Local Mathematical Competitions

The ratio of the areas of the equilateral triangle $\Delta$ of side $n+\frac{1}{2n}$ and that of the equilateral triangle ${\Delta }_1$ of side $1$ is the square of the ratio of the lengths of their sides, i.e. $$(n+\frac{1}{2n}{)}^2>n^2+1,$$ hence at least $n^2+2$ triangles ${\Delta }_1$ are needed.

For $n=1$ we can do that by placing three ${\Delta }_1$ triangles at the corners.

Assume now this proven until $n$, and prove by induction for $n+1$. A $\Delta$ of side $n+1$ triangle placed at the top corner will use $n^2+2$ triangles ${\Delta }_1$, according with the induction hypothesis. It remains a trapezoidal strip at the bottom, of length of the nonparallel sides $$n+1+\frac{1}{2(n+1)}-n-\frac{1}{2n}=1-\frac{1}{2n(n+1)},$$ and basis lengths $n+\frac{1}{2n}$ and $n+1+\frac{1}{2(n+1)}$, with $(n+1{)}^2+2-n^2-2=2n+1$ triangles ${\Delta }_1$ available to cover it.

Place $2n+1$ triangles ${\Delta }_1$ one next to another, every second one "slid" downwards by $\frac{1}{2n(n+1)}$. They will cover a trapezoidal strip of exactly the dimensions of the above, since $$(n+1)\cdot 1+n\cdot \frac{1}{2n(n+1)}=n+1+\frac{1}{2(n+1)}.$$