给定边长 圆外切多边形的唯一性

hbghlyj

维基百科:
If the number of sides $n$ is odd, then for any given set of sidelengths $a_{1},\dots ,a_{n}$ satisfying the existence criterion above there is only one tangential polygon. But if n is even there are an infinitude of them. For example, in the quadrilateral case where all sides are equal we can have a rhombus with any value of the acute angles, and all rhombi are tangential to an incircle.

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由这帖,
\begin{multline*}r^4 (t_1+t_2+t_3+t_4+t_5)
\\-r^2 (t_1 t_2 t_3+t_1 t_2 t_4+t_1 t_3 t_4+t_2 t_3 t_4+t_1 t_2 t_5+t_1 t_3 t_5+t_2 t_3 t_5+t_1 t_4 t_5+t_2 t_4 t_5+t_3 t_4 t_5)
\\+t_1 t_2 t_3 t_4 t_5=0\end{multline*}
关于$r^2$是二次方程, 它的判别式为正, 两根之积为正, 所以$r^2$有两个正根, 而$r$是$r^2$的正平方根, 所以$r$有两个解.
这两个解分别对应于凸五边形和五角星.

hbghlyj

推广: 设$a_1,\dots,a_{2n+1}$满足存在性条件. 则给定边长$a_1,\dots,a_{2n+1}$的圆外切$2n+1$边形有$n$个解, 它们关于内切圆圆心的环绕数分别为$I=1,2,\dots,n$.
证明: 因为$a_1,\dots,a_{2n+1}$满足存在性条件, 且$2n+1$为奇数, 所以\[t_1+t_2=a_1, t_2+t_3=a_2, \ldots, t_{2n+1}+t_1=a_{2n+1}\]存在唯一的正数解$t_1,\dots,t_{2n+1}$.
\[\tan^{-1}\frac{t_1}r+\dots+\tan^{-1}\frac{t_{2n+1}}r=πI\]当$r\to0^+$时, 左边$=\fracπ2(2n+1)$
当$r\to+\infty$时, 左边$=0$
根据介值定理, 关于$r$的方程有解, 对$I=1,2,\dots,n$.
根据这帖, $r^2$满足一个$n$次方程, 所以$r$有$n$个正根, 但上面已得到$n$个正根, 故为全部解.