Dynamic Geometry: P19

Let the radii of the cyan, green, and orange circle be $r_1$ , $r_2$ , and $r_3$ respectively, and the height of the red point to the base of the square be $h$ . From the equation $A = sr$ , where $A$ , $s$ , and $r$ are the area, semiperimeter, and the inradius of a triangle. Then we have:

$\begin{aligned} r_1 & = \frac h{2\sqrt{h^2+\frac 14}+1} = \frac h{\sqrt{4h^2+1}+1} & \small \blue{\text{Replace }h \text{ with }1-h.} \\ r_2 & = \frac {1-h}{\sqrt{4(1-h)^2+1}-1} \\ r_3 & = \frac 1{\sqrt{4h^2+1} + \sqrt{4(1-h)^2+1}+2} \end{aligned}$

We note that the area of the purple quadrilateral is half of the red rectangle shown. Therefore we need to find the $h$ , when

$\frac {(1-r_1-r_2)(1-2r_3)}2 = \frac {7\sqrt{41}-3}{240}$

Solving the equation above, we get $h=\dfrac 58$ . Then $r_1 = \dfrac 5{\sqrt{164}+8} \approx 0.240312424$ , $r_2 = \dfrac 16 \approx 0.166666667$ , and $r_3 = \dfrac 4{\sqrt{41}+13} \approx 0.206152368$ . Therefore $r_2 = \dfrac 16$ is the smallest and $a+b = \boxed 7$ .