Estonian Mathematical Olympiad

Let the radii of ${\omega }_1$ and ${\omega }_2$ be $r_1$ and $r_2$ respectively. Let $E$ be the second intersection of $AK$ and ${\omega }_2$ (Fig. 40). As $AK$ and $KE$ are diameters of ${\omega }_1$ and ${\omega }_2$ respectively, the angles $ABK$ and $KDE$ must be right angles. In triangle $AKC$, the segment $KB$ is both a median and an altitude, thus $KA=KC$ and $\angle KAC=\angle ACK$. As $CDEK$ is cyclic, we have $$\angle EAD=\angle KAC=\angle ACK=\angle DEA.$$ Thus $ADE$ is isosceles, also $KBA$ and $KDE$ are similar. Therefore $$\frac{r_1}{r_2}=\frac{AK}{KE}=\frac{AB}{DE}=\frac{AB}{AD}=\frac{1}{3}.$$

Fig. 40

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Alternative solution.

Let the radii of ${\omega }_1$ and ${\omega }_2$ be $r_1$ and $r_2$ respectively. Let $P$ be the midpoint of $CD$ and $Q$ the center of ${\omega }_2$ (Fig. 41). Then $PQ$ is perpendicular to $CD$, as it connects the midpoint of the chord $CD$ of ${\omega }_2$ and the center of ${\omega }_2$. Thus $\angle APQ=\angle CPQ=90^{\circ }$. As $AK$ is a diameter of ${\omega }_1$, we have $\angle ABK=90^{\circ }$. Therefore $BK\parallel PQ$. Together with $AB=BC=CD$, we get $\frac{AQ}{AK}=\frac{AP}{AB}=2.5$ or $2AQ=5AK$. From here $2(2r_1+r_2)=5\cdot 2r_1$, from which $\frac{r_1}{r_2}=\frac{1}{3}$.

Fig. 41