## Task 1 - 070821

If squares are constructed outward on the sides of an equilateral triangle, the outer vertices of the squares form the vertices of a convex hexagon. We denote the area of the triangle by $A_{3}$; that of each of the squares by $A_{4}$, and that of the hexagon by $A_{6}$.

We are looking for integers $n$ and $m$ such that the equation $A_{6}=n A_{3}+m A_{4}$ holds.
In the following drawing, it can be seen that the area of the hexagon $A_{6}$ is composed as follows (the area $A_{x}$ denotes the area of the triangle between the squares as shown in the drawing):

$$
A_{6}=A_{3}+3 \cdot A_{4}+3 \cdot A_{x}
$$

The red lines divide the triangles $A_{3}$ and $A_{x}$ into congruent triangles, each with a right angle, a hypotenuse of length $a$, and another equal angle $\gamma$. This can be justified as follows: around point $C$, the full angle $(=360^{\circ})$ is composed of two angles of $90^{\circ}$ (in the squares) and two angles $\gamma=30^{\circ}$ and $\alpha$, i.e., $\alpha+\gamma=90^{\circ}$ or, in the triangle with $\alpha$, it follows that $\delta=\gamma$.

Thus, and using the area formula for an equilateral triangle, we have:

$$
A_{3}=A_{x}=\frac{\sqrt{3}}{4} a^{2}
$$

Furthermore, for the square, we have:

$$
A_{4}=a^{2}
$$

Substituting (2) and (3) into (1) yields: $A_{6}=4 \cdot A_{3}+3 \cdot A_{4}$, which shows that $n=4$ and $m=3$ are possible solutions.