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'construction of central proportional'
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| Title of object: |
construction of central proportional |
| Canonical Name: |
ConstructionOfCentralProportion |
| Type: |
Algorithm |
| Created on: |
2007-10-04 16:44:31 |
| Modified on: |
2007-10-05 02:13:33 |
| Classification: |
msc:51M15 |
| Defines: |
cathetus, catheti |
Revision comment (for changes between this and next version):
| removing center of circle (which was at (0,0)) |
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
\usepackage{pstricks}
% define commands here
\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
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Content:
\textbf{Task.} Given two line segments $p$ and $q$. Using compass and straightedge, construct the central proportional (the geometric mean) of the line segments.
\PMlinkescapetext{\textbf{Solution.}} Set the line segments\, $AD = p$\, and\, $DB = q$\, on a line so that $D$ is between $A$ and $B$. Draw a half-circle with diameter $AB$. Let $C$ be the point where the normal line of $AB$ passing through $D$ intersects the arc of the half-circle. The line segment $CD$ is the required central proportional. Below is a picture that illustrates this solution:
\begin{center}
\begin{pspicture}(-3,-3)(3,3)
\rput[r](3,0){.}
\rput[a](0,2.5){.}
\psline{<->}(-3,0)(3,0)
\psarc(0,0){2.5}{0}{180}
\psline(-0.5,0)(-0.5,2.45)
\psdots(-2.5,0)(-0.5,0)(0,0)(2.5,0)(-0.5,2.45)
\rput[a](-2.5,-0.3){$A$}
\rput[a](-0.8,0.2){$D$}
\rput[a](2.5,-0.3){$B$}
\rput[b](-0.5,2.65){$C$}
\end{pspicture}
\end{center}
(For more details on the procedure to create this picture, see compass and straightedge construction of geometric mean.)
{\em Proof.} By Thales' theorem, the triangle $ABC$ is a right triangle. Its height $CD$ divides this triangle into two smaller right triangles which have equal angles with the triangle $ABC$ and thus are \PMlinkname{similar}{SimilarityInGeometry}. Accordingly, we can write the proportion equation concerning the catheti of the smaller triangles
$$p:CD = CD:q.$$
The equation shows that $CD$ is the central proportional of $p$ and $q$.
\textbf{Note.} The word {\em catheti} (in sing. {\em cathetus}) \PMlinkescapetext{means} the two shorter sides of a right triangle. |
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