An isosceles trapezoid has sides of lengths $1,1,1$ and $\sqrt{2}$. Calculate the radius of the circle circumscribed around the trapezoid!
Let's label the vertices of the trapezoid such that $\overline{A B}=\overline{B C}=\overline{D C}=1, \overline{A D}=\sqrt{2}$. Denote the center of the circumscribed circle by $O$, the angle $\angle B A D$ by $\alpha$, the projection of $B$ onto the side $A D$ by $B^{\prime}$, and the foot of the perpendicular from $O$ to $B D$ by $F$.

![](https://cdn.mathpix.com/cropped/2024_05_02_7a3b989207595e5b93e5g-1.jpg?height=532&width=637&top_left_y=257&top_left_x=733)

Since $\triangle B O F \sim \triangle B A B^{\prime}$ and $\angle B O D = 2 \alpha$ by the inscribed angle theorem, and since $\triangle B O D$ is isosceles,

$$
\angle B O F = \alpha
$$

Thus,

$$
r = \overline{B O} = \overline{B F} \cdot \frac{\overline{B A}}{\overline{B B^{\prime}}}
$$

Let's calculate the necessary segment lengths. $\overline{B A} = 1$, the distance $\overline{B B^{\prime}}$ can be found from the triangle $\triangle B B^{\prime} A$, and $\overline{B F} = \frac{1}{2} \overline{B D}$ can be calculated from the right triangle $\triangle B B^{\prime} D$ using the Pythagorean theorem.

$$
\begin{gathered}
\overline{B B^{\prime}} = \sqrt{\overline{B A}^{2} - \overline{B^{\prime} A}^{2}} = \sqrt{1 - \left(\frac{\sqrt{2} - 1}{2}\right)^{2}} = \sqrt{\frac{2 \sqrt{2} + 1}{4}} \\
\overline{D B} = \sqrt{\overline{B B^{\prime}}^{2} + \overline{D B^{\prime}}^{2}} = \sqrt{\sqrt{\frac{2 \sqrt{2} + 1}{4}} + \left(\frac{2 \sqrt{2} - (\sqrt{2} - 1)}{2}\right)^{2}} = \sqrt{\sqrt{2} + 1}
\end{gathered}
$$

Substituting into (1),

$$
r = \frac{\sqrt{\sqrt{2} + 1}}{\sqrt{2 \sqrt{2} + 1}} = \sqrt{\frac{\sqrt{2} + 3}{7}} = 0.794
$$