1. Let $ABCD$ be an isosceles trapezoid ($AD \| BC$) with an angle $\frac{\pi}{3}$ at the longer base $AD$ and diagonal $\overline{AC}=\sqrt{3}$. A point $M$ is at a distance of 1 from $A$ and at a distance of 3 from $D$. Find $\overline{MC}$.
Solution. One

$$
\begin{gathered}
\overline{A D} \geq \overline{M D}-\overline{A M}=3-1=2 \text { and } \\
\overline{A D} \sin \frac{\pi}{3} \leq \sqrt{3}
\end{gathered}
$$

it follows that $\overline{A D}=2$ and $\measuredangle A C D=\frac{\pi}{2}$. Therefore, point $M$ lies on line $A D$ and is at a distance of 1 and 3 from $A$ and $D$ respectively. From the right triangle

![](https://cdn.mathpix.com/cropped/2024_06_05_47c254b2217fdab00c02g-4.jpg?height=342&width=647&top_left_y=985&top_left_x=818)
$A C D$ it follows that $\overline{C D}=1$. From this, it also follows that $\overline{B C}=1$ and

$$
\overline{M C}=\sqrt{\left(1+2-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\sqrt{7}
$$