We have an equilateral triangle of side $n$. It's been cut into $m$ equilateral triangles of side 1 and some rhombi whose acute angles is $60^\circ$. I have to show that $m\geq n$. So, it's straightforward when I draw it, that there are less than $(n^2-n)/2$ pairs of different smaller triangles with one common side. I could describe the situation as I see it, which is trying to stick in another rhombus step-by-step to show when it eventually becomes impossible. But is there a more formal way to do that? If not, what's the most formal way of presenting what I have described?
We have an equilateral triangle of side $n$. It's been cut into $m$ equilateral triangles of side 1 and some rhombi whose acute angles is $60^\circ$. I have to show that $m\geq n$. So, it's straightforward when I draw it, that there are less than $(n^2-n)/2$ pairs of different smaller triangles with one common side. I could describe the situation as I see it, which is trying to stick in another rhombus step-by-step to show when it eventually becomes impossible. But is there a more formal way to do that? If not, what's the most formal way of presenting what I have described?
Divide the triangle into smaller triangles of side 1, and colour them in a chequerboard pattern, as in the attachment. There are $\tfrac12n(n+1)$ black triangles and $\tfrac12n(n-1)$ white triangles. Each rhombus must contain a white triangle, and there must be at least n left-over black triangles.