数学星空 发表于 2014-1-5 11:37
对于最后一种情况：(已设定)
只需求解
，
我得到的方程还没这么长求极值软件直接不算:L 或者无限运行没结果
m= 4; n= 3; r= 2
\beta = \arccos((m^2+n^2-r^2)/(2*m*n))
w = \arccos((m^2-n^2+r^2)/(2*m*r))
v =\arccos((-m^2+n^2+r^2)/(2*r*n))
\phi= \arctan(\sin(\beta)/(x/y-\cos(\beta)))
\theta = \arctan(\sin(\beta)/(y/x-\cos(\beta)))
s={\sin(\beta)*m*n}/2-(m-x)^2/{2*(1/\tan(w)-1/\tan(2*\phi))}-(n-y)^2/{2*(1/\tan(v)-1/\tan(2*\theta))}-{\sin(\beta)*x*y}/2
难怪你算不出来的，你的式子有理化展开
-2*m*n*q*y-m^2*y+2*m*n*x-n^2*y+r^2*y＝0 （1）
-m^4+2*m^2*n^2+2*m^2*r^2-n^4+2*n^2*r^2-r^4-p^2＝0（2）
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y^4+4*p^3*x^5*y+4*p^3*x*y^5-16*p*q^2*x^5*y-64*p*q^2*x^3*y^3-16*p*q^2*x*y^5+16*p*r^2*x^4*y^2+16*p*r^2*x^2*y^4-32*m*n*p*q*x^3*y-32*m*n*p*q*x*y^3+128*m*n*q*s*x^3*y+128*m*n*q*s*x*y^3+32*p*q*x^4*y^2+32*p*q*x^2*y^4+16*m*n*p*x^2*y^2-64*m*n*s*x^2*y^2-16*p*x^3*y^3=0(3)
若需要化为{m,n,r,s,x,y}的方程，对以上（1~3）消元{p,q}得到的消元结果14444项之多，你想数值求解也是不太可能的！
在东论里我留了两个可以达到极值的条件，也可以作为方程，有时候可以得到数值解有时候不能，似乎得到的数值解不全面，这高次方程怎么会只一组解。可惜得到的这一组解也不是想要的，复制到Maple里
m := 3;
n := 4;
r := 2;
beta := arccos((m^2+n^2-r^2)/(2*m*n));
w := arccos((m^2-n^2+r^2)/(2*m*r));
v := arccos((-m^2+n^2+r^2)/(2*r*n));
varphi := arctan(sin(beta)/(x/y-cos(beta)));
theta := arctan(sin(beta)/(y/x-cos(beta)));
a := sqrt(x^2+y^2-2*x*y*cos(beta));
b := simplify((n-y)*sin(v)/sin(2*theta-v));
c := simplify((m-x)*sin(w)/sin(2*varphi -w));
eq1 := b*cos(theta)+c*cos(varphi )-(1/2)*a;
eq2 := b^2+(a*cos(varphi ))^2-(1/2)*a^2-(c-a*cos(varphi ))^2;
fsolve({eq1 = 0., eq2 = 0.}, {x, y});
{x = 2.109601080, y = 2.386627598}
其实可以算折痕与边夹角与重叠面积的函数，可以发现函数图像有几个拐点，达到最大值时候角度要等于90度
针对最后一种情况:我将计算的附件贴上来供电脑里装有MAPLE 软件的网友，运行一下哈：
只需要输入红色部份的数据（即需要计算的三角形三边,依次为a>b>c）,从第一个冒号开始依次往下敲回车键即可
最终将输出图像及相关的数据（k为面积重叠率，其余参数含义见6#说明）
数学星空 发表于 2014-1-6 21:12
针对最后一种情况:我将计算的附件贴上来供电脑里装有MAPLE 软件的网友，运行一下哈：
只需要输入红色部份 ...
有没有最优折叠还小于0.414213562的情况:o
星空得出的表达式看起来很复杂，让人看着眼花。我试着手工推导了一下，k应该可以表示成lambda和mu的分式形式，分子六次，分母四次。然后如果求两偏微分让之为零，相当于得到2元9次方程组
对于星空6楼的标记，我们设AE/AB为s,AF/AC为t,A1B1/AB为u,A1C1/AC为v,那么k=st-uv
设角AEF为h,那么s*c*sin(A+h)=t*b*sin(h)。左边展开然后移项可得出tan(h)表示成s,t的一次分式。于是sin(2h),cos(2h)都可以表示成s,t的二次分式
$$\tan(h) = \frac{\sin(C)*\sin(A)}{\frac{t}{s}*\sin(B) - \sin(C)*\cos(A)}$$
$$\sin(2h)=\frac{2\sin(C)\sin(A)(\sin(B)\frac{t}{s} - \sin(C)\cos(A))}{\sin^2(C)\sin^2(A)+(\sin(B)\frac{t}{s}-\sin(C)\cos(A))^2} =\frac{u_1(\frac{t}{s})}{v(\frac{t}{s})}$$
$$\cos(2h)=\frac{(\sin(B)\frac{t}{s}-\sin(C)\cos(A))^2-\sin^2(C)\sin^2(A)}{\sin^2(C)\sin^2(A)+(\sin(B)\frac{t}{s}-\sin(C)\cos(A))^2}=\frac{u_2(\frac{t}{s})}{v(\frac{t}{s})}$$