Consider the following image that shows a centre circle with radius $=1$ and $n=7$ smaller circles with radius $r$; enter image description here

Calculate $r(n)$.

Initial try

Let $\alpha$ be the angle between the vertical (larger) radius and the “first” circle; enter image description here and let $\beta$ be the angle between the radii for 2 tangent circles; enter image description here

We get $$\cos(\alpha)=\frac{1-r}{1+r}, \qquad \sin(\beta/2)=\frac{r}{1+r}.$$

Summing up “full circle” for the larger circle we get $$(n-1)\beta+2\alpha=2\pi$$ or $$ (n-1)2\arcsin\left(\frac{r}{1+r}\right)+2\arccos\left(\frac{1-r}{1+r}\right)=2\pi $$

$$ \iff (n-1)\arcsin\left(\frac{r}{1+r}\right)+\arccos\left(\frac{1-r}{1+r}\right)=\pi. $$

I don't get further than this (if it is even the right approach) and the equation looks hard to solve.

Any other approach that gives a “neater” (and explicit) result for $r(n)$? TIA.

Here are some additional graphs for $n=21$; enter image description here and $n=100$; enter image description here

It is not hard to let Mathematica (numerically) solve the equation for any $n$ and draw these images, but an explicit expression would be much nicer.

If we let $$S(n,r)=(n-1)2\arcsin\left(\frac{r}{1+r}\right)+2\arccos\left(\frac{1-r}{1+r}\right)$$ we get $$S'_r(n,r)=\frac{2 \left(n+\frac{\sqrt{\frac{2 r+1}{(r+1)^2}}}{\sqrt{\frac{r}{(r+1)^2}}}- 1\right)}{(r+1)^2 \sqrt{\frac{2 r+1}{(r+1)^2}}}>0$$ for all $r>0$ and $n\ge2$. Hence $S(n,r)$ is increasing.

Note that $S(n,0)=0$.

For $n=\{2,3\}$ we have a special case where the outer circles are larger than the centre circle with $r(2)=4$ and $r(3)=\phi$ (the Golden Mean) (an interesting property in its own).

For $n=4$ we have a part of a regular hexagon hence $r(4)=1$.

For $n\ge5$ we have $S(n,1)=\frac{1}{3}(2+n)\pi\ge\frac73\pi>2\pi$.

Since $S(n,r)$ is increasing and $S(n,0)=0$ and $S(n,1)>2\pi$ there is a solution $r_0\in(0,1)$ for the equation $S(n,r)=2\pi$.

  • 1
    $\begingroup$ A nice problem! Where did it come from? Do you have any reason to think it has an explicit solution for $r(n)$? $\endgroup$
    – TonyK
    Commented Jul 16, 2023 at 12:52
  • $\begingroup$ Your problem is not specific, in fact there are an infinite number of different possible solutions, it might be worth making it a complete closed loop to become a specific problem $\endgroup$ Commented Jul 16, 2023 at 13:18
  • $\begingroup$ The problem can be solved (numerically) with Mathematica. The problem is well-defined. Each $n$ has a unique $r$. The question if is there is a more "classical" explicit solution. As the length of the polygon made up by the smaller circles centres is $(n+1)r$ it is reasonable to think that $r\sim 2\pi/n$. $\endgroup$
    – mf67
    Commented Jul 16, 2023 at 14:29
  • $\begingroup$ @TonyK: I got it from Instagram and since it was 6 days old (now 7 days) without any reply or solution I got interested in it. It seems tricky. A quick look at the graph in Mathematica shows that it appears to be a monotone(?) function that, sooner or later, crosses $2\pi$ so an unique solution exist for every $n$. $\endgroup$
    – mf67
    Commented Jul 16, 2023 at 14:33
  • $\begingroup$ It crosses $2\pi$? That sounds unlikely. $\endgroup$
    – TonyK
    Commented Jul 16, 2023 at 15:58

2 Answers 2


We have $$ \cos(\beta) = \cos^2(\beta/2) - \sin^2(\beta/2) = 1-2\sin^2(\beta/2) = 1 - \frac{2r^2}{(1 + r)^2} $$ and $$ \cos(2\alpha) = \cos^2(\alpha) - \sin^2(\alpha) = 2\cos^2 (\alpha) - 1 = \frac{2(1-r)^2}{(1+r)^2} - 1 $$ From your relation $(n-1)\beta + 2\alpha = 2\pi$, $$ \cos((n-1)\beta) = \cos(2\pi - 2\alpha) = \cos(2\alpha) $$ By definition, $\cos((n-1)\beta) = T_{n-1}(\cos\beta)$, where $T_n$ is the $n$-th Chebyshev polynomial. Thus $$ T_{n-1}(\cos\beta) = \cos(2\alpha) $$ or, in terms of $r$: $$ T_{n-1}\left(1 - \frac{2r^2}{(1+r)^2}\right) = \frac{2(1-r)^2}{(1 + r)^2} - 1 $$ In other words, determining $r(n)$ amounts to finding positive real roots of some big rational function of $r$. I have very limited hope that there will be any nice expression for these roots, but maybe by some miracle it ends up being easy to write down an expression for them.

  • $\begingroup$ Thanks. It seems the Instagram poster made up a difficult problem. No wonder there were no replies. The $r(3)=\phi$ was also a problem from the same author. It must be easier to prove. I have not looked into it. Probably more ‘basic’ geometry. $\endgroup$
    – mf67
    Commented Jul 16, 2023 at 17:42
  • 1
    $\begingroup$ That will not be so difficult because $T_{3 - 1}(x)$ is not a very complicated polynomial. Somewhere along the way, you'll get the polynomial $r^2 - r-1$, the positive root of which is $\varphi$. $\endgroup$ Commented Jul 16, 2023 at 17:43
  • $\begingroup$ I'll just add that you can of course clear denominators in my final expression and shift everything over to the left so that you get a polynomial equation $P_n(r) = 0$, but, again, this will be a pretty complicated polynomial that I have limited hope will be analytically solvable for general $n$. $\endgroup$ Commented Jul 16, 2023 at 21:19
  • $\begingroup$ OK, Thanks. I will look into this tomorrow. What looked to be a pretty standard problem turned out to be rather complex. $\endgroup$
    – mf67
    Commented Jul 16, 2023 at 22:58
  • $\begingroup$ If $n$ is odd, the distance between the centres of the middle outer circle and the end one is $2\sqrt{1+r}$. When $n=3$ this is also $2r$ so that's how you get $r^2 = r+1$. $\endgroup$ Commented Jul 17, 2023 at 9:18

enter image description here

Comment: For rough estimation of $\beta$, we may assume the centers of small circles are coincident with the vertices of a decagon, i.e $\frac{\beta}2\approx 18^o$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .