Kundabuffer

Let the radius of the congruent circle be $r$ and $\angle AEB = \theta$ . The the length $BC = 1$ is given by:

$\begin{aligned} r + r \cot \frac \theta 2 + r \cot \left(90^\circ - \frac \theta 2\right) + r & = 1 \\ r (2 + \cot \frac \theta 2 + \tan \frac \theta 2) & = 1 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \implies 2 + \frac 1t + t & = \frac 1r \end{aligned}$

Similarly for $AB$ :

$\begin{aligned} r \cot \left(45^\circ - \frac \theta 2\right) + r & = 1 \\ \implies \frac {1+t}{1-t} + 1 & = \frac 1r \end{aligned}$

Therefore.

$\begin{aligned} 2 + \frac 1t + t & = \frac {1+t}{1-t} + 1 \\ \frac {t^2 + t + 1}t & = \frac {1+t}{1-t} \\ (1-t)(t^2 + t + 1) & = t + t^2 \\ 1 - t^3 & = t + t^2 \\ t^3 + t^2 + t - 1 & = 0 \end{aligned}$

Solving for $t$ using Cardano's method , we get $t = \tan \frac \theta 2 = \sqrt[3]{\frac {17}{27}+\sqrt{\frac {11}{27}}} + \sqrt[3]{\frac {17}{27}-\sqrt{\frac {11}{27}}} - \frac 13 \approx 0.543689013$ . Note that $EB = \cot \theta = \dfrac {1-t^2}{2t} \approx 0.647798871 \implies \lfloor 10^5 \cdot EB \rfloor = \boxed{64779}$ .

Let $F$ , $G$ , $H$ , $J$ be contact points as seen in the figure. Let $K$ , $L$ be the centers of the circles and $M$ the intersection of $AE$ with $JF$ . Denote $\angle KAF$ by $\theta$ .

Then, $\angle KAG=\theta$ , $\angle GKM=\angle MLH=2\theta$ .
Hence, we have

$\begin{aligned} JF& =JL+LM+MK+KF \\ \Rightarrow 1 & =r+\dfrac{r}{\cos 2\theta }+\dfrac{r}{\cos 2\theta }+r \\ \Rightarrow \cos 2\theta & =\dfrac{2r}{1-2r} \ \ \ \ \ (1) \\ \end{aligned}$ Moreover, on $\triangle KAF$ ,

$\tan \left( \angle KAF \right)=\dfrac{KF}{AF}\Rightarrow \tan \theta =\dfrac{r}{1-r} \ \ \ \ \ (2)$ Furthermore, $\cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \ \ \ \ \ (3)$ .

Combining $(1)$ , $(2)$ and $(3)$ ,

$\dfrac{2r}{1-2r}=\dfrac{1-{{\left( \dfrac{r}{1-r} \right)}^{2}}}{1+{{\left( \dfrac{r}{1-r} \right)}^{2}}}\Leftrightarrow \ldots \Leftrightarrow 4{{r}^{3}}-8{{r}^{2}}+6r-1=0$ By Cardano's method , this cubic equation gives

$r=\frac{1}{6}\left( 4-\frac{2}{\sqrt[3]{3\sqrt{33}-17}}+\sqrt[3]{3\sqrt{33}-17} \right)\approx 0.228155493653961$ Finaly, on $\triangle EAB$ ,

$\tan \left( \angle EAB \right)=\frac{EB}{AB}\Rightarrow \tan 2\theta =\frac{EB}{1}\Rightarrow EB=\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }\Rightarrow EB=\frac{2\frac{r}{1-r}}{1-{{\left( \frac{r}{1-r} \right)}^{2}}}\Rightarrow EB=\frac{2r\left( 1-r \right)}{1-2r}$ Substituting the value we found previously, we get $EB\approx 0.647798871261039$ .
For the answer, $\left\lfloor {{10}^{5}}EB \right\rfloor =\boxed{64779}$ .

Let $r$ denote the radius of the two identical circles. And for simplicities sake, define $0<h<1$ as the length $EB$ .

The radius of the incircle of a triangle can be expressed as $A = r \cdot s,$ where $A$ denotes the area of the triangle, and $s$ as the semiperimeter of the triangle.

Let us focus on the bottom right-triangle, its area $A$ is simply $\frac h2$ . By Pythagorean theorem , the hypotenuse of this right-triangle is $\sqrt{1+h^2}$ , and so its semiperimeter is $s = \frac12(1 + h + \sqrt{1+h^2})$ . Thus, we can express $r$ in terms of $h$ alone.

$A = rs \quad \Leftrightarrow \quad r = \frac As = \dfrac h{1 + h + \sqrt{1+h^2}}$

Let us rotate the graph picture and plot these points on the Cartesian plane as such:

The equation of the straight line $EA$ can be determined as $y = hx + (1-h)$

And because the bottom left circle (it was previously a top right circle) is tangent to the $x$ -axis, $y$ -axis, and the straight line $EA$ , the equation of this circle of radius $r$ satisfy $(x - r)^2 + (y-r)^2 = r^2$ . Substituting $y = hx + (1-h)$ , we get $(x-r)^2 + ( hx + 1-h-r)^2 = r^2 \quad\Leftrightarrow \quad x^2 (h^2 + 1) + x\big (-2r + 2h(1-h-r) \big) + (1-h-r)^2 = 0.$

The above is a quadratic equation whose discriminant (in $x$ ) is 0, $(-2r)^2 - 4(h^2 + 1)(1-h-r)(2h + (1-h-r)) = 0.$ Lastly, we substitute $r = \dfrac h{1 + h + \sqrt{1+h^2}}$ , and we can get $2h^3 -2h^2 + 2h - 1 = 0$ . By Cardano's method , $\Large EB = h = \frac13 \left( \, 1 - \sqrt[3]{\tfrac{32}{13 + 3 \sqrt{33}}} + \sqrt[3]{\tfrac{{13 + 3\sqrt{33}} }{{4}}} \, \right ) \approx 0.6477988.$ Hence, $\lfloor{10^5\cdot{EB}}\rfloor = \boxed{64779}.$

Let two circles radii equal to x, CE=b+x and EB=c+x. Write the following relations from figure

b+c+2x=1, on line CB

[(c-b)/2]^2+x^2=[(1/2)-x]^2, on tangent AE between two circles

(1-x+c)^2-(c+x)^2=1, on right triangle AEB

Solve with WolframAlpha, x=0.228155, c=0.419643,

EB=c+x=0.647799

Answer = 64779