Trapezoid $A B C D$ with bases $B C=2$ and $A D=10$ is such that a circle can be inscribed in it and a circle can be circumscribed around it. Determine where the center of the circumscribed circle is located, i.e., whether it is inside, outside, or on one of the sides of trapezoid $A B C D$. Also, find the ratio of the radii of the circumscribed and inscribed circles.
Prove that angle $A C D$ is obtuse. Find the radius of the circumscribed circle using the formula: $R=\frac{a}{2 \sin \alpha}$.

## Solution

Let $R$ and $r$ be the radii of the inscribed and circumscribed circles, and $K$ be the foot of the perpendicular dropped from vertex $C$ to side $A D$. Since the trapezoid is inscribed, it is isosceles. Then $A K=6, K D=4$, and since $2 C D=B C+A D$, we have $C D=6$. From this, we find that

$$
C K=2 \sqrt{5}, A C=2 \sqrt{14}
$$

Using the cosine theorem, we verify that angle $\angle A C D$ is obtuse. Therefore, the center of the circumscribed circle lies outside the trapezoid. Additionally,

$$
\sin \angle D=\frac{C K}{C D}=\frac{\sqrt{5}}{3}
$$

Thus,

$$
R=\frac{A C}{2 \sin \angle D}=\frac{3 \sqrt{14}}{5}, r=\frac{1}{2} C K=\sqrt{5}
$$

Therefore,

$$
\frac{R}{r}=\frac{3 \sqrt{14}}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-45.jpg?height=392&width=420&top_left_y=1576&top_left_x=819)

## Answer

Outside; $\frac{3 \sqrt{14}}{5}$