|
|
|
Viewing Version
7
of
'symmetry'
|
[ view 'symmetry'
|
back to history
]
| Title of object: |
symmetry |
| Canonical Name: |
Symmetry2 |
| Type: |
Definition |
| Created on: |
2007-06-04 21:39:42 |
| Modified on: |
2007-06-07 17:28:41 |
| Classification: |
msc:51A15, msc:51A10, msc:15A04 |
| Defines: |
symmetry about, symmetric, symmetric about, rotational symmetry, point symmetry, symmetry about a point, symmetric about a point, reflectional symmetry, line symmetry, symmetry about a line, symmetric about a line |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{pstricks}
\usepackage{psfrag}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{xypic}
|
Content:
\PMlinkescapeword{terms}
\PMlinkescapetext{This entry is not yet complete.}
Let $V$ be a Euclidean vector space, $F \subseteq V$, and $E \colon V \to V$ be a Euclidean transformation that is not the identity map.
The following terms are used to indicate that $E(F)=F$ if $E$ is a rotation:
\begin{itemize}
\item $F$ has \emph{rotational symmetry};
\item $F$ has \emph{point symmetry};
\item $F$ has \emph{symmetry about a point};
\item $F$ is \emph{symmetric about a point}.
\end{itemize}
If $V=\mathbb{R}^2$, then the last two terms may be used to indicate the specific case in which $E$ is conjugate to $\displaystyle \left( \begin{array}{rr}
-1 & 0 \\
0 & -1 \end{array} \right)$, \PMlinkname{i.e.}{Ie} the angle of rotation is $180^{\circ}$.
For example, let $\displaystyle F=\cup_{k=1}^4 P_4$, where $\displaystyle P_1=\left\{ (x,y) : 0 \le x \le \frac{4}{1+\sqrt{3}} \text{ and } (2-\sqrt{3})x \le y \le x \right\}$, $\displaystyle P_2=\left\{ (x,y) : \frac{4}{1+\sqrt{3}} \le x \le 2 \text{ and } x \le y \le (2+\sqrt{3})x-4 \right\}$, $\displaystyle P_3=\left\{ (x,y) : 2 \le x \le \frac{4 \sqrt{3}}{1+\sqrt{3}} \text{ and } (-2+\sqrt{3})x+8-4\sqrt{3} \le y \le (-2-\sqrt{3})x+4+4\sqrt{3} \right\}$, and $\displaystyle P_4=\left\{ (x,y) : \frac{4 \sqrt{3}}{1+\sqrt{3}} \le x \le 4 \text{ and } (-2+\sqrt{3})x+8-4\sqrt{3} \le y \le -x+4 \right\}$ has point symmetry with respect to the point $\displaystyle \left( 2, \frac{2}{\sqrt{3}} \right)$. The valid angles of rotation for $F$ are $120^{\circ}$ and $240^{\circ}$. The boundary of $F$ and the point $\displaystyle \left( 2, \frac{2}{\sqrt{3}} \right)$ are shown in the following picture.
\begin{center}
\begin{pspicture}(0,0)(4,3.5)
\pspolygon(0,0)(2,0.536)(4,0)(2.5359,1.4641)(2,3.4641)(1.4641,1.4641)
\psdot(2,1.1547)
\end{pspicture}
\end{center}
As another example, the figure $\{ (x,y) : -3 \le x \le -1 \text{ and } (x+1)^2+y^2 \le 4 \} \cup \big( [-1,1] \times [-2,2] \big) \cup \{ (x,y) : 1 \le x \le 3 \text{ and } (x-1)^2+y^2 \le 4 \}$ is symmetric about the origin. The boundary of this figure and the point $(0,0)$ are shown in the following picture.
\begin{center}
\begin{pspicture}(-3,-2)(3,2)
\psarc(-1,0){2}{180}{270}
\psline(-1,-2)(1,-2)(1,0)(3,0)
\psarc(1,0){2}{0}{90}
\psline(1,2)(-1,2)(-1,0)(-3,0)
\psdot(0,0)
\end{pspicture}
\end{center}
\PMlinkescapetext{How to determine $E$ for a specific point} $p \in \mathbb{R}^2$ \PMlinkescapetext{will be added.}
If $E(F)=F$ and $E$ is a reflection, then $F$ has \emph{reflectional symmetry}. In the special case that $V=\mathbb{R}^2$, the following terms are used:
\begin{itemize}
\item $F$ has \emph{line symmetry};
\item $F$ has \emph{symmetry about a line};
\item $F$ is \emph{symmetric about a line}.
\end{itemize}
\PMlinkescapetext{Picture in the case of} $V=\mathbb{R}^2$ \PMlinkescapetext{will be added.}
\PMlinkescapetext{How to determine $E$ for a specific line} $\ell \subseteq \mathbb{R}^2$ \PMlinkescapetext{will be added.}
\PMlinkescapetext{Feel free to add these before I do if you wish!} |
|
|
|
|
|
