\(x^2+y^2-2c_iy+1=0, i=0,1,2,3\)
这时我们可以得到
\(u_2=\frac{c_0^2+2c_0c_1-3}{c_0^2-1}, u_1=\frac{2c_0c_1+c_1^2-3}{c_0^2-1}, u_0=\frac{1-c_1^2}{c_0^2-1}\)
而\(h_1,h_2,h_3\)同上一楼相同。
可以得到约束表达式
4*c0^6 + ((-4*c3*c2 - 12)*c1 + (-4*c2 - 4*c3))*c0^5 + ((c2^2 + 10*c3*c2 + (c3^2 + 12))*c1^2 + (2*c3*c2^2 + (2*c3^2 + 14)*c2 + 14*c3)*c1 + (c3^2*c2^2 + 6*c3*c2 - 3))*c0^4 + ((-2*c2^2 - 8*c3*c2 + (-2*c3^2 - 4))*c1^3 + (-6*c3*c2^2 + (-6*c3^2 - 18)*c2 - 18*c3)*c1^2 + ((-4*c3^2 - 2)*c2^2 - 16*c3*c2 + (-2*c3^2 + 8))*c1 + (-2*c3*c2^2 + (-2*c3^2 + 2)*c2 + 2*c3))*c0^3 + ((c2^2 + 2*c3*c2 + c3^2)*c1^4 + (6*c3*c2^2 + (6*c3^2 + 10)*c2 + 10*c3)*c1^3 + ((6*c3^2 + 4)*c2^2 + 16*c3*c2 + (4*c3^2 - 6))*c1^2 + (6*c3*c2^2 + (6*c3^2 - 6)*c2 - 6*c3)*c1 + (c2^2 - 2*c3*c2 + c3^2))*c0^2 + ((-2*c3*c2^2 + (-2*c3^2 - 2)*c2 - 2*c3)*c1^4 + ((-4*c3^2 - 2)*c2^2 - 8*c3*c2 - 2*c3^2)*c1^3 + (-6*c3*c2^2 + (-6*c3^2 + 6)*c2 + 6*c3)*c1^2 + (-2*c2^2 + 4*c3*c2 - 2*c3^2)*c1)*c0 + ((c3^2*c2^2 + 2*c3*c2 + 1)*c1^4 + (2*c3*c2^2 + (2*c3^2 - 2)*c2 - 2*c3)*c1^3 + (c2^2 - 2*c3*c2 + c3^2)*c1^2)=0
只是我们现在不能选择简单的\(c_0=0\),我们可以选择\(c_0=2,c_1=1.8,c_2=1.6\),然后解得\(c_3\approx 1.2029\)
对应图 也是那个视频号发的，结论比较简洁
本帖最后由 creasson 于 2025-6-5 20:37 编辑
是比较好的发现，不过证明都很简单。第一个，运行下面程序，取不含参数$c$的因式即得。
points = {P -> 0, A -> (2 r)/(I + a),B -> (2 r)/(I + b),
C -> (2 r)/(I + c),
Subscript -> (2 Subscript)/(I + u),
Subscript -> (2 Subscript)/(I + u),
Subscript -> (2 Subscript)/(I + u)};
eqs = Factor[
Discriminant]] // Numerator,
u]] & /@ ({(A - Subscript)/(A - B), (
B - Subscript)/(B - C), (C - Subscript)/(C - A)} /.
points);
Eliminate // Factor
第二个只需将各点的表示改为:
points = {P -> 0, Q -> 1, A -> (1 + I Tan[\])/(1 - I a),
B -> (1 + I Tan[\])/(1 - I b),
C -> (1 + I Tan[\])/(1 - I c),
Subscript -> (1 + I Tan[\])/(1 - I u),
Subscript -> (1 + I Tan[\])/(1 - I u),
Subscript -> (1 + I Tan[\])/(1 - I u)};
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