4. Variant 1.

Diagonals $A C$ and $B D$ of an isosceles trapezoid $A B C D$ intersect at point $O$. It is known that $A D$ : $B C=3: 2$. Circle $\omega$ with center $O$, passing through vertices $A$ and $D$, intersects the extension of base $B C$ beyond point $B$ at point $K$. It turns out that $B K=B O$. Find the ratio of the base $A D$ to the radius of circle $\omega$.
Answer: $1 / 4$.

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_fc73715326e960a104bag-3.jpg?height=707&width=1022&top_left_y=1186&top_left_x=497)

Let $B C=2 a, O B=b$, then $A D=3 a, O A=3 b / 2$. Drop a perpendicular $O M$ from point $O$ to line $B C$ (from the isosceles property of the trapezoid, it follows that $M$ is the midpoint of $B C$). Next, we will find $O M^{2}$ in two ways. From triangle $O B M$ by the Pythagorean theorem, $O M^{2}=O B^{2}-B M^{2}=b^{2}-a^{2}$. From triangle $O K M$ by the Pythagorean theorem, $O M^{2}=O K^{2}-K M^{2}=\frac{9 b^{2}}{4}-(b+a)^{2}$. Equating the right sides, we have: $b^{2}-a^{2}=\frac{9 b^{2}}{4}-(b+a)^{2}, b^{2}=8 a b \cdot \frac{A D}{A O}=\frac{2 a}{b}=\frac{1}{4}$.