![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-45.jpg?height=232&width=895&top_left_y=1159&top_left_x=456)

In triangle $ABC$, the cosine of angle $BAC$ is $\frac{1}{2}$, $AB=2$, $AC=3$. Point $D$ lies on the extension of side $AC$, and $C$ is between $A$ and $D$, with $CD=3$. Find the ratio of the radius of the circumcircle of triangle $ABC$ to the radius of the incircle of triangle $ABD$.
Apply the Law of Cosines.

## Solution

By the Law of Cosines in triangle $ABC$, we find that

$$
BC = \sqrt{AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos \angle BAC} = \sqrt{4 + 9 - 6} = \sqrt{7}
$$

If $R$ is the radius of the circumcircle of triangle $ABC$, then

$$
R = \frac{BC}{2 \sin \angle BAC} = \frac{\sqrt{7}}{\sqrt{3}}
$$

By the Law of Cosines in triangle $ABD$, we find that $BD = 2\sqrt{7}$. Let $O$ be the center of the incircle of this triangle, $r$ be its radius, and $K$ be the point of tangency with side $AD$. Then

$$
r = OK = AK \cdot \tan \angle OAK = \left(\frac{AB + BD + AD}{2} - BD\right) \cdot \tan 30^\circ = \frac{4 - \sqrt{7}}{\sqrt{3}}
$$

Therefore, $\frac{R}{r} = \frac{7 + 4\sqrt{7}}{9}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-46.jpg?height=363&width=740&top_left_y=-1&top_left_x=662)

## Answer

$\frac{7 + 4\sqrt{7}}{9}$.