# Heron’s principle

In the Euclidean plane, let $l$ be a line and $A$ and $B$ two points not on $l$.  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 Heron’s principle, concerning the reflection of light, is a special case of Fermat’s principle in optics.

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 $l$.

Justification:  From two pairs of similar right triangles we get the proportion equations

 $AP:CX\;=\;PQ:XQ,\qquad BQ:CX\;=\;PQ:PX,$

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.

We still state that the route $AXB$ is the shortest.  If $X_{1}$ is another point of the line $l$, then  $AX_{1}\,=\,A^{\prime}X_{1}$,  and thus we obtain

 $AX_{1}B\;=\;A^{\prime}X_{1}B\;=\;A^{\prime}X_{1}+X_{1}B\;\geqq\;A^{\prime}B\;=% \;A^{\prime}XB\;=\;AXB.$

## References

• 1 Tero Harju: Geometria. Lyhyt kurssi.  Matematiikan laitos. Turun yliopisto (University of Turku), Turku (2007).
Title Heron’s principle HeronsPrinciple 2014-09-15 15:38:36 2014-09-15 15:38:36 pahio (2872) pahio (2872) 13 pahio (2872) Theorem msc 51M04 Catacaustic PropertiesOfEllipse HeronianMeanIsBetweenGeometricAndArithmeticMean