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<record version="8" id="519">
 <title>Heron's principle</title>
 <name>HeronsPrinciple</name>
 <created>2009-02-13 13:25:55</created>
 <modified>2026-09-05 23:09:49</modified>
 <type>Theorem</type>
 <creator id="21" name="pahio"/>
 <modifier id="1" name="bloftin"/>
 <comment>commenting out pictures until we can upload them</comment>
 <author id="21" name="pahio"/>
 <classification>
	<category scheme="pacs" code="02.40.Dr"/>
	<category scheme="pacs" code="42.15.-i"/>
 </classification>
 <related>
	<object name="FermatsPrinciple"/>
 </related>
 <keywords>
	<term>reflection</term>
 </keywords>
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\newtheorem*{thmplain}{Theorem}</preamble>
 <content>\textbf{Theorem.}
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 \emph{Heron's principle}, concerning the reflection of light, is a
special case of \emph{Fermat's principle} in optics.

\emph{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 to $l$
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 to $l$ through $C$ intersects
$l$.

\begin{center}
\includegraphics{HeronsPrinciple_construction.png}

\vspace{0.5em}

\textbf{Figure.}
Construction used in the proof of Heron's principle.
\end{center}

Justification:
From two pairs of similar right triangles we get the proportion equations
\[
AP:CX = PQ:XQ,
\qquad
BQ:CX = PQ:PX,
\]
which imply
\[
AP:PX = BQ:XQ.
\]
From this we can infer that
\[
\triangle AXP \sim \triangle BXQ.
\]
Thus the corresponding angles $AXP$ and $BXQ$ are equal.

\begin{center}
\includegraphics{HeronsPrinciple_reflection.png}

\vspace{0.5em}

\textbf{Figure.}
Reflection construction showing why the path through $X$ is shortest.
\end{center}

It remains to show that the route $AXB$ is the shortest.
If $X_1$ is another point of the line $l$, then
$AX_1=A'X_1$, and therefore
\[
AX_1B
=
A'X_1B
=
A'X_1+X_1B
\geq
A'B
=
A'XB
=
AXB.
\]

\begin{thebibliography}{8}

\bibitem{th}
Tero Harju,
\emph{Geometria. Lyhyt kurssi}.
Matematiikan laitos. Turun yliopisto, Turku (2007).

\end{thebibliography}
</content>
</record>
