[ Pythagorean Theorem (direct and inverse). ] [ Common tangent to two circles ]

In the right sector $A O B$, from point $B$ as the center, an arc $O C$ ( $C$ - the point of intersection of this arc with the arc $A B$ ) of radius $B O$ is drawn. Circle $S_{1}$ is tangent to the arc $A B$, the arc $O C$, and the line $O A$, while circle $S_{2}$ is tangent to the arc $A B$, the line $O A$, and circle $S_{1}$. Find the ratio of the radius of circle $S_{1}$ to the radius of circle $S_{2}$.
Let $R$ be the radius of the sector, $P$ the center of circle $S_{1}$, $r$ its radius, $M$ and $D$ its points of tangency with the arc $AB$ and the line $OA$ respectively, $x$ the radius of circle $S_{2}$, $Q$ its center, $K$ and $N$ its points of tangency with the arc $AB$ and the line $OA$ respectively. According to problem $\underline{52700}$ a) $OD=2 \sqrt{r R}$. By the Pythagorean theorem, $OP^{2}=OD^{2}+DP^{2}$, or $(R-r)^{2}=4 R r+r^{2}$. From this, $R=6 r$. Consider two cases.

1) Point $N$ lies on the extension of segment $OD$ beyond point $D$ (left figure). Then $ON=OD+DN=2 \sqrt{r R}+2 \sqrt{r x}$.

By the Pythagorean theorem, $OQ^{2}=QN^{2}+ON^{2}$, that is, $(6-x)^{2}=x^{2}+4(r \sqrt{6}+\sqrt{r x})^{2}$, or $3 \frac{r}{x}-2 \sqrt{\frac{6 r}{x}}-4=0$. From this, $\frac{r}{x}=\left(\frac{\sqrt{6}+3 \sqrt{2}}{3}\right)^{2}=\frac{4(2+\sqrt{3})}{3}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-19.jpg?height=656&width=1446&top_left_y=1438&top_left_x=314)

2) Point $N$ lies between points $O$ and $D$ (right figure). Then $ON=OD-DN=2 \sqrt{r R}-2 \sqrt{r x}$. This leads to the equation $\frac{3 r}{x}+2 \sqrt{\frac{6 r}{x}}-4=0$, from which we find that $\frac{r}{x}=\frac{4(2-\sqrt{3})}{3}$.

## Answer

$\frac{4(2 \pm \sqrt{3})}{3}$.