Question

Let $ABCD$ be a square of side of unit length. Let a circle $C_1$ centered at $A$ with unit radius is drawn. Another circle $C_2$ which touches $C_1$ and the lines $AD$ and $AB$ are tangent to it, is also drawn. Let a tangent line from the point C to the circle $C_2$ meet the side $AB$ at $E$. If the length of $EB$ is $\alpha +\sqrt{3}\beta$ where $\alpha ,\beta$ are integers, then $\alpha +\beta$ is equal to

Answer 1

$\left(i\right)\sqrt{2r}+r=1$
$r=\frac{1}{\sqrt{2}+1}r=\sqrt{2}-1\left(ii\right)C{C}_2=2\sqrt{2}-2=2(\sqrt{2}-1)$
From $△C{C}_{2}N=\sin \varphi =\frac{\sqrt{2}-1}{2(\sqrt{2}-1)}$
$\Rightarrow \varphi ={30}^{\circ }$
$\left(iii\right)$ In $△ACE,$ from sine law,
$\frac{AE}{\sin \varphi }=\frac{AC}{\sin {105}^{\circ }}\Rightarrow AE=\frac{1}{2}\times \frac{\sqrt{2}}{\sqrt{3}+1}.2\sqrt{2}\Rightarrow AE=\frac{2}{\sqrt{3}+1}=\sqrt{3}-1\therefore EB=1-(\sqrt{3}-1)=2-\sqrt{3}$
Hence $\alpha =2,\beta =-1$
$\Rightarrow \alpha +\beta =1$