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佩特-诺伊曼-道格拉斯定理:若一任意的n边形A0,每边上往外画顶角为$\frac{2k\pi}n$(1≤k≤n−2)的等腰三角形,再针对各等腰三角形的顶角形成的n边形进行类似的作法,但需用不同的数字k,一直用到用完所有满足1≤k≤n−2的数值为止(以任意顺序),最后形成的n边形An−2会是正多边形,且其形心和原n边形A0的形心重合.佩特诺-伊曼-道格拉斯定理最早是由卡瑞尔·佩特诺(1868–1950)1908年在布拉格提出.1940年及1941年时也分别被杰西·道格拉斯(1897–1965)和伯恩哈德·诺伊曼(1909–2002)独立证明.
当n=3时定理为拿破仑定理:因为k=1,以A0的边向外作顶角为2π/3的等腰三角形,其顶角形成A1.这些点是以A0的边向外作的等边三角形的中心.
The theorem can be proved using some elementary concepts from linear algebra.
The proof begins by encoding an $n$-gon by a list complex numbers representing the vertices of the $n$-gon. This list can be thought of as a vector in the $n$-dimensional complex linear space $C_n$.
Take an $n$-gon $A$ and let it be represented by the complex vector $A = ( a_1, a_2, ... , a_n )$.
Let the polygon $B$ be formed by the free vertices of similar triangles built on the sides of $A$ and let it be represented by the complex vector $B = ( b_1, b_2, ... , b_n )$.
Then we have $α( a_r - b_r ) = a_{r+1}- b_r$, where $α =\exp( i θ )$ for some $θ$ (here $i$ is the square root of $-1$).
\begin{tikzpicture}
\fill[green!40] (-105:.5) arc (-105:-75:.5)--(0,0)--cycle;
\draw (-105:.5) arc (-105:-75:.5);
\node at(-90:.5)[below,font=\scriptsize]{$\theta$};
\draw(0,0)node(br)[above]{$b_r$}--(-75:1.5)node[below]{$a_r$}--(-105:1.5)node[below]{$a_{r+1}$}--cycle;
\end{tikzpicture}
This yields the following expression to compute the $b_r$'s:$$b_r = (1−α)^{−1} ( a_{r+1}− αa_r )$$
In terms of the linear operator $S :\mathbb C^{n} →\mathbb C^{n}$ that cyclically permutes the coordinates one place, i.e.
$$S(a_1,a_2,\cdots,a_{n-1},a_n)=(a_2,a_3,\cdots,a_n,a_1)$$
we have$$B = (1−α)^{−1}( S − αI )A$$
where $I$ is the identity matrix.
This means that the polygon $A_{n−2}$ that we need to show is regular is obtained from $A_0$ by applying the composition of the following operators:$( 1 − ω^{k} )^{−1}( S − ω^{k} I )$ for $k = 1, 2, ... , n − 2$, where $ω = \exp( 2πi/n )$.
(These commute because they are all polynomials in the same operator $S$.)
A polygon $P = ( p_1, p_2, ..., p_n )$ is a regular $n$-gon if each side of $P$ is obtained from the next by rotating through an angle of $2π/n$, that is, if$$p_{r+1} − p_r = ω( p_{r+2}−p_{r+1})$$
This condition can be formulated in terms of $S$ as follows:$$( S − I )( I − ωS ) P = 0$$
Or equivalently as$( S − I )( S - ω^{n - 1} I ) P = 0$, since $ω^{n} = 1$.
Petr–Douglas–Neumann theorem now follows from the following computations.
$( S − I )( S - ω^{n - 1} I ) A_{n - 2}$
$= ( S − I )( S - ω^{n - 1} I ) ( 1 − ω )^{−1} ( S - ω I ) ( 1 − ω^{2} )^{−1} ( S - ω^{2} I ) ... ( 1 − ω^{n - 2} )^{−1} ( S - ω^{n - 2} I ) A_0$
$ = ( 1 − ω )^{−1}( 1 − ω^{2} )^{−1} ... ( 1 − ω^{n - 2} )^{−1} ( S - I ) ( S - ω I ) ( S - ω^{2} I ) ... ( S - ω^{n - 1} I)A_0$
$ = ( 1 − ω )^{−1}( 1 − ω^{2} )^{−1} ... ( 1 − ω^{n - 2} )^{−1} ( S^{n} - I ) A_0$
$ = 0$, since $S^{n} = I$. |
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