# construction of contraharmonic mean of two segments

Let $a$ and $b$ two line segments (and their ).  The contraharmonic mean

 $x\;:=\;\frac{a^{2}\!+\!b^{2}}{a\!+\!b}\;=\;\frac{\left(\sqrt{a^{2}\!+\!b^{2}}% \,\right)^{2}}{a\!+\!b},$

satisfying the proportion equation

 $\frac{a\!+\!b}{\sqrt{a^{2}\!+\!b^{2}}}\;=\;\frac{\sqrt{a^{2}\!+\!b^{2}}}{x},$

can be constructed geometrically (http://planetmath.org/GeometricConstruction) as the third proportional of the segments $a\!+\!b$ and $\sqrt{a^{2}\!+\!b^{2}}$, the latter of which is gotten as the hypotenuse of the right triangle with catheti $a$ and $b$.  See the construction of fourth proportional.

(The dotted lines are parallel, with declivity $45^{\circ}$.)

Title construction of contraharmonic mean of two segments ConstructionOfContraharmonicMeanOfTwoSegments 2013-03-22 19:12:34 2013-03-22 19:12:34 pahio (2872) pahio (2872) 8 pahio (2872) Example msc 51M15 msc 51-00 ContraharmonicProportion HarmonicMeanInTrapezoid