3. Let a regular n-gon be denoted as $A_{1} A_{2} \ldots A_{n}$. For which $n \geqq 5$ does the image of point $A_{3}$ under the axial symmetry with respect to the line $A_{1} A_{2}$ lie on the line $A_{4} A_{5}$? (Josef Tkadlec)
SOLUTION. Let $S$ be the center of the regular $n$-gon $A_{1} A_{2} \ldots A_{n}$ and $P$ the intersection of the rays $A_{1} A_{2}$ and $A_{5} A_{4}$. The point $P$ will exist for every $n>6$, the remaining cases $n=5$ and $n=6$ will be discussed at the end of the solution. Now, let us assume that $n>6$.

![](https://cdn.mathpix.com/cropped/2024_04_17_e8a9d2201357bccae2b6g-5.jpg?height=577&width=642&top_left_y=594&top_left_x=707)

Fig. 1

By symmetry with respect to the line $S A_{3}$, it is clear that the point $P$ lies on the ray $S A_{3}$. Let us denote by $A_{3}^{\prime}$ the image of the point $A_{3}$ under the reflection in the line $A_{1} A_{2}$. Further, let $\alpha=\left|\angle A_{1} S A_{2}\right|=\left|\angle A_{2} S A_{3}\right|=360^{\circ} / n$. In the isosceles triangle $S A_{1} A_{2}$, we have $\left|\angle S A_{2} A_{1}\right|=90^{\circ}-\frac{1}{2} \alpha$, so for its exterior angle, we have $\left|\angle S A_{2} P\right|=180^{\circ}-\left|\angle S A_{2} A_{1}\right|=90^{\circ}+\frac{1}{2} \alpha$, and thus from the triangle $S A_{2} P$ we get $\left|\angle A_{3} P A_{2}\right|=\left|\angle S P A_{2}\right|=180^{\circ}-\alpha-\left(90^{\circ}+\frac{1}{2} \alpha\right)=90^{\circ}-\frac{3}{2} \alpha$. Due to the symmetric placement of the points $A_{2}, A_{4}$ with respect to the line $S A_{3}=S P$, we also have $\left|\angle S P A_{4}\right|=90^{\circ}-\frac{3}{2} \alpha$. Similarly, due to the symmetric placement of the points $A_{3}, A_{3}^{\prime}$ with respect to the line $A_{1} A_{2}$, we also have $\left|\angle A_{2} P A_{3}^{\prime}\right|=90^{\circ}-\frac{3}{2} \alpha$. Together, this means that the (not necessarily convex) angle $A_{3}^{\prime} P A_{4}$ with the interior point $S$ has a size of $3 \cdot\left(90^{\circ}-\frac{3}{2} \alpha\right)=270^{\circ}-\frac{9}{2} \alpha$. Our task is to find all $n>6$ for which the last-determined size is $180^{\circ}$ (precisely then the point $A_{3}^{\prime}$ lies on the line $A_{4} A_{5}$). The equality $270^{\circ}-\frac{9}{2} \alpha=180^{\circ}$ occurs precisely when $\alpha=20^{\circ}$. Since we know $\alpha=360^{\circ} / n$, $n=18$ is the only solution to the problem in the domain of all numbers $n>6$.

It remains to discuss the cases $n=6$ and $n=5$. For $n=6$, the lines $A_{1} A_{2}$ and $A_{4} A_{5}$ are parallel and the line $A_{4} A_{5}$ lies entirely in the half-plane opposite to the half-plane $S A_{3} A_{3}^{\prime}$, and therefore the point $A_{3}^{\prime}$ does not lie on the line $A_{4} A_{5}$. For $n=5$, the point $A_{3}^{\prime}$ and the entire line $A_{4} A_{5}$ lie in opposite half-planes with the boundary line $A_{1} A_{3}$ (which is parallel to the line $A_{4} A_{5}$), so in this case, the point $A_{3}^{\prime}$ does not lie on the line $A_{4} A_{5}$ either.

Conclusion. The only solution to the problem is $n=18$.