Four Circles in a Box

Let the width of the rectangle be $1$ . Then the height of the rectangle is $q$ and the radius of the big circle is $\frac 12$ . Let the radius of the two red circles be $r_1$ and the radius of the small circle be $r_2$ . Then the (green) line bisecting the rectangle gives

$2r_2 + 1 = q \quad ...(1)$

This length is equal to the vertical component of the green dash line.

$\begin{aligned} r_1 + \sqrt{\left(\frac 12 + r_1^2\right)^2 - \left(\frac 12 - r_1^2\right)^2 } + \frac 12 &= q \\ r_1 + \sqrt{2r_1} + \frac 12 & = q \\ \left(\sqrt{r_1}+\frac 1{\sqrt 2} \right)^2 & = q & ...(2) \end{aligned}$

We can get another equation considering the length of the blue line.

$\begin{aligned} 2r_1 + 2\sqrt{(r_1+r_2)^2 - (r_1-r_2)^2} & = 1 \\ 2r_1 + 4\sqrt{r_1r_2} & = 1 & ...(3) \end{aligned}$

With three unknowns and three equations we can solve for $q$ numerically, which I did. But to follow @Pi Han Goh 's tradition ... From $(3)$ :

$\begin{aligned} r_1 + 2\sqrt{r_1r_2} & = \frac 12 \\ r_1 + 2\sqrt{r_1r_2} + r_2 & = r_2 + \frac 12 \\ \left(\sqrt{r_1} + \sqrt{r_2}\right)^2 & = \frac \blue{2r_2 + 1}2 & \small \blue{\text{Note }(1): 2r_2 +1 = q} \\ \left(\sqrt{r_1} + \sqrt{r_2}\right)^2 & = \frac \blue q2 & \small \blue{\text{and }(2): \left(\sqrt{r_1}+\frac 1{\sqrt 2} \right)^2 = q} \\ \left(\sqrt{r_1} + \sqrt{r_2}\right)^2 & = \frac 12 \blue {\left(\sqrt{r_1}+\frac 1{\sqrt 2} \right)^2} \\ \sqrt{r_1} + \sqrt{r_2} & = \sqrt{\frac {r_1}2} + \frac 12 & \small \blue{\text{Multiply both sides by }4\sqrt{r_1}} \\ 4r_1 + \blue{4\sqrt{r_1r_2}} & = 2\sqrt 2r_1 + 2\sqrt{r_1} & \small \blue{\text{From }(3): 2r_1 + 4\sqrt{r_1r_2} = 1} \\ 4r_1 + \blue{1-2r_1} & = 2\sqrt 2r_1 + 2\sqrt{r_1} \\ 2(\sqrt 2-1)r_1 + 2\sqrt{r_1} - 1 & = 0 & \small \blue{\text{Substitute }\sqrt{r_1}= \sqrt q - \frac 1{\sqrt 2}} \\ 2(\sqrt 2 -1)\left(q - \sqrt{2q} + \frac 12\right) + 2\sqrt q - \sqrt 2 - 1 & = 0 \\ (\sqrt 2 -1)q + (\sqrt 2-1) \sqrt q - 1 & = 0 & \small \blue{\text{Multiply both sides by }\sqrt 2 + 1} \\ q + \sqrt q - \sqrt 2 - 1 & = 0 & \small \blue{\text{A quadratic of }\sqrt q} \\ \implies \sqrt q & = \frac {\sqrt{4\sqrt 2+5}-1}2 \\ q & = \frac {2\sqrt 2 + 3 - \sqrt{4\sqrt 2+5}}2 \\ & \approx 1.28197168 \\ \implies \lfloor 10^5 q \rfloor & = \boxed{128197} \end{aligned}$

Label the top left of the diagram as follows:

Let the width of the rectangle be $1$ , so that the large circle is $\frac{1}{2}$ , and let the height of the rectangle be $h$ , so that the ratio is $q = \frac{h}{1} = h$ , and the radius of the little circle is $r = \frac{1}{2}(h - 1)$ . Also, let the radius of a red circle be $R$ .

Since $HF = HD + DF = R + r$ and $EH = BH - BE = BH - CF = R - r$ , by the Pythagorean Theorem on $\triangle EHF$ we have $EF = \sqrt{HF^2 - EH^2} = \sqrt{(R + r)^2 - (R - r)^2} = 2\sqrt{Rr} = 2\sqrt{\frac{1}{2}R(h - 1)}$ .

Similarly, since $KH = KL + LH = \frac{1}{2} + R$ and $JK = IK - IJ = IK - GH = \frac{1}{2} - R$ , by the Pythagorean Theorem on $\triangle HJK$ we have $HJ = \sqrt{KH^2 - JK^2} = \sqrt{(\frac{1}{2} + R)^2 - (\frac{1}{2} - R)^2} = \sqrt{2R}$ .

That means $AC = AB + BC = GR + EF = R + 2\sqrt{\frac{1}{2}R(h - 1)} = \frac{1}{2}$ , and $AI = AG + GI = BH + HJ = R + \sqrt{2R} = h - \frac{1}{2}$ .

These two equations solve to $h = q = \frac{1}{2}(3 + 2\sqrt{2} - \sqrt{5 + 4 \sqrt{2}})$ , so $\lfloor 10^5q \rfloor = \boxed{128197}$ .