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Tangential polygon
Dušan Djukić, Vladimir Janković, Ivan Matić, Nikola Petrović, The IMO Compendium 1995-2004, p. 561.
存在圆外切 $n$ 边形 $A_1 A_2 \ldots A_n$, 边长为 $A_1 A_2=a_1$, $A_2 A_3=a_2, \ldots, A_n A_1=a_n$ 当且仅当
\begin{equation}\label1
x_1+x_2=a_1, x_2+x_3=a_2, \ldots, x_n+x_1=a_n
\end{equation}
有正数解 $\left(x_1, \ldots, x_n\right)$.
证明
⇒:若存在圆外切 $n$边形 $A_1 A_2 \ldots A_n$. 设 $A_i A_{i+1}$ 上的切点为 $P_i$. 则$x_1=A_1 P_n=A_1 P_1, x_2=A_2 P_1=A_2 P_2, \ldots, x_n=A_n P_{n-1}=A_n P_n$ 是\eqref{1}的解.
⇐:若\eqref{1}有正数解 $\left(x_1, \ldots, x_n\right)$.
Let us draw a polygonal line $A_1A_2 \dots A_{n+1}$ touching a circle of radius $r$ at points $P_1, P_2,\dots, P_n$ respectively such that $A_1P_1 =A_{n+1},P_n = x_1$ and $A_iP_i = A_iP_{i-1} = x_i$ for $i = 2, \dots , n$. Observe that $O A_1=O A_{n+1}=\sqrt{x_1^2+r^2}$ and the function $f(r)=\angle A_1 O A_2+$ $\angle A_2 O A_3+\cdots+\angle A_n O A_{n+1}=$ $2\left(\arctan \frac{x_1}{r}+\cdots+\arctan \frac{x_n}{r}\right)$ is continuous. Thus $A_1 A_2 \ldots A_{n+1}$ is a closed simple polygonal line if and only if $f(r)=360^{\circ}$. But such an $r$ exists, since $f(r) \rightarrow 0$ when $r \rightarrow \infty$ and $f(r) \rightarrow \infty$ when $r → 0$. |
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