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Is there a geometric way of looking at the relationship between the positive real numbers $a$, $b$ and $c$ if $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$?

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rearranging you get $(a+b)c = ab$, this might be easier to see geometricly... – the L Feb 13 '11 at 8:36
up vote 24 down vote accepted

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:

Similar triangles using the altitude

(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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(should be a comment to Billare's answer) In your pole-and-wire example, the crossing height is correctly given in the formula, but $c$ is half the harmonic mean of $a$ and $b$. See – John Bentin Feb 13 '11 at 11:42
Right, right. Fixed, thank you. – Uticensis Feb 13 '11 at 11:52

Not necessarily geometric, but if $a$ and $b$ are electrical resistances placed in parallel, then their combined resistance is $c$.

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Given three circles of areas $a\geq b\geq c >0$, they can be put in a position such that they are tangent to each other and to a straight line if and only if $$ \frac{1}{c} = \frac{1}{a} + \frac{1}{b}. $$

Tree tangent circles

Note that this a special case of Descartes' theorem.

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You find this kind of “addition” happening frequently. For instance, if you have lenses of focal length $a$ and $b$ and put them together, the combined focal length is $c$. What's happening in this case is that the “strength” of the first lens is $1/a$, of the second is $1/b$, and the strengths add.

Another example from optics is that if you have a lens of focal length $c$ and place something a distance of $a$ in front of it, the image forms at a distance of $b$ behind the lens.

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I don't know what its origin is. I learned that from a book long time ago. And I thought it was used to calculate electrical resistances placed in parallel, which Tpofofn also mentioned. enter image description here

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Another non-mathematical example is Sum Frequency Generation in non-linear optics, e.g.

...generation of red light (→ red lasers), e.g. by mixing the outputs of a 1064-nm Nd:YAG laser and a 1535-nm fiber laser, resulting in an output at 628 nm

Here we have 1/1064 nm + 1/1525 nm = 1/628.41 nm

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In a right-angled triangle (with legs a and b, and hypotenuse c) you can inscribe a square in two "obvious" ways:

  • The square shares the right angle with the triangle and touches c. Then 1/s = 1/a + 1/b
  • One side of the square lies on c, and the square touches a and b. Then 1/t = 1/c + 1/h (where h is the height on c)
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