, 2015

Denote by $r_1, r_2, r_3$, and $r_4$ the radii of the circles $K_1, K_2, K_3$, and $K_4$. We know that $r_3 = r_4$. Since the triangle $ABC$ is equilateral the circles $K_1$ and $K_2$ have a common center which we denote by $S$. The triangle $BSD$ is a half of an equilateral triangle, thus $r_1 = |SB| = 2|SD| = 2r_2$. Since $GE$ is the diameter of the circle $K_2$ we have $|GE| = 2r_2 = r_1$. Let's express the length of $|GE|$ in another way in terms of $r_4$. Let $T$ be the point where the circles $K_3$ and $K_4$ touch. Denote the center of the circle $K_4$ by $R$ and its contact points with sides $DE$ and $EF$ of the square $DEFG$ by $U$ and $V$. Since the circles $K_3$ and $K_4$ are of equal size, $T$ is the center of the square $DEFG$ and thus $|GE| = 2|TE|$. The quadrilateral $RUEV$ is a square with side length $r_4$ therefore its diagonal is of length $|RE| = \sqrt{2}r_4$. From this we deduce $|GE| = 2|TE| = 2(|TR| + |RE|) = 2(r_4 + \sqrt{2}r_4) = (2 + 2\sqrt{2})r_4$. It follows $r_1 = (2 + 2\sqrt{2})r_4$ and thus $\frac{r_1}{r_4} = (2 + 2\sqrt{2})$.