|
|
|
Viewing Version
3
of
'midpoint'
|
[ view 'midpoint'
|
back to history
]
| Title of object: |
midpoint |
| Canonical Name: |
Midpoint3 |
| Type: |
Definition |
| Created on: |
2008-11-04 17:36:11 |
| Modified on: |
2008-11-05 14:49:23 |
| Classification: |
msc:51-00, msc:51M15 |
| Synonyms: |
midpoint=center |
Revision comment (for changes between this and next version):
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
|
Content:
\PMlinkescapeword{degree}
The concept of \PMlinkname{midpoint of line segment}{Midpoint} is a special case of the midpoint of a curve or arbitrary figure in $\mathbb{R}^2$ or $\mathbb{R}^3$.
A point $T$ is a {\em midpoint} of the figure $f$, if for each point $A$ of $f$ there is a point $B$ of $f$ such that $T$ is the midpoint of the line segment $AB$.\, One says also that $f$ is symmetric about the point $T$.\\
Given the equation of a curve in $\mathbb{R}^2$ or of a surface $f$ in $\mathbb{R}^3$, one can, if \PMlinkescapetext{necessary}, take a new point $T$ for the origin by using the linear substitutions of the form
$$x := x'\!+\!a, \quad y := y'\!+\!b \quad \mbox{etc.}$$
Thus one may test whether the origin is the midpoint of $f$ by checking whether $f$ always contains along with any point\, $(x,\,y,\,z)$\, also the point\, $(-x,\,-y,\,-z)$.\\
It is easily verified the
\textbf{Theorem.}\, If the origin is the midpoint of a quadratic curve or a quadratic surface, then its equation has no \PMlinkname{terms of degree}{BasicPolynomial} 1.
Similarly one can verify the generalisation, that if the origin is the midpoint of an algebraic curve or surface of degree $n$, the equation has no terms of degree $n\!-\!1$,\, $n\!-\!3$\, and so on.\\
\textbf{Note.}\, Some curves and surfaces have infinitely many midpoints (see \PMlinkname{quadratic surfaces}{QuadraticSurfaces}).
\begin{thebibliography}{8}
\bibitem{IF}{\sc Felix Iversen}: {\em Analyyttisen geometrian oppikirja}. Tiedekirjasto Nr. 19.\, Second edition.\, Kustannusosakeyhti\"o Otava, Helsinki (1963).
\end{thebibliography}
|
|
|
|
|
|
