At least how many acute-angled triangles must be fitted together without gaps or overlaps to form an isosceles triangle with a $120^{\circ}$ vertex angle?
Solution. The 1st figure shows that seven triangles are sufficient to solve the problem. Here, $O$ is the center of the inscribed circle, which touches the sides at points $E, F, G$. Points $X, Y, Z, V$ were chosen such that the smaller acute angles of the congruent right triangles $O E X$, $O G Z$, $O G V$, $O F Y$ are $40^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_f342ef71066986ca9fa7g-1.jpg?height=366&width=845&top_left_y=305&top_left_x=629)

In the following, we will show that in general, there is no obtuse triangle that can be decomposed into fewer than seven acute triangles.

Consider an obtuse triangle that we have assembled from the minimum number, $h$, of acute triangles. At least two triangles must meet at the obtuse vertex. One of these small triangles cannot have a vertex in the interior of the side opposite the obtuse angle, because then one side of the small triangle would either cut off an obtuse triangle (which would require at least $h$ acute triangles to decompose) or it would cut the obtuse triangle into two right triangles, and a right triangle clearly cannot be assembled from fewer than four acute triangles. (Indeed, three triangles can only form a triangle in the two ways shown in the 2nd figure.)
![](https://cdn.mathpix.com/cropped/2024_05_02_f342ef71066986ca9fa7g-1.jpg?height=160&width=458&top_left_y=1126&top_left_x=824)

Figure 2

Therefore, there is at least one interior point that is a vertex of one of the small triangles. Such an interior point is either a vertex of at least 5 small triangles (otherwise, the $360^{\circ}$ angle would be divided into at most 4 parts, resulting in a non-acute angle), or it lies on the side of one small triangle and is a vertex of at least 3 more small triangles. In the first case, the small triangles meeting at this point form a convex polygon with at least 5 sides (each angle of the polygon is the sum of two acute angles of two small triangles). If an acute triangle is added to such a polygon, the number of sides of the resulting shape decreases by at most 1 (Figure 3). Therefore, at least 2 triangles must be added to form a triangle. In the second case, there are at least 4 small triangles, which again form a polygon with at least 5 sides (Figure 4). However, in this case, the angle marked $\alpha$ is acute, so at least 2 more small triangles must meet at this vertex, meaning that it cannot be completed into a triangle with fewer than 3 acute triangles.

![](https://cdn.mathpix.com/cropped/2024_05_02_f342ef71066986ca9fa7g-1.jpg?height=263&width=415&top_left_y=1842&top_left_x=844)

Figure 3

![](https://cdn.mathpix.com/cropped/2024_05_02_f342ef71066986ca9fa7g-1.jpg?height=279&width=353&top_left_y=2235&top_left_x=867)

Figure 4

Therefore, $h \geq 7$, so an isosceles triangle with a $120^{\circ}$ angle can only be assembled from at least seven acute triangles.