PlanetPhysics/Heron's Principle

\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 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 $$l$$. \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} 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{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} 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 $$AX_1B \;=\; A'X_1B \;=\; A'X_1+X_1B \;\geqq\; A'B \;=\; A'XB \;=\; AXB.$$