A square, a rhombus, and five tangent circles

Label the bottom left as follows, and let $r$ be the radius of one of the circles and let $\theta = \angle IGJ = \angle JGK$ .

Since $\angle IGL$ is a right angle, $\angle KGL = 90° - 2\theta$ , and since $\angle LKG$ is a right angle, $\angle KLG = 2\theta$ .

As half the length of the square, $AG = GL = 4$ , and since $AI = r$ , $IG = 4 - r$ .

By the Pythagorean Theorem on $\triangle IGJ$ , $JG = \sqrt{(4 - r)^2 + r^2} = \sqrt{2r^2 - 8r + 16}$ , so that $\cos \theta = \cfrac{4 - r}{\sqrt{2r^2 - 8r + 16}}$ , $\sin \theta = \cfrac{r}{\sqrt{2r^2 - 8r + 16}}$ , and $\cos 2\theta = \cos^2 \theta - \sin^2 \theta = \cfrac{8 - 4r}{r^2 - 4r + 8}$ .

From $\triangle KLG$ , $\cos 2\theta = \cfrac{r}{4}$ .

Therefore, $\cos 2\theta = \cfrac{8 - 4r}{r^2 - 4r + 8} = \cfrac{r}{4}$ , which rearranges to $r^3 - 4r^2 + 24r - 32 = 0$ , and solves numerically to $r \approx \boxed{1.5864}$ .

Let the radius of the five circles be $r$ , centers of the center circle and the top right circle be $O$ and $P$ respectively, $PQ$ be perpendicular to $CE$ and $\angle OEF = \theta$ . Then we note that $\sin \theta = \dfrac r4$ . And

$\begin{aligned} PQ \cot \angle PEQ + QC & = CE \\ r \cot \left(\frac {90^\circ - \theta}2 \right) + r & = 4 & \small \blue{\text{Divide both sides by }r} \\ \tan \left(45^\circ + \frac \theta 2\right) + 1 & = \frac 4r = \frac 1{\sin \theta} & \small \blue{\text{Note that }\sin \theta = \frac r4} \\ \frac {1+t}{1-t} + 1 & = \frac {1+t^2}{2t} & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \frac 2{1-t} & = \frac {1+t^2}{2t} \\ t^3 - t^2 + 5t - 1 & = 0 & \small \blue{\text{By Cardano's method}} \\ \implies t & = \frac 13 + \sqrt[3]{-\frac 8{27}+\frac {2\sqrt{78}}9} + \sqrt[3]{-\frac 8{27}-\frac {2\sqrt{78}}9} \\ & \approx 0.206783495 \\ r & = 4\sin \theta = \frac {8t}{1+t^2} \approx \boxed{1.59} \end{aligned}$

The radius of the incircle of rhombus $EHGF$ is, $r=\frac{HF\cdot EG}{2\sqrt{HF^2+EG^2}}=\frac{8x}{\sqrt{4x^2+64}} \;\;\;\;\;\;\;\;\;\;\;\; \color{#D61F06} (1)$ Let $\angle HGO=2\alpha\implies \angle AGH=90^\circ-2\alpha$ . Since the center of the incircle ( $X)$ lies on the angle bisector of $\angle AGH$ , $\angle AGX=\angle YGX=45^\circ-\alpha$ In $\triangle HOG$ , $\tan \angle HGO=\frac{HO}{OG}\implies \tan(2\alpha)=\frac{x}{4}$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\implies \frac{2\tan \alpha}{1-\tan^2\alpha}=\frac{x}{4}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \color{#D61F06}(2)$ In $\triangle XYG$ , $\tan \angle YGX=\frac{XY}{YG}\implies \tan (45^\circ-\alpha)=\frac{r}{4-r}$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\implies \frac{1-\tan\alpha}{1+\tan\alpha}=\frac{r}{4-r}$ Solving for $\tan\alpha$ in the above equation, $\tan\alpha =\dfrac{1-\dfrac{r}{4-r}}{1+\dfrac{r}{4-r}}=1-\frac{r}{2}$ Substituting this in $\color{#D61F06}(2)$ , $\frac{2\tan \alpha}{1-\tan^2\alpha}=\dfrac{2-r}{r-\dfrac{r^2}{4}}=\frac{x}{4}$ $\implies x=\frac{32-16r}{4r-r^2}$ Substituting this in $\color{#D61F06}(1)$ , $r^2=\frac{64x^2}{4x^2+64}=\dfrac{64\left(\dfrac{32-16r}{4r-r^2}\right)^2}{4\left(\dfrac{32-16r}{4r-r^2}\right)^2+64}$ Solving this equation in $r$ ,we have, the radius of the five circles is $\boxed{r\approx 1.586}$

As seen in the figure, we place the square on a coordinate plane. Point $A$ is on the origin and point $B$ lies on the positive x-semiaxis, i.e. $B\left( 8,0 \right)$ . The center of the rhombus’ incircle is $K\left( 4,4 \right)$ and the center of the lower-left circle is $L\left( r,r \right)$ , where $r$ is the radius of the circles.
$LN$ is perpendicular to $AB$ . Let $\theta =\angle BGH$ and $\varphi =\angle LGN$ . Then,

$\tan \frac{\theta }{2}=\cot \left( \frac{\pi -\theta }{2} \right)=\cot \varphi =\frac{NG}{LN}=\frac{4-r}{r} \ \ \ \ \ (1)$ The line $l$ through points $G$ and $H$ has equation $y=m\left( x-4 \right)\Leftrightarrow mx-y-4m=0$ , where $m$ is the slope of $l$ . Thus,

$m=\tan \theta =\frac{2\tan \frac{\theta }{2}}{1-{{\tan }^{2}}\frac{\theta }{2}}\overset{\left( 1 \right)}{\mathop{=}}\,\frac{2\frac{4-r}{r}}{1-{{\left( \frac{4-r}{r} \right)}^{2}}}\Rightarrow m=\frac{4r-{{r}^{2}}}{4r-8} \ \ \ \ \ (2)$

$l$ is tangent to both circles with centers $K$ and $L$ , hence these two points are equidistant from $l$ . Using the distance formula found here we have

$\begin{aligned} \frac{\left| mr-r-4m \right|}{\sqrt{{{m}^{2}}+1}}=\frac{\left| 4m-4-4m \right|}{\sqrt{{{m}^{2}}+1}} & \Leftrightarrow \left| mr-r-4m \right|=\left| 4m-4-4m \right| \\ & \Leftrightarrow \left| m\left( r-4 \right)-r \right|=4 \\ & \Leftrightarrow m\left( r-4 \right)-r=-4\ \ \ \text{ or }\ \ \ m\left( r-4 \right)-r=4 \\ \end{aligned}$ Asume $m\left( r-4 \right)-r=-4$ . Combining with $(2)$ ,

$\frac{4r-{{r}^{2}}}{4r-8}\left( r-4 \right)-r=-4\Leftrightarrow \ldots \Leftrightarrow {{r}^{2}}=8\overset{r>0}{\mathop{\Leftrightarrow }}\,r=2\sqrt{2}$ This value is rejected, because, obviously, $4r<8\Rightarrow r<2$ . Hence, the relation that holds is $m\left( r-4 \right)-r=4$ . Again, with $(2)$ ,

$\frac{4r-{{r}^{2}}}{4r-8}\left( r-4 \right)-r-4=0\Leftrightarrow \ldots \Leftrightarrow {{r}^{3}}-4{{r}^{2}}+24r-32=0$ The unique real solution of this equation gives the answer: $r\approx \boxed{1.5864}$ .