An Octagon and a Circle

From the problem, the area of the circle can be expressed as $A=πr^2=(a+b\sqrt{c})π$ , where $r^2=a+b\sqrt{c}$

Considering one triangle from the octagon and by applying cosine law,

$2^2=r^2+r^2-2(r)(r)(cos45)$

$4=2r^2-2r^2\frac{\sqrt{2}}{2}$

$4=r^2(2-\sqrt{2})$

$r^2=\frac{4}{2-\sqrt{2}}$

rationalize the denominator by multiplying the fraction by $\frac{2+\sqrt{2}}{2+\sqrt{2}}$ , we obtain

$r^2=4 + 2\sqrt{2}$

we can see that $a=4,b=2$ $and$ $c=2$

$a+b+c=4+2+2=8$