Task B-2.3. Four brothers live on an estate in the shape of an isosceles trapezoid with the length of the larger base $a$, the angle at the base $50^{\circ}$, and the angle between the diagonals $76^{\circ}$. Their houses are at the vertices of the trapezoid. They have decided to build a well equally distant from all four houses, that is, from the vertices of the trapezoid. How far should the well be from the vertices of the trapezoid? Can the well be built on their estate?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_4f4efb97ca3723b253dag-08.jpg?height=794&width=870&top_left_y=644&top_left_x=606)

For the well to be equidistant from each house, it must be at the center of the circumscribed circle of the trapezoid.

It is given that $\angle C E B=76^{\circ}$ or $\angle A E B=76^{\circ}$. However, $\angle A E B$ cannot be $76^{\circ}$ because otherwise, the angles $\angle E A B$ and $\angle E B A$ would be $52^{\circ}$ (triangle $A B E$ is isosceles). This is impossible because these angles are inside the trapezoid, within the angles $\angle D A B$ or $\angle C B A$, and must be less than $50^{\circ}$. Therefore,

$$
\begin{aligned}
& \angle C E B=76^{\circ} \Rightarrow \angle A E B=180^{\circ}-76^{\circ}=104^{\circ} \\
& \angle A B E=\frac{180^{\circ}-\angle A E B}{2}=38^{\circ} \\
& \angle A D B=180^{\circ}-(\angle D A B+\angle A B E)=92^{\circ}(\text { inscribed angle over chord } A B)
\end{aligned}
$$

Then the corresponding central angle $\angle A S B=2 \angle A D B=184^{\circ}>180^{\circ}$

It follows that the center of the circumscribed circle $S$ is outside the trapezoid.

This means that the well cannot be built on their property and be equidistant from all houses.

It is true that $\angle B A S=2^{\circ}$, so $\cos 2^{\circ}=\frac{\frac{a}{2}}{d}$, from which it follows that $d=\frac{a}{2 \cos 2^{\circ}}$. The well should be located at a distance from the houses of

$$
d=\frac{a}{2 \cos 2^{\circ}}
$$