4[Pythagorean Theorem (direct and inverse)]

In the right sector $A O B$, from point $B$ as the center, an arc $O C$ ($C$ is the intersection point of this arc with arc $A B$) of radius $B O$ is drawn. Circle $\omega$ is tangent to arc $A B$, arc $O C$, and line $O A$, while circle $\omega'$ is tangent to arc $O C$, line $O A$, and circle $\omega$. Find the ratio of the radius of circle $\omega$ to the radius of circle $\omega'$.
We can assume that the radius $\omega$ is 1. Let $R$ be the radius of the sector, $O_{1}$ be the center of $\omega$, and $D$ be the projection of $O_{1}$ onto $OB$. Then, $O_{1}B^{2} - BD^{2} = O_{1}D^{2} = O_{1}O^{2} - OD^{2}$, which means $(R+1)^{2} - (R-1)^{2} = (R-1)^{2} - 1^{2}$, from which we get $R=6$, and $O_{1}D^{2} = 24$.

Consider the case when $\omega^{\prime}$ is located between $\omega$ and point $O$. Let $O^{\prime}$ be the center of $\omega^{\prime}$, $r$ be its radius, $E$ be the projection of $O^{\prime}$ onto $OB$, and $F, G$ be the projections of $O^{\prime}$ and $O_{1}$ onto $OA$. Then,

$FG^{2} = (1+r)^{2} - (1-r)^{2} = 4r, O^{\prime}B^{2} - BE^{2} = O^{\prime}E^{2}$, which means $(6+r)^{2} - (6-r)^{2} = (\sqrt{24} - 2\sqrt{r})^{2}$. From this, we get
$r = \frac{6}{25}(\sqrt{6} - 1)^{2}, \frac{1}{r} = \frac{7 + 2\sqrt{6}}{6}$.

The second case of the position of the circle $\omega^{\prime}$ is considered similarly.

## Answer

$\frac{7 \pm 2\sqrt{6}}{6}$.