[Theorem of tangent and secant; product of the entire secant and its external part Problem $\underline{52984}{ }_{\text {Topics: }}$ Law of Cosines Law of Sines

On the side of an angle with vertex $O$, points $A$ and $B$ are taken ( $A$ is between $O$ and $B$ ), such that $O A=3 A B$. A circle is drawn through points $A$ and $B$, touching the other side of the angle at point $D$. On the ray $O D$, a point $E$ is taken ( $D$ is between $O$ and $E$). It is known that $O E=m, \angle B O E=\alpha, \angle B E O=\beta$. Find the radius of the circle.
Apply the tangent and secant theorem, the sine and cosine theorems. The radius of the circumcircle of triangle $ABD$ is found using the formula $R=\frac{AD}{2 \sin \angle ABD}$.

## Solution

Let $AB=x, AO=3x, \angle ODA=\angle OBD=\gamma$. Then

\[
\begin{gathered}
OD^2 = OB \cdot OA = 12x^2, \quad AD^2 = 3x^2(7 - 4\sqrt{3} \cos \alpha), \\
\sin \gamma = \frac{3x \sin \alpha}{AD}, \quad x = \frac{\sin \beta}{\sin (\alpha + \beta)}.
\end{gathered}
\]

Therefore, the desired radius is

\[
\frac{AD}{2 \sin \gamma} = \frac{2m(7 - 4\sqrt{3} \cos \alpha) \sin \beta}{\sin \alpha \sin (\alpha + \beta)}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_c317f3f1f50d1d30304bg-46.jpg?height=340&width=829&top_left_y=1877&top_left_x=617)

## Answer

 $\frac{2m(7 - 4\sqrt{3} \cos \alpha) \sin \beta}{\sin \alpha \sin (\alpha + \beta)}$