Whole Circle in a Quarter Circle

Radius of the quarter circle $=R$

Radius of full circle $=r$

$(R-r) ^2=2r^2$

$\implies R=r(\sqrt 2 +1)$

$\implies$ the required ratio of areas is

$4(\frac rR) ^2=\dfrac {4}{(\sqrt 2 +1)^2}$

So $a=4,b=2,c=1,a+b+c=\boxed 7$ .

Let the radius of the quarter circle be $a$ and the radius of the inscribed circle be $b$ . Line segment $\overline{AE}$ is equal to $a$ . Line segments $\overline{ED}$ , $\overline{FD}$ , $\overline{GD}$ , and $\overline{AF}$ are all equal to $b$ .

We see that $\overline{AD}$ is equal to $a - b$ . Since it is the hypotenuse of an isosceles right triangle, we can set up an equality.

$\begin{aligned} \overline{AD}^{2} &= \overline{AF}^{2} + \overline{DF}^{2} \\ (a-b)^{2} &= b^{2} + b^{2} \\ a^{2} - 2ab + b^{2} &= 2b^{2} \\[0.3em] \implies a &= b + b\sqrt{2} \end{aligned}$

The area of the quarter circle is equal to $\frac{πa^{2}}{4}$ or $\frac{π(b + b\sqrt{2})^{2}}{4}$ while the area of the inscribed circle is $πb^{2}$ . The needed ratio would then be:

$\small \dfrac{πb^{2}}{\frac{π(b + b\sqrt{2})^{2}}{4}} = \dfrac{4πb^{2}}{π(b + b\sqrt{2})^{2}} \implies \dfrac{4}{(1 + \sqrt{2})^{2}}$

Thus, $\frac{a}{(\sqrt{b} + c)^{2}} \implies a=4, b=2, c=1$ . The sum of the variables would then be $\boxed{7}$ .

Let the radius of the circular quadrant be $1$ and that of the red circle be $r$ . Then we note that

$\begin{aligned} \sqrt 2 r + r & = 1 \\ \implies r & = \frac 1{\sqrt 2 + 1} \end{aligned}$

Then the ratio of the area of the red circle to the area of the quadrant

$\frac {\pi r^2}{\frac \pi 4} = \frac 4{(\sqrt 2+1)^2}$

Therefore $a+b+c = 4+2+1 = \boxed 7$ .