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Revision difference : construction of central proportional |
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Version 7 |
| \PMlinkescapeword{solution} |
\PMlinkescapeword{solution} |
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| \textbf{Task.} Given two line segments $p$ and $q$. Using compass and straightedge, construct the central proportional (the geometric mean) of the line segments. |
\textbf{Task.} Given two line segments $p$ and $q$. Using compass and straightedge, construct the central proportional (the geometric mean) of the line segments. |
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\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 $AB$ (for finding the centre, see the entry perpendicular bisector). 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:
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\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:
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| \begin{center} |
\begin{center} |
| \begin{pspicture}(-3,-3)(3,3) |
\begin{pspicture}(-3,-3)(3,3) |
| \rput[r](3,0){.} |
\rput[r](3,0){.} |
| \rput[a](0,2.5){.} |
\rput[a](0,2.5){.} |
| \psline{<->}(-3,0)(3,0) |
\psline{<->}(-3,0)(3,0) |
| \psarc(0,0){2.5}{0}{180} |
\psarc(0,0){2.5}{0}{180} |
| \psline(-0.5,0)(-0.5,2.45) |
\psline(-0.5,0)(-0.5,2.45) |
| \psline[linestyle=dashed](-2.5,0)(-0.5,2.45) |
\psline[linestyle=dashed](-2.5,0)(-0.5,2.45) |
| \psline[linestyle=dashed](2.5,0)(-0.5,2.45) |
\psline[linestyle=dashed](2.5,0)(-0.5,2.45) |
| \psdots(-2.5,0)(-0.5,0)(2.5,0)(-0.5,2.45) |
\psdots(-2.5,0)(-0.5,0)(2.5,0)(-0.5,2.45) |
| \rput[a](-2.5,-0.3){$A$} |
\rput[a](-2.5,-0.3){$A$} |
| \rput[a](-0.5,-0.3){$D$} |
\rput[a](-0.5,-0.3){$D$} |
| \rput[a](2.5,-0.3){$B$} |
\rput[a](2.5,-0.3){$B$} |
| \rput[b](-0.5,2.63){$C$} |
\rput[b](-0.5,2.63){$C$} |
| \rput[b](-1.4,0.11){$p$} |
\rput[b](-1.4,0.11){$p$} |
| \rput[b](0.8,0.11){$q$} |
\rput[b](0.8,0.11){$q$} |
| \end{pspicture} |
\end{pspicture} |
| \end{center} |
\end{center} |
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| (For more details on the procedure to create this picture, see compass and straightedge construction of geometric mean.) |
(For more details on the procedure to create this picture, see compass and straightedge construction of geometric mean.) |
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{\em Proof.} By Thales' theorem, the triangle $ABC$ is a right triangle. Its height $CD$ \PMlinkescapetext{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
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{\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
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$$p:CD\, = \,CD:q.$$
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$$p:CD = CD:q.$$
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| The equation shows that $CD$ is the central proportional of $p$ and $q$. |
The equation shows that $CD$ is the central proportional of $p$ and $q$. |
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| \textbf{Note.} The word {\em catheti} (in sing. {\em cathetus}) \PMlinkescapetext{means} the two shorter sides of a right triangle. |
\textbf{Note.} The word {\em catheti} (in sing. {\em cathetus}) \PMlinkescapetext{means} the two shorter sides of a right triangle. |
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