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# construction of contraharmonic mean of two segments

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

Related:

ContraharmonicProportion, HarmonicMeanInTrapezoid

Type of Math Object:

Example

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Reference

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## Mathematics Subject Classification

51M15*no label found*51-00

*no label found*

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