = Pi, x = 4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = Pi, t2 = Pi, t3 = 0, x = -4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = Pi, t2 = Pi, t3 = Pi, x = -4, x0 = 5, y = 5, y0 = 0, z = 0}, {t1 = Pi, t2 = Pi, t3 = 0, x = -4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = Pi, t2 = Pi, t3 = Pi, x = -4, x0 = 5, y = -5, y0 = 0, z = 0}, {t1 = 0, t2 = Pi, t3 = Pi, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = 4}, {t1 = 0, t2 = Pi, t3 = Pi, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = -4}, {t1 = 0, t2 = Pi, t3 = Pi, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = 4}, {t1 = 0, t2 = Pi, t3 = Pi, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = -4}, {t1 = Pi, t2 = Pi, t3 = Pi, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = 4}, {t1 = Pi, t2 = Pi, t3 = Pi, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = -4}, {t1 = Pi, t2 = Pi, t3 = Pi, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = 4}, {t1 = Pi, t2 = Pi, t3 = Pi, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = -4}, {t1 = t1, t2 = 0, t3 = t3, x = 3, x0 = 0, y = 0, y0 = 0, z = 5}, {t1 = t1, t2 = 0, t3 = t3, x = 3, x0 = 0, y = 0, y0 = 0, z = -5}, {t1 = t1, t2 = 0, t3 = t3, x = -3, x0 = 0, y = 0, y0 = 0, z = 5}, {t1 = t1, t2 = 0, t3 = t3, x = -3, x0 = 0, y = 0, y0 = 0, z = -5}, {t1 = t1, t2 = Pi, t3 = t3, x = 3, x0 = 0, y = 0, y0 = 0, z = 5}, {t1 = t1, t2 = Pi, t3 = t3, x = 3, x0 = 0, y = 0, y0 = 0, z = -5}, {t1 = t1, t2 = Pi, t3 = t3, x = -3, x0 = 0, y = 0, y0 = 0, z = 5}, {t1 = t1, t2 = Pi, t3 = t3, x = -3, x0 = 0, y = 0, y0 = 0, z = -5}, {t1 = 0, t2 = 0, t3 = t3, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = 4}, {t1 = 0, t2 = 0, t3 = t3, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = -4}, {t1 = 0, t2 = 0, t3 = t3, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = 4}, {t1 = 0, t2 = 0, t3 = t3, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = -4}, {t1 = Pi, t2 = 0, t3 = t3, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = 4}, {t1 = Pi, t2 = 0, t3 = t3, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = -4}, {t1 = Pi, t2 = 0, t3 = t3, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = 4}, {t1 = Pi, t2 = 0, t3 = t3, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = -4}, {t1 = 0, t2 = t2, t3 = 0, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = -4}, {t1 = 0, t2 = t2, t3 = 0, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = 4}, {t1 = 0, t2 = t2, t3 = 0, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = 4}, {t1 = 0, t2 = t2, t3 = 0, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = -4}, {t1 = Pi, t2 = t2, t3 = 0, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = 4}, {t1 = Pi, t2 = t2, t3 = 0, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = -4}, {t1 = Pi, t2 = t2, t3 = 0, x = 0, x0 = 9/5, y = -3, y0 = 12/5, z = 4}, {t1 = Pi, t2 = t2, t3 = 0, x = 0, x0 = 9/5, y = 3, y0 = 12/5, z = -4}

mathe 发表于 2019-6-18 19:59:41

11#对10#图中数据可以向右稍微旋转使得B处于纵轴，于是坐标变为
B(0,1),C(0,0.1092675364520465140352283787768904246535113380440663906763348446700065850023732609300919223564997222)
D(0.2206878378576897663569673248112808422866921116956422331187067862126079562531909204630138717256754545, -0.01115691966896547183626964551952168282785040319960129201709452306440172872203567137867115305352654314)
I(0.9379126148744844910385797339873341343788882774991827035898869657385926713080225693953147982993571409, 0.3468716287898261267227998511542798623527515370442986039409660232570123275013181931872817051193589951)
F(0.4547700088856559616111703127201680254505237443541252792063195172758473943153882024128356229047011978, -0.8906089147421220360537224362745621948518967999090101949369497842495659945986607479557887773005302673)
三个圆心坐标
C1(2.610368436367958404491284525187488486313546545982811104271424444318817351354460679268433961355551524, 4.630558294715476348667266375922776916044921485666528588048711196265485096214318316314736350816818757)
C2(2.218722662691959343725748235434943440093694446686795092887541130499745103718051219155795347965168932, -3.116334356943520698030665541027137812969097066322487497901618589349170714734464224947952171197020806)
C3(2.380969347099767820499509576705035847020807724547277542936172293918429808539028384921439770937359424, 0.09296274691008334347865263062126008379189849264522718286596198002390310003383896041356608598290436626)
于是圆弧C1半径长度为5.220736872735916808982569050374976972627093091965622208542848888637634702708921358536867922711103048
圆周角0.02407999971322311193944800249462247151284442940032043723155351790056746300947915211226570987886739062
圆弧C2的半径为 3.692461249384952539304889799710501292434674355165793762723516798839728663406426958340264090496172891
圆周角为0.1087625993287105023487822084588600318639191552613317169251111017622426231545391099031120010746490995
圆弧C3的半径为2.162789195493116676321516345655395632975270692489854914315852918361864905407340901408415264230120203
圆周角为0.2119769786848744897526495675933652469834680423803871501060879542428614271059734202664127331939619636

查看完整版本: 反演会保持田埂最短的特性吗？