Circles in a rectangle - question 1

Let the radius of the red circle be $r$ and the radius of the yellow circle be $R$ and label the diagram as follows:

From right $\triangle JBD$ with sides $JB = BD = 1$ , $JD = \sqrt{2}$ , and from right $\triangle LCD$ with sides $LC = CD = r$ , $LD = \sqrt{2}r$ . Since $JK + KL + LD = JD$ , $1 + r + \sqrt{2}r = \sqrt{2}$ , which solves to $r = (\sqrt{2} - 1)^2$ .

By the kissing circles theorem on the blue, yellow, and red circles, $\cfrac{1}{\sqrt{JB}} + \cfrac{1}{\sqrt{ME}} = \cfrac{1}{\sqrt{LC}}$ , or $\cfrac{1}{\sqrt{1}} + \cfrac{1}{\sqrt{R}} = \cfrac{1}{\sqrt{(\sqrt{2} - 1)^2}}$ , which solves to $R = \frac{1}{2}$ .

By the Pythagorean Theorem on $\triangle LMN$ , $LN = \sqrt{LM^2 - MN^2} = \sqrt{(R + r)^2 - (R - r)^2} = 2 \sqrt{Rr} = 2 \sqrt{\frac{1}{2} \cdot (\sqrt{2} - 1)^2} = 2 - \sqrt{2} = CE$ .

So $AF = AD - CD + CE + EF = 2 - (\sqrt{2} - 1)^2 + (2 - \sqrt{2}) + \frac{1}{2} = \frac{3}{2} + \sqrt{2}$ , and the area of the rectangle is $AI \cdot AF = 2(\frac{3}{2} + \sqrt{2}) = 3 + 2\sqrt{2}$ .

Therefore, $a = 3$ , $b = 2$ , and $a + b = \boxed{5}$ .

Let the radii of the blue, yellow, and red circles be $r_1$ , $r_2$ , and $r_3$ respectively. Then we note that the length of the purple (full) line is given by $\ell_{\purple{\rm purple}} = \sqrt{(r_1+r_2)^2 - (r_1-r_2)^2} = 2\sqrt{r_1r_2}$ . Similarly, those of the red and green lines are $\ell_{\red{\rm red}} = 2\sqrt{r_3r_1}$ and $\ell_{\green{\rm green}} = 2\sqrt{r_2r_3}$ . Since $\ell_{\purple{\rm purple}} = \ell_{\red{\rm red}} + \ell_{\green{\rm green}}$ , we have:

$\begin{aligned} 2\sqrt{r_1r_2} & = 2\sqrt{r_3r_1} + 2\sqrt{r_2r_3} \\ \implies \sqrt{r_2} & = \frac {\sqrt{r_3r_1}}{\sqrt{r_1}-\sqrt{r_3}} \end{aligned}$

$\implies \boxed{\frac 1{\sqrt{r_3}} = \frac 1{\sqrt{r_1}} + \frac 1{\sqrt{r_2}}}$

Given that $r_1 = 1$ , and we can find $r_3$ by consider the red circle on the left. We note that the length of the red dash line is $\sqrt 2 = 1 + r_3 + \sqrt 2 r_3 \implies r_3 = \dfrac {\sqrt 2 - 1}{\sqrt 2 + 1} = \left(\sqrt 2-1\right)^2$ . Therefore $\sqrt{r_2} = \dfrac {\sqrt 2 -1}{1 - \sqrt 2 + 1} = \dfrac 1{\sqrt 2}$ .

Let the width of the rectangle be $w$ . We note that $w$ is the length of the purple dash/full line, which is $w = r_1 + 2\sqrt{r_1r_2} + r_2 = \left(\sqrt{r_1} + \sqrt{r_2}\right)^2 = \left(1 + \dfrac 1{\sqrt 2}\right)^2 = \dfrac {3+2\sqrt 2}2$ . The area of the rectangle is $2w = 3 + 2\sqrt 2$ . Therefore $a+b = 3+2 = \boxed 5$ .