<s>[INST] A geometric look at $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$? Is there a geometric way of looking at the relationship between the positive real numbers $a$, $b$ and $c$ if $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$?
 [/INST] If you have two poles of length $a$ and $b$, and string wire from the top of one pole to the bottom of the other, and vice versa, the height $c$ at which the wires  intersect will be given by half the harmonic mean of the height of the two poles, $\displaystyle \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}$, no matter how far apart they are.  This can be easily shown using similar triangles.
I can also think of one more place where the harmonic mean is commonly found in a geometric context:  Given a right triangle with legs $a$, $b$, the squared height of the altitude to the hypotenuse is given by half the harmonic mean of the squared lengths. Wikipedia provides a handy diagram so I'll just quickly write it out:

(Here $A$, $B$, $C$ are the angles opposite the lowercase sides, as is usual, and $P$ is the point where the altitude intersects the hypotenuse.) Then, since $\bigtriangleup ABC \; \sim \; \; \bigtriangleup CBP \;$,
$$\dfrac{a}{f} = \dfrac{c}{b}  \Rightarrow \dfrac{a^2}{f^2} = \dfrac{c^2}{b^2} = \dfrac{a^2 + b^2}{b^2}$$
So, by cross-multiplying we get,
$$\dfrac{1}{f^2} = \dfrac{a^2 + b^2}{a^2b^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}$$
half the harmonic mean of the squared lengths.
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