$\underline{\text { Sengerov B.A. }}$

Given a semicircle with diameter $A B$. Using a compass and a straightedge, construct a chord $M N$, parallel to $A B$, such that the trapezoid $A M N B$ is circumscribed.
Let $P$ be the projection of point $N$ onto $A B$. Express $A P$ in terms of $A B$.

## Solution

Assume the required trapezoid $A M N B$ is constructed. Let $P$ be the projection of vertex $N$ onto $A B$ (Fig.1). Then

$$
A P=\frac{M N+A B}{2}=\frac{A M+B N}{2}=B N
$$

(trapezoid $A M N B$ is isosceles, as it is inscribed in a circle).

On the other hand, since triangle $A N B$ is right-angled, we have

$$
A P^{2}=B N^{2}=A B \cdot B P=A B(A B-A P), \text{ or } A P^{2}+A P \cdot A B-A B^{2}=0
$$

From this, we find that $A P=A B \cdot \frac{\sqrt{5}-1}{2}$.

To construct the required segment $A P$ (Fig.2), on the perpendicular to the given segment $A B$ passing through point $B$, lay off segment $B D$ equal to $\frac{1}{2} A B$, and on ray $D A$, lay off segment $D E$ equal to $B D$. Then

$$
A E=A D-D E=\sqrt{A B^{2}+\frac{A B^{2}}{4}}-\frac{1}{2} A B=A B \cdot \frac{\sqrt{5}}{2}-\frac{1}{2} A B=A B \cdot \frac{\sqrt{5}-1}{2} .
$$

On segment $A B$, mark a point $P$ such that $A P=A E$. Then the perpendicular to $A B$ through point $P$ intersects the given semicircle at point $N$ - the vertex of the required trapezoid.

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-47.jpg?height=295&width=552&top_left_y=1213&top_left_x=407)

Fig. 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-47.jpg?height=338&width=554&top_left_y=1160&top_left_x=1115)

Fig. 2.

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