Approximating $\pi$ with hexagons

Let $d$ be the diameter of the circle.

Let's look at the smaller hexagon first. Drawing the diagonals between opposite vertices of the hexagon yields six equilateral triangles. I've left out the larger hexagon from this picture to keep it from getting cluttered:

Each of these equilateral triangles has side length $\frac{d}{2}$ , and so the perimeter of the smaller hexagon is $3d$ .

Now let's look at the larger hexagon. We'll keep the red dotted segments from before, but we'll also draw segments connecting the opposite vertices of the larger hexagon. The smaller hexagon is removed from the picture:

We can now see that $30-60-90$ triangles are formed with the segments we've drawn.

We can use these special right triangles to determine that the large hexagon has side length ${\large\frac{\sqrt{3}}{3}}d$ . Therefore, the perimeter of the large hexagon is $2\sqrt{3}d$ .

Now that we have both perimeters, we can approximate the circumference of the circle:

$C\approx{\Large\frac{P_1+P_2}{2}}={\Large\left(\frac{2\sqrt{3}+3}{2}\right)}d=\left({\large\frac{3}{2}}+\sqrt{3}\right)d$

$\pi={\large\frac{C}{d}}$ , and so $\pi\approx{\large\frac{3}{2}}+\sqrt{3}$ .

$a=3$ , $b=2$ , and $c=3$ . Thus, $a+b+c=\boxed{8}$ .

$\text{For a 6-gon, the relations between circumradius R, inradius r and the side S are,}\\ \color{#3D99F6}{ S=2RSin30=R,\ \ \ \ \ r=RCos30=\ \frac{\sqrt3} 2 }. \\ C=2\pi*radius. \ \therefore ~\text{If radius }=\color{#3D99F6}{ \frac 1 2}, ~ C=\pi.\\ Small\ hexagon:- \\ R_s\ =\ \frac 1 2.\ \ \therefore\ P_1= 6*\frac 1 2 = \color{#3D99F6}{ 3}\\ Big\ hexagon:- \\ {\Large r}_B =\ \frac 1 2.\ \ \implies\ R_B=\ \dfrac 2 {\sqrt3}*{\Large r}_B=\ \frac 1 {\sqrt3} \\ \therefore\ P_2\ =\ 6*\frac 1 {\sqrt3} = \color{#3D99F6}{2*\sqrt 3}\\ \pi \ \approx \dfrac{P_1+P_2} 2=\dfrac{3 + 2*\sqrt3} 2=\frac 3 2 +\sqrt3.\\ \therefore\ a+b+c= \ \ \ \ \ \Large \color{#D61F06}{8}$