Seven Circles in a Hexagon

This problem (as well as this one ) is a variation of of “Suzuki’s problem of four congruent incircles in an equilateral triangle” . In this paper, the following generalisation theorem is stated and proven:

If ${{a}_{0}}$ and ${{a}_{i}}$ are the lengths of the sides of the big and the small n-gon respectively, then ${{a}_{0}}={{a}_{i}}\left( 1+\sqrt{1+{{\left( \sin \dfrac{\pi }{n} \right)}^{-2}}} \right)$

In our case, ${{a}_{0}}=1$ and $n=6$ , so for the side of the inner hexagon we get $1={{a}_{i}}\left( 1+\sqrt{1+{{\left( \sin \dfrac{\pi }{6} \right)}^{-2}}} \right)\Rightarrow {{a}_{i}}=\dfrac{1}{1+\sqrt{1+{{\left( \dfrac{1}{2} \right)}^{-2}}}}=\dfrac{1}{1+\sqrt{5}}$ The radius $r$ of the inscribed circle is the apothem of the regular hexagon, thus we have $r=\frac{\sqrt{3}}{2}{{a}_{i}}=\frac{\sqrt{3}}{2}\cdot \frac{1}{1+\sqrt{5}}=\frac{\sqrt{15}-\sqrt{3}}{8}$ For the answer, $a=15$ , $b=3$ , $c=8$ , hence $a+b+c=\boxed{26}$ .

Let the radius of the circle be $r$ . Let us consider one of the six triangles and I have chosen the bottom one as shown ( @David Vreken , if you have used thiner lines, it would be easy for me to reuse the figure). We note the top angle of the triangle is always $60^\circ$ . Let the smaller base angle be $\theta$ . Then the other base angle is $120^\circ - \theta$ . For the length of the bottom side of the hexagon, we have:

$r \cot \frac \theta 2 + r \cot \left(60^\circ - \frac \theta 2 \right) = 1$

Now consider the radius of the center circle. We note that $r = \sin (60^\circ - \theta)$ . Substitute into the equation above:

$\begin{aligned} \sin (60^\circ - \theta) \left(\cot \frac \theta 2 + \cot \left(60^\circ - \frac \theta 2 \right)\right) & = 1 \\ \left(\frac {\sqrt 3}2\cos \theta - \frac 12 \sin \theta \right) \left(\frac 1{\tan \frac \theta 2} + \cot \left(60^\circ - \frac \theta 2\right) \right) & = 1 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \left(\sqrt 3 \cdot \frac {1-t^2}{1+t^2} - \frac {2t}{1+t^2} \right) \left(\frac 1t + \frac {1+\sqrt 3t}{\sqrt 3-t} \right) & = 2 \\ \left(\frac {\sqrt 3 - 2t - \sqrt 3 t^2}{1+t^2} \right) \left(\frac {\sqrt 3(1+t^2)}{\sqrt 3t-t^2} \right) & = 2 \\ 3 - 2\sqrt 3 t - 3t^2 & = 2\sqrt 3 t - 2t^2 \\ \implies t^2 + 4\sqrt 3 t - 3 & = 0 \\ \implies t & = \sqrt{15}-\sqrt{12} \end{aligned}$

Then we have:

$\begin{aligned} r & = \frac {\sqrt 3 - 2t - \sqrt 3 t^2}{2(1+t^2)} \\ & = \frac {\sqrt 3-2(\sqrt{15} - 2\sqrt 3) - \sqrt 3(27-12\sqrt 5)}{2(1+27-12\sqrt 5)} \\ & = \frac {10\sqrt{15} - 22\sqrt 3}{8(7-3\sqrt 5)} \\ & = \frac {(5\sqrt{15} - 11 \sqrt 3)(7+3\sqrt 5)}{4(7-3\sqrt 5)(7+3\sqrt 5)} \\ & = \frac {\sqrt{15}-\sqrt 3}8 \end{aligned}$

Therefore $a+b+c = 15+3+8 = \boxed{26}$ .

The six triangles formed have sidelengths $1,a,b$ , where (say) $a<b$ , and the angle between the sides of length $a,b$ is $60^\circ$ . The area of one of these triangles is therefore $\Delta=\frac12 ab \sin 60 = \frac{\sqrt3}{4}ab$

and the semiperimeter is $1+a+b$ . Hence the inradius is $r=\frac{\sqrt3}{2} \cdot \frac{ab}{1+a+b}$

Also, by the cosine rule, $1=a^2+b^2-ab$

The central hexagon has sidelength $a-b$ ; so the radius of the central circle is $\frac{\sqrt3}{2}(b-a)$ .

Equating these radii, we have to solve $\frac{\sqrt3}{2}(b-a)=\frac{\sqrt3}{2} \cdot \frac{ab}{1+a+b}$

and $1=a^2+b^2-ab$

These are messy but not hard to solve; we find $a=\frac14 \left(1+\sqrt5\right)$ and $b=\frac12 \sqrt5$ , from which we find $r=\frac{\sqrt{15}-\sqrt3}{8}$

giving the answer $\boxed{26}$ .

The length of each side of the inner hexagon is $\dfrac {\sqrt 5-1}{4}$

Radius of each circle is

$\dfrac {\sqrt 3}{2}\times \dfrac {\sqrt 5-1}{4}=\dfrac {\sqrt {15}-\sqrt 3}{8}$

So, $a=15,b=3,c=8$ , and $a+b+c=\boxed {26}$ .