How many circles of radius $\frac{\pi}{4}$ can be drawn within a circle of radius $\pi$ such that they do not intersect one another?

I have seen a solution where it is stated to use the expression below :

$$\sin \frac{360}{2n}=\frac{\frac{\pi}{4}}{\pi-\frac{\pi}{4}}$$

where $n$=number of circles possible.

But I'm unable to understand the geometry behind it.


The answer is eleven (proved by Melissen) [thanks @Rushabh Mehta]

(And why is this problem posed with the radii in multiples of $\pi$ rather than just $1$?)

  • 3
    $\begingroup$ @Entrepreneur the citation for a proof was already linked in the answer. $\endgroup$ – JMoravitz Aug 17 '18 at 19:00
  • 1
    $\begingroup$ Isn't it actually 11? I'm pretty sure 12 is not possible... $\endgroup$ – Don Thousand Aug 17 '18 at 19:03
  • $\begingroup$ Also, sorry to nitpick, but it was proved by Melissen $\endgroup$ – Don Thousand Aug 17 '18 at 19:06

enter image description here enter image description here

I previously answered this question, of course, I did not prove that 11 is maximal but I showed OP that at least you can make 11 circles inside and 9 circles touching the outer circle

but @rushabhamehta said that it is not possible to draw 9 circles touching inner part of the outer circle.

So I decided to plot them one by one and you can clearly see that drawing 9 circles touching the inner part of the outer circle is possible

however here I am not claiming that the circles themselves must touch they may not touch themselves.

Then I decided to draw 2 extra inner circles which also seems easy

and by looking at the picture it is clear that you can not draw an extra circle, however, I did not prove it.


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