B1.
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Let $r_{1}, r_{2}, r_{3}$, and $r_{4}$ be the radii of the circles $\mathcal{K}_{1}, \mathcal{K}_{2}, \mathcal{K}_{3}$, and $\mathcal{K}_{4}$, respectively, with $r_{3}=r_{4}$. Since triangle $A B C$ is equilateral, the circles $\mathcal{K}_{1}$ and $\mathcal{K}_{2}$ share a common center, which we denote by $S$. Triangle $B S D$ is half of an equilateral triangle, so $r_{1}=|S B|=2|S D|=2 r_{2}$. Since $G E$ is the diameter of circle $\mathcal{K}_{2}$, we have $|G E|=2 r_{2}=r_{1}$. Express the length $|G E|$ in another way using $r_{4}$. Let $T$ be the point of tangency of circles $\mathcal{K}_{3}$ and $\mathcal{K}_{4}$. Denote the center of circle $\mathcal{K}_{4}$ by $R$, and its points of tangency with sides $D E$ and $E F$ of square $D E F G$ by $U$ and $V$. Since circles $\mathcal{K}_{3}$ and $\mathcal{K}_{4}$ are of equal size, $T$ is the center of square $D E F G$, so $|G E|=2|T E|$. Quadrilateral $R U E V$ is a square with side length $r_{4}$, so its diagonal is $|R E|=\sqrt{2} r_{4}$. From this, we express $|G E|=2|T E|=2(|T R|+|R E|)=2\left(r_{4}+\sqrt{2} r_{4}\right)=(2+2 \sqrt{2}) r_{4}$. From both expressions for the length $|G E|$, we get $r_{1}=(2+2 \sqrt{2}) r_{4}$, so $\frac{r_{1}}{r_{4}}=(2+2 \sqrt{2})$.

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B1. Circles $\mathcal{K}_{1}$ and $\mathcal{K}_{2}$ in the figure are the circumcircle and incircle of an equilateral triangle $A B C$. Circle $\mathcal{K}_{2}$ has a square $D E F G$ inscribed in it, with point $D$ lying on side $A B$. Circles $\mathcal{K}_{3}$ and $\mathcal{K}_{4}$ are of equal size and touch each other, and each touches two sides of the square $D E F G$. Determine the ratio of the radii of circles $\mathcal{K}_{1}$ and $\mathcal{K}_{4}$.

(6 points)

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