What is the maximum number of sides a convex polygon can have if it can be divided into right-angled triangles with acute angles of 30 and 60 degrees?
Solution. The polygon is convex, so each of its angles is less than $180^{\circ}$. In the decomposition of the polygon into right-angled triangles, every angle is a multiple of $30^{\circ}$, so the angles of the polygon are also multiples of $30^{\circ}$, and thus each angle is at most $150^{\circ}$.

We know that the sum of the angles of an $n$-sided polygon is $(n-2) \cdot 180^{\circ}$, so in our case,

$$
(n-2) \cdot 180^{\circ} \leq n \cdot 150^{\circ}, \quad \text { i.e., } \quad n \leq 12
$$

A 12-sided polygon that meets the conditions does exist, as shown in Figure 1. This was created by first combining two congruent triangles into a rectangle. Then, on the longer sides of the rectangle, we placed symmetric trapezoids made from 4 larger triangles, each with two angles of $150^{\circ}$ and $30^{\circ}$. Finally, similar symmetric trapezoids were placed on the shorter sides of the rectangle. The resulting 12-sided polygon, made from 18 triangles, has every angle at $150^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9b6b3149f3d835518fddg-1.jpg?height=347&width=393&top_left_y=667&top_left_x=844)

Remarks. 1. The problem can also be solved using exterior angles. We know that the sum of the exterior angles of a convex polygon is $360^{\circ}$. If our polygon can be decomposed into triangles where every angle is a multiple of $30^{\circ}$, then every exterior angle is at least $30^{\circ}$. Thus, the polygon can have at most $\frac{360}{30}=12$ vertices.

2. If we do not require the polygon to be convex, the number of vertices can be arbitrarily large. This is easily seen from Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_9b6b3149f3d835518fddg-1.jpg?height=214&width=613&top_left_y=1319&top_left_x=737)