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[几何] 从大圆上任一点开始作内部小圆的切线最终恰能回到起点?.

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TSC999 Posted 2022-7-12 23:12 |Read mode
能否作出一大一小两个不同心的圆,满足:从大圆上任一点开始连续作小圆的切线,作五条后就恰好回到起始点?

求作两个圆.png

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kuing Posted 2022-7-13 02:58

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hbghlyj Posted 2022-7-13 04:04

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 Author| TSC999 Posted 2022-7-15 09:23
【数学中国】上的 creasson  先生给出了以下公式:

求作两个圆0.png

两个例子:

求作两个圆 1.png

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hbghlyj Posted 2022-7-15 14:47
mathworld.wolfram.com/PonceletsPorism.html
In fact, there is a general analytic expression relating the circumradius $R$, inradius $r$, and offset between the circumcenter and incenter $d$ for a bicentric polygon. Given $R$, $r$, and $d$, define
\begin{align}
a&={1\over R+d}\\
b&={1\over R-d}\\
c&={1\over r}
\end{align}Note that since $r, R,$ and $d$ are positive quantities with $d<R, 0<a<b$.

Now let
\begin{align}
\lambda&=1+{2c^2(a^2-b^2)\over a^2(b^2-c^2)}\\
\omega&=\cosh^{-1}\lambda,
\end{align}and define the elliptic modulus $k$ via
\begin{align}
k^2=1-e^{-2\omega}.
\end{align}Then the condition for an $n$-gon to be bicentric is
\begin{align}
\operatorname{sc}\left(\frac{K(k)}n,k\right)={c\sqrt{b^2-a^2}+b\sqrt{c^2-a^2}\over a(b+c)},
\end{align}where $\operatorname{sc}(x,k)$ is a Jacobi elliptic function and $K(k)$ is a complete elliptic integral of the first kind (Richelot 1830, Kerawala 1947).
x^2-y^2 x^2 y^2.gif


In[]:= PonceletClosed[n_,d_,r_]:=Module[{R,a,b,c,lambda,omega,m},a=1/(R+d);b=1/(R-d);c=1/r;
lambda=1+2c^2(a^2-b^2)/a^2/(b^2-c^2);
omega=ArcCosh[lambda];
m=1-Exp[-2omega];
R/.FindRoot[JacobiSC[EllipticK[m]/n,m]==(c Sqrt[b^2-a^2]+b Sqrt[c^2-a^2])/a/(c+b),{R,3}]]
In[]:= PonceletClosed[2.5,.5,1.]
Out[]= 3.29906
In[]:= PonceletClosed[2.5,1.,1.]
Out[]= 3.4768

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青青子衿 Posted 2023-6-6 20:29
hbghlyj 发表于 2022-7-15 14:47
https://mathworld.wolfram.com/PonceletsPorism.html


Poncelet Polygons
More About Areas and Centers of Poncelet Polygons
armj.math.stonybrook.edu/pdf-Springer-final/020-0154.pdf

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