I have 12 circles of radius r arranged in hexagonal pattern inside a container circle of radius R.
Picture of 12 equal circles packed inside a container circle in hexagonal pattern
This is not the densest packing, but an expression of the relation between the radii can be evaluated:
$$r = \frac{1}{25}(2\sqrt{21}-3)R$$
I'm looking for a proof of this expression. I tried different ways, trigonometry, symmetries, Pythagoras, secant-tangent theorem, quadrilaterals, but I don't seem to find the right approach. Can someone provide a demonstration or point me to a relevant reference?
Thanks!
Alberto
Following on from Jaap’s comments,
Taking the radius of the small circles to be $1$, the radius of the big circle is $a+1$ as shown in the diagram.
The distance $x$ is $\frac23$ of the height of the equilateral triangle formed by the centres of three neighbouring small circles, so $$x=\frac23\sqrt{3}$$
The distance $y$ is the height of the same triangle just referred to, so $$y=\sqrt{3}$$
Now, by Pythagoras, $$a^2=(x+y)^2+1$$ $$\implies a^2=\frac{28}{3}$$ $$\implies a=\frac23\sqrt{21}$$
Hence the large circle has radius $$\frac23\sqrt{21}+1$$