Place two equilateral triangles with side lengths $12$ on the square in such a way that both have one vertex lie on the side of the square and the opposite sides of these vertices partially coincide with the other side of the square and with each other (Fig. 16). The area common to both triangles forms an equilateral triangle of side length $1$. Position the third

Fig. 16

equilateral triangle with side length $12$ between the two triangles, turned $180^\circ$, such that the lowermost vertex of that triangle coincides with the uppermost vertex of the small triangle. To show that the square is fully covered by these triangles, we must show that the sum of the heights of the large and the small triangle is at least $11$, which is equivalent to showing $\frac{\sqrt{3}}{2} \cdot (12+1) > 11$. As we can simplify this equation to $13\sqrt{3} > 22$ and $3 \cdot 169 > 484$, we see that it holds. Therefore at least a third of the $2012$ points or at least $671$ points lie in one of the three chosen triangles.