5. Let a regular n-gon be denoted as $A_{1} A_{2} \ldots A_{n}$. The point $A_{3}$ is reflected over the axis $A_{2} A_{4}$, resulting in the point $A_{3}^{\prime}$. Then, the point $A_{3}^{\prime}$ is reflected over the axis $A_{1} A_{3}$, resulting in the point $A_{3}^{\prime \prime}$. For which $n \geqq 4$ is the point $A_{3}^{\prime \prime}$ identical to the intersection of the lines $A_{1} A_{2}$ and $A_{3} A_{4}$?

(Jaroslav Zhouf)
SOLUTION. Let $S$ be the center of the circle circumscribed around the $n$-gon $A_{1} A_{2} \ldots A_{n}$. Since the central angle $A_{i} S A_{i+1}$ is equal to $\frac{360^{\circ}}{n}$ for each $i=1, \ldots, n-1$, all the acute inscribed angles over the chords $A_{i} A_{i+1}$ have a size of $\frac{180^{\circ}}{n}$. Four of them are marked in red in the diagram. For $n=4$, the sum of the sizes of these 4 red angles is $180^{\circ}$, and thus the lines $A_{1} A_{2}$ and $A_{3} A_{4}$ are parallel, which excludes the smallest value $n=4$. For further considered $n \geq 5$, the sum of the sizes of the four red angles is less than $180^{\circ}$, so the intersection of the lines $A_{1} A_{2}$ and $A_{3} A_{4}$ lies on the "converging" rays $A_{1} A_{2}$ and $A_{4} A_{3}$ as shown in Figure 1.

According to the two congruent red angles at vertex $A_{4}$, the ray $A_{4} A_{2}$ is the angle bisector of $\angle A_{3} A_{4} A_{1}$, and thus the point $A_{3}^{\prime}$ (the image of $A_{3}$ under reflection over $A_{4} A_{2}$) lies on the ray $A_{4} A_{1}$, and, as suggested by Figure 1 and as will be further utilized, inside the segment $A_{4} A_{1}$. Indeed, the necessary inequality $\left|A_{3} A_{4}\right| < \left|A_{1} A_{4}\right|$ follows from triangle $A_{1} A_{3} A_{4}$, in which, due to our assumption $n \geq 5$, we have

$$
\left|\angle A_{4} A_{1} A_{3}\right| = \frac{180^{\circ}}{n} < (n-3) \cdot \frac{180^{\circ}}{n} = 180^{\circ} - \frac{3 \cdot 180^{\circ}}{n} = \left|\angle A_{1} A_{3} A_{4}\right| .
$$

Similarly, from the congruent angles at vertex $A_{1}$, it follows that the point $A_{3}^{\prime \prime}$ (the image of $A_{3}^{\prime}$ under reflection over $A_{1} A_{3}$) lies on the ray $A_{1} A_{2}$. Since this holds for any $n \geq 5$, our task is to determine when the point $A_{3}^{\prime \prime}$ also lies on the ray $A_{4} A_{3}$, or when the angle $\angle A_{3}^{\prime \prime} A_{3} A_{4}$ is a right angle.

![](https://cdn.mathpix.com/cropped/2024_04_17_018d606fe9591a6bd18ag-10.jpg?height=483&width=782&top_left_y=1472&top_left_x=640)

From the isosceles triangle $A_{3} A_{3}^{\prime} A_{4}$, we have $\left|\angle A_{3}^{\prime} A_{3} A_{4}\right| = 90^{\circ} - \frac{180^{\circ}}{n}$, and thus

$$
\begin{aligned}
\left|\angle A_{1} A_{3} A_{3}^{\prime}\right| & = \left|\angle A_{1} A_{3} A_{4}\right| - \left|\angle A_{3}^{\prime} A_{3} A_{4}\right| = \\
& = \left(180^{\circ} - \frac{3 \cdot 180^{\circ}}{n}\right) - \left(90^{\circ} - \frac{180^{\circ}}{n}\right) = 90^{\circ} - \frac{360^{\circ}}{n} .
\end{aligned}
$$

The determined angle $\angle A_{1} A_{3} A_{3}^{\prime}$ is congruent to the symmetrically associated angle $\angle A_{1} A_{3} A_{3}^{\prime \prime}$, so the angle $\angle A_{3}^{\prime \prime} A_{3} A_{3}^{\prime}$ has twice the size. We now have everything prepared for the necessary expression of the angle $\angle A_{3}^{\prime \prime} A_{3} A_{4}$:

$$
\begin{aligned}
& \left|\angle A_{3}^{\prime \prime} A_{3} A_{4}\right| = \left|\angle A_{3}^{\prime \prime} A_{3} A_{3}^{\prime}\right| + \left|\angle A_{3}^{\prime} A_{3} A_{4}\right| = 2 \cdot \left|\angle A_{1} A_{3} A_{3}^{\prime}\right| + \left|\angle A_{3}^{\prime} A_{3} A_{4}\right| = \\
& \quad = 2 \cdot \left(90^{\circ} - \frac{360^{\circ}}{n}\right) + \left(90^{\circ} - \frac{180^{\circ}}{n}\right) = 3 \cdot 90^{\circ} - \frac{5 \cdot 180^{\circ}}{n} = \frac{(3 n - 10) \cdot 180^{\circ}}{2 n} .
\end{aligned}
$$

The last expression clearly has the required value of $180^{\circ}$ if and only if $3 n - 10 = 2 n$ or $n = 10$. Thus, only for this $n$ does the point $A_{3}^{\prime \prime}$ lie on both lines $A_{1} A_{2}$ and $A_{3} A_{4}$.

Conclusion. The only number $n$ that satisfies the condition is 10.