In a semicircle constructed over the diameter $A A^{\prime}=2 r$, the trapezoid $A M M^{\prime} A^{\prime}$ must be placed such that

$$
A M+M M^{\prime}+M^{\prime} A^{\prime}=2 l
$$

is satisfied. The distance $x=O P$ must be calculated, where $O$ is the center of the circle and $P$ is the projection of $M$ on the diameter $A A^{\prime}$. Discussion!
If $I$ is the midpoint of $M M^{\prime}$, then $M M^{\prime}=2 M I=2 O P=2 x$. From the right triangle $A M A^{\prime}$,

![](https://cdn.mathpix.com/cropped/2024_05_02_a8737651e017f2e818fbg-1.jpg?height=363&width=767&top_left_y=198&top_left_x=665)

Since $M^{\prime} A^{\prime}=A M$, our conditional equation can be written as

$$
2 \sqrt{2 r(2 r-x)}+2 x=2 l \quad \text { or } \quad \sqrt{2 r(r-x)}=l-x
$$

From this, we can see that $x \leq l$ and $x>0$. This implies that $r$ always lies outside the roots, and is greater than the roots, because it is greater than the half-sum of the roots, $(l-r)$. Indeed,

$$
r>l-r, \quad \text { because } 2 r>l, \text { provided } l \leq \frac{3 r}{2}
$$

$f(l)=2 r(l-r)>0$, if $l>r$. In this case, $l$ is greater than the roots, because $l>l-r, r>0$. If $l>0$, if $l>r \sqrt{2}$. With these values, both the product and the sum of the roots are positive, so both roots are positive and each is less than $l$ (provided $l>r!$). If $l<r$, there are no solutions.

Accordingly,

$$
\begin{aligned}
& \text { if } r \sqrt{2}<l<\frac{3 r}{2}, \text { there are two solutions; } \\
& \text { if } r<l<r \sqrt{2}, \text { there is one solution; } \\
& \text { if } l<r, \text { there are no solutions. }
\end{aligned}
$$

Let's examine the limiting cases. If $l=r$, then $x_{1}=x_{2}=r$; the trapezoid collapses into the diameter $A A^{\prime}$.

$$
l=r \sqrt{2} \quad \text { case } \quad x_{1}=0, \quad x_{2}=2 r(\sqrt{2}-1) \sim 0.83 r
$$

For $l=\frac{3 r}{2}$, $x_{1}=x_{2}=\frac{r}{2}$. In this case, the trapezoid is half of a regular hexagon! $\left(A M=M M^{\prime}=M^{\prime} A^{\prime}=r\right)$.