Talk:PlanetPhysics/Heron's Principle

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: Heron's principle %%% Primary Category Code: 02.40.Dr %%% Filename: HeronsPrinciple.tex %%% Version: 7 %%% Owner: pahio %%% Author(s): pahio %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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\textbf{\htmladdnormallink{theorem}{http://planetphysics.us/encyclopedia/Formula.html}.}\, Let $A$ and $B$ be two points and $l$ a line of the Euclidean plane.\, If $X$ is a point of $l$ such that the sum $AX\!+\!XB$ is the least possible, then the lines $AX$ and $BX$ form equal angles with the line $l$.

This {\em Heron's principle}, concerning the \htmladdnormallink{reflection}{http://planetphysics.us/encyclopedia/FluorescenceCrossCorrelationSpectroscopy.html} of light, is a special case of {\em \htmladdnormallink{Fermat's principle}{http://planetphysics.us/encyclopedia/GravitationalLensing.html}} in optics.\\

{\em Proof.}\, If $A$ and $B$ are on different sides of $l$, then $X$ must be on the line $AB$, and the assertion is trivial since the vertical angles are equal.\, Thus, let the points $A$ and $B$ be on the same side of $l$.\, Denote by $P$ and $Q$ the points of the line $l$ where the normals of $l$ set through $A$ and $B$ intersect $l$, respectively.\, Let $C$ be the intersection point of the lines $AQ$ and $BP$.\, Then, $X$ is the point of $l$ where the normal line of $l$ set through $C$ intersects $l$. \begin{center} \begin{pspicture}(-3,-1)(3,3) \psline(-2.6,0)(2.6,0) \psdots[linecolor=blue](-2,2.5)(2,1.6) \psline[linestyle=dashed](-2,2.5)(-2,0) \psline[linestyle=dashed](2,1.6)(2,0) \psline(-2,2.5)(2,0) \psline(2,1.6)(-2,0) \psline(0.439,0.976)(0.439,0) \psdot[linecolor=red](0.439,0) \rput(-2.2,2.75){$A$} \rput(2,1.83){$B$} \rput(-2,-0.25){$P$} \rput(2,-0.25){$Q$} \rput(0.44,1.3){$C$} \rput(0.44,-0.25){$X$} \rput(2.8,0){$l$} \end{pspicture} \end{center} Justification:\, From two pairs of similar right triangles we get the proportion equations $$AP:CX \;=\; PQ:XQ, \quad BQ:CX \;=\; PQ:PX,$$ which imply the equation $$AP:PX \;=\; BQ:XQ.$$ From this we can infer that also $$\Delta AXP \sim \Delta BXQ.$$ Thus the corresponding angles $AXP$ and $BXQ$ are equal. \begin{center} \begin{pspicture}(-3,-3)(3,3) \psline(-2.6,0)(2.6,0) \psdots[linecolor=blue](-2,2.5)(2,1.6) \psline[linestyle=dashed](-2,2.5)(-2,-2.5)(0.439,0) \psline[linecolor=blue](-2,2.5)(0.439,0) \psline[linecolor=blue](0.439,0)(2,1.6) \psdot[linecolor=red](0.439,0) \rput(-2,2.75){$A$} \rput(2,1.83){$B$} \rput(-2.2,-0.25){$P$} \rput(0.44,-0.27){$X$} \rput(2.8,0){$l$} \psline[linestyle=dotted](-2,2.5)(-0.7,0)(2,1.6) \psdots(-0.7,0)(-2,-2.5) \rput(-0.7,-0.29){$X_1$} \rput(-2.3,-2.5){$A'$} \psline(-2.15,1.15)(-1.85,1.15) \psline(-2.15,1.05)(-1.85,1.05) \psline(-2.15,-1.2)(-1.85,-1.2) \psline(-2.15,-1.1)(-1.85,-1.1) \end{pspicture} \end{center} We still state that the route $AXB$ is the shortest.\, If $X_1$ is another point of the line $l$, then\, $AX_1\,=\,A'X_1$,\, and thus we obtain $$AX_1B \;=\; A'X_1B \;=\; A'X_1+X_1B \;\geqq\; A'B \;=\; A'XB \;=\; AXB.$$

\begin{thebibliography}{8} \bibitem{th}{\sc Tero Harju}: {\em Geometria. Lyhyt kurssi}.\, Matematiikan laitos. Turun yliopisto, Turku (2007). \end{thebibliography}

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