[ Isosceles, Inscribed, and Circumscribed Trapezoids ]

A circle is inscribed in trapezoid $A B C D$. The extensions of the lateral sides $A D$ and $B C$ of the trapezoid beyond points $D$ and $C$ intersect at point $E$. The perimeter of triangle $D C E$ and the base of the trapezoid $A B$ are 60 and 20, respectively, and angle $A D C$ is $\beta$. Find the radius of the circle.
Let $M$ be the point of tangency of the given circle with the segment $A D$. Express the segments $A M$ and $D M$ in terms of the radius of the circle and the angle $\beta$. Find the perimeter of triangle $A B E$ and use the similarity of triangles $E D C$ and

$E A B$.

## Solution

Let $O$ be the center of the given circle, $r$ its radius, and $M, N, K$ the points of tangency with segments $A D, D C, B C$ respectively. Since

$$
E M=E K, E M+E K=E D+D N+N C+C E=E D+D C+C E=60
$$

then $E M=30$.

Let $P$ be the perimeter of triangle $A B E$. Then $E M=\frac{P}{2}-A B$. From this, we find that

$$
P=2(E M+A B)=2(30+20)=100
$$

Therefore, the similarity coefficient of triangles $E D C$ and $E A B$ is $\frac{3}{5}$ (the ratio of the perimeters).

From the right triangles $A M O$ and $D M O$, we find that

$$
D M=\frac{O M}{\operatorname{tg} \angle O D M}=\frac{r}{\operatorname{tg} \frac{\beta}{2}}, A M=O M \lg \angle A O M=\operatorname{rtg} \frac{\beta}{2}
$$

Therefore,

$$
\frac{E D}{E A}=\frac{E M-M D}{E M+A M}=\frac{\left(30-\frac{r}{\operatorname{tg} \frac{\beta}{2}}\right)}{30+r \operatorname{tg} \frac{\beta}{2}}=\frac{3}{5}
$$

From this, we find that

$$
r=\frac{60 \operatorname{tg} \frac{\beta}{2}}{5+3 \operatorname{tg}^{2} \frac{\beta}{2}}=\frac{30 \sin \beta}{4+\cos \beta}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-26.jpg?height=440&width=686&top_left_y=1206&top_left_x=686)

Answer

$\frac{60 \operatorname{tg} \frac{\beta}{2}}{5+3 \cdot \operatorname{tg}^{2} \frac{\beta}{2}}=\frac{30 \sin \beta}{4+\cos \beta}$