双心五边形 不知应该取哪个因式

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相关：mathworld.wolfram.com/PonceletsPorism.html

hejoseph 发表于 2017-6-13 09:13
把 $R$、$r$、$d$ 代入那个式子，因式分解后，简单的那个因式是三角形的情形，复杂的那个因式才是五边形的 ...

在Cayley closure conditions中，
取$n=3$得$a_2=0$
取$C$为圆$x^2+y^2=R^2$，$D$为圆$(x-d)^2+y^2=r^2$

1. SeriesCoefficient[Sqrt[Det[t{{1,0,0},{0,1,0},{0,0,-R^2}}+{{1,0,-d},{0,1,0},{-d,0,-r^2+d^2}}]],{t,0,2}]

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$4 r^2 R^2-R^4+2 R^2 d^2-d^4=0$
分解得$d^2=R^2-2 r R$，即Euler公式

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在Cayley closure conditions中，
取$n=4$得$a_3=0$
取$C$为圆$x^2+y^2=R^2$，$D$为圆$(x-d)^2+y^2=r^2$

1. SeriesCoefficient[Sqrt[Det[t{{1,0,0},{0,1,0},{0,0,-R^2}}+{{1,0,-d},{0,1,0},{-d,0,-r^2+d^2}}]],{t,0,3}]

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$d^6-2 d^4 r^2-3 d^4 R^2+3 d^2 R^4+2 r^2 R^4-R^6=0$
分解得$d^4-2 d^2 r^2-2 d^2 R^2-2 r^2 R^2+R^4$，即$2r^{2}(R^{2}+d^{2})=(R^{2}-d^{2})^{2}$，即Fuss公式

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在Cayley closure conditions中，
取$n=5$得$a_2a_4-a_3^2=0$
取$C$为圆$x^2+y^2=R^2$，$D$为圆$(x-d)^2+y^2=r^2$

1. {a2,a3,a4}=SeriesCoefficient[Sqrt[Det[t{{1,0,0},{0,1,0},{0,0,-R^2}}+{{1,0,-d},{0,1,0},{-d,0,-r^2+d^2}}]],{t,0,#}]&/@{2,3,4};
2. Factor[a2 a4 - a3^2]

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$0=\left(d^6-2 d^4 r R-3 d^4 R^2+8 d^2 r^3 R-4 d^2 r^2 R^2+4 d^2 r R^3+3 d^2 R^4+4 r^2 R^4-2 r R^5-R^6\right) \left(-d^6-2 d^4 r R+3 d^4 R^2+8 d^2 r^3 R+4 d^2 r^2 R^2+4 d^2 r R^3-3 d^2 R^4-4 r^2 R^4-2 r R^5+R^6\right)$
有两个因式，不知应该取哪个

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在Cayley closure conditions中，
取$n=6$得$a_3a_5-a_4^2=0$
取$C$为圆$x^2+y^2=R^2$，$D$为圆$(x-d)^2+y^2=r^2$

1. {a3,a4,a5}=SeriesCoefficient[Sqrt[Det[t{{1,0,0},{0,1,0},{0,0,-R^2}}+{{1,0,-d},{0,1,0},{-d,0,-r^2+d^2}}]],{t,0,#}]&/@{3,4,5};
2. Factor[a3 a5 - a4^2]

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$\left(d^2+2 d r-R^2\right) \left(d^2+2 r R-R^2\right) \left(-d^2+2 d r+R^2\right) \left(-d^2+2 r R+R^2\right) \left(-3 d^8+4 d^6 r^2+12 d^6 R^2-4 d^4 r^2 R^2-18 d^4 R^4+16 d^2 r^4 R^2-4 d^2 r^2 R^4+12 d^2 R^6+4 r^2 R^6-3 R^8\right)=0$
取最后一个因式，$-3 d^8+4 d^6 r^2+12 d^6 R^2-4 d^4 r^2 R^2-18 d^4 R^4+16 d^2 r^4 R^2-4 d^2 r^2 R^4+12 d^2 R^6+4 r^2 R^6-3 R^8=0$