reduces to constructing an $n$-sided polygon inscribed in a given circle, the sides of which are respectively parallel to $n$ given lines.

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Suppose the required polygon $A_{1} A_{2} \ldots A_{n}$ is constructed. Let $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ be the angles at its vertices $A_{1}, A_{2}, \ldots, A_{n}$ respectively. Through the center $O$ of the given circle, draw lines $l_{1}, l_{2}, \ldots, l_{n}$, respectively perpendicular to the sides $A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{n} A_{1}$. Then the angles between the lines $l_{n}$ and $l_{1}, l_{1}$ and $l_{2}, \ldots, l_{n-1}$ and $l_{n}$ are respectively $\pi-\alpha_{1}, \pi-\alpha_{2}, \ldots, \pi-\alpha_{n}$, and their sum is $\pi n-\pi(n-2)=2 \pi$.

Under the composition of reflections with respect to the lines $l_{1}, l_{2}, \ldots, l_{n}$, the vertex $A_{1}$ is mapped to itself.

If $n$ is odd, then such a composition is a reflection with respect to some line $l$, and the point $A_{1}$ (a fixed point of this reflection) lies on the axis of symmetry. In this case, the following construction is suitable.

Through the center $O$ of the given circle, draw lines $l_{1}, l_{2}, \ldots, l_{n}$, successively forming angles $\pi-\alpha_{2}, \pi-\alpha_{3}, \ldots, \pi-\alpha_{n}, \pi-\alpha_{1}$ between them. Construct the image $M_{1}$ of an arbitrary point $M$ on the given circle under the composition of reflections with respect to the lines $l_{1}, l_{2}, \ldots, l_{n}$. The perpendicular bisector of the segment $M M_{1}$ intersects the circle at the desired vertex $A_{1}$.

The other vertices of the desired $n$-gon are constructed using reflections with respect to the lines $l_{1}, l_{2}, \ldots, l_{n-1}$. In this case, the problem has a unique solution (up to a motion).

If $n$ is even, then the composition of reflections with respect to the lines $l_{1}, l_{2}, \ldots, l_{n}$ is a rotation by an angle

$$
2\left(\pi-\alpha_{2}\right)+2\left(\pi-\alpha_{4}\right)+\ldots+2\left(\pi-\alpha_{n}\right)=\pi n-2\left(\alpha_{2}+\alpha_{4}+\ldots+\alpha_{n}\right)
$$

around the point $O$. Since the point $A_{1}$, distinct from the center $O$ of the rotation, remains in place under this rotation, it is an identity transformation. Therefore,

$$
\pi n-2\left(\alpha_{2}+\alpha_{4}+\ldots+\alpha_{n}\right)=2 \pi
$$

$$
\alpha_{2}+\alpha_{4}+\ldots+\alpha_{n}=\frac{\pi(n-2)}{2}
$$

Then

$$
\alpha_{1}+\alpha_{3}+\ldots+\alpha_{n-1}=\pi(n-2)-\frac{\pi(n-2)}{2}=\frac{\pi(n-2)}{2}
$$

Thus,

$$
\alpha_{1}+\alpha_{3}+\ldots+\alpha_{n-1}=\alpha_{2}+\alpha_{4}+\ldots+\alpha_{n}
$$

and the problem has infinitely many solutions. The vertex $A_{1}$ can be any point on the circle.

If

$$
\alpha_{1}+\alpha_{3}+\ldots+\alpha_{n-1} \neq \alpha_{2}+\alpha_{4}+\ldots+\alpha_{n}
$$

then there are no solutions.

## Answer

The construction is possible: for odd $n$ - always, for even $n$ - under the condition that

$$
\alpha_{1}+\alpha_{3}+\ldots+\alpha_{n-1}=\alpha_{2}+\alpha_{4}+\ldots+\alpha_{n}
$$