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There was an interesting puzzle by Presh Talwalker in 'MindYourDecisions' about finding the radius of a circle that was cotangent to two larger circles.

https://www.youtube.com/watch?v=i0dZukEw1JY

I extended the problem by considering adding a second small circle adjacent to the circle of radius 2, (now 16 in my diagram) and wondered what the radius of the two large circles would have to be to make the two small circles have equal radii. The attached diagram will make this more clear.

enter image description here

I was surprised to find that the ratio of the two circles must be equal to 2.61803398875 which is, of course, the Golden Ratio + 1 The straight lines tangent to the larger circles subtend an angle of 53.130102 degrees. The series of circles can be extended to the right indefinitely and will give pairs of circles of equal radii. See second image.

enter image description here

Has this been found before? I have looked on the Internet without finding anything.

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I don't know if this has been seen before, although I suspect it's known. Here's a quick proof.

enter image description here

Let $\bigcirc P$, $\bigcirc Q$, $\bigcirc R$ (of radii $p$, $q$, $r$) be successively-tangent circles with common external tangent lines. Note that similarity guarantees a common proportionality factor, $k$, such that $$\frac{q}{p} = k = \frac{r}{q}$$

Now, suppose that $\bigcirc P^\prime$, tangent to $\bigcirc Q$, $\bigcirc R$, and one of those lines, has radius $p$. By Descartes' Theorem for "kissing circles" (one of which is a line, hence a circle of zero curvature), we have $$\left(\;\frac1{p}+\frac1{q}+\frac1{r}\;\right)^2 = 2\left(\;\frac1{p^2}+\frac1{q^2}+\frac1{r^2}\;\right) \quad\to\quad \left(\;1+\frac1k+\frac1{k^2}\;\right)^2= 2\left(\;1+\frac1{k^2}+\frac1{k^4}\;\right)$$ so that

$$(k^2 - 3 k + 1) (k^2 + k + 1) = 0$$

The latter factor has no real roots, and the former gives $k = \frac12\left(3\pm\sqrt{5}\right) = 1 + \phi^{\pm 1}$, where $\phi$ is the Golden Ratio. The smaller of these turns out to be extraneous, so we conclude $k = 1 + \phi$. $\square$

There's probably a less-algebraic "picture proof" demonstration of the relation, but I'm not seeing it at the moment.


The fact that "[t]he series of circles can be extended to the right indefinitely and will give circles of equal radii" is a simple consequence of similarity.

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    $\begingroup$ Thanks, yours is a much more elegant proof than mine which ran to dozens of lines of algebra. I did not spot the common proportionality so had to use brute force. I also used Descartes' theorem after a hint from Presh Talwalker. It was fun! $\endgroup$ – George Carey Oct 13 '18 at 15:09

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