Which are the regular polygons that can be dissected into regular polygons?
For an $n$-sided regular polygon, one of its angles is $\frac{n-2}{n} \cdot 180^{\circ}$. Therefore, three regular polygons cannot meet at a vertex during the dissection, because even the triple of the smallest angle of a regular polygon is already $180^{\circ}$. Thus, we only need to consider the cases where one or two of the angles of the polygons formed during the dissection cover one angle of the original regular polygon.

![](https://cdn.mathpix.com/cropped/2024_05_02_84a8d7d92293b06af44bg-1.jpg?height=211&width=483&top_left_y=369&top_left_x=802)

Figure 1

First, let's consider the case where one angle of the regular polygon is covered by a single angle of the same size. The side of the regular polygon formed during the dissection will obviously be smaller than the side of the original polygon, so, as shown in Figure 1, a $\frac{2}{n} \cdot 180^{\circ}$ angle will also form next to such a polygon, which must be filled with angles of regular polygons. If $n=3$ or $n=4$, then $\frac{2}{n} \cdot 180^{\circ}$ will be $120^{\circ}$ or $90^{\circ}$, respectively, and as shown in Figures 2 and 3, the dissection is possible (in multiple ways). For $n=5$, $\frac{2}{n} \cdot 180^{\circ}=72^{\circ}$, which cannot be filled with angles of regular polygons. The same situation occurs if $n>6$, because then $\frac{2}{n} \cdot 180^{\circ}<60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_84a8d7d92293b06af44bg-1.jpg?height=195&width=223&top_left_y=1087&top_left_x=932)

Figure 2

![](https://cdn.mathpix.com/cropped/2024_05_02_84a8d7d92293b06af44bg-1.jpg?height=208&width=198&top_left_y=1414&top_left_x=947)

Figure 3

![](https://cdn.mathpix.com/cropped/2024_05_02_84a8d7d92293b06af44bg-1.jpg?height=409&width=458&top_left_y=1747&top_left_x=817)

Figure 4

For $n=6$, the dissection can occur as shown in Figure 4. If we continue the procedure on the hexagons containing vertex $O$, we can find additional solutions. In this case, the angle at vertex $A$ is covered by a single new polygon's angle, while the angle at vertex $B$ is covered by two.

With our last solution, we transition to the case where two equilateral triangles meet at a vertex of the polygon. The angle of the polygon is thus $120^{\circ}$, which is the angle of a regular hexagon, which can clearly be dissected (in multiple ways) into equilateral triangles.

An equilateral triangle and a square can also meet at a vertex. In this case, the angle of the regular polygon is $60^{\circ}+90^{\circ}=150^{\circ}$, and

$$
\frac{n-2}{n} \cdot 180^{\circ}=150^{\circ} \quad \text { based on } \quad n=12
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_84a8d7d92293b06af44bg-2.jpg?height=458&width=460&top_left_y=121&top_left_x=819)

Figure 5

As shown in Figure 5, the regular dodecagon can indeed be constructed from 6 equilateral triangles, 6 squares, and one regular hexagon.

We only need to check if an equilateral triangle and a regular pentagon can meet at a vertex.

In this case, the angle of the polygon is $60^{\circ}+108^{\circ}=168^{\circ}$, and

$$
\frac{n-2}{n} \cdot 180^{\circ}=168^{\circ} \quad \text { based on } \quad n=30
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_84a8d7d92293b06af44bg-2.jpg?height=395&width=401&top_left_y=1079&top_left_x=843)

Figure 6

If the side of the pentagon and the 30-gon are equal, then the $84^{\circ}$ angle shown in Figure 6 cannot be covered by angles of regular polygons. If the side of the pentagon is smaller than that of the 30-gon, then, similar to the case shown in Figure 1, a $\frac{2}{5} \cdot 180^{\circ}=72^{\circ}$ angle will form next to the pentagon, which cannot be covered by angles of regular polygons.

Therefore, the $3, 4, 6$, and 12-sided regular polygons meet the requirements of the problem.