# Equilateral triangle

• Mar 10th 2010, 10:00 AM
atreyyu
Equilateral triangle
We have an equilateral triangle of side \$\displaystyle n\$. It's been cut into \$\displaystyle m\$ equilateral triangles of side 1 and some rhombi whose acute angles is \$\displaystyle 60^\circ\$. I have to show that \$\displaystyle m\geq n\$. So, it's straightforward when I draw it, that there are less than \$\displaystyle (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?
• Mar 10th 2010, 11:14 AM
Opalg
Quote:

Originally Posted by atreyyu
We have an equilateral triangle of side \$\displaystyle n\$. It's been cut into \$\displaystyle m\$ equilateral triangles of side 1 and some rhombi whose acute angles is \$\displaystyle 60^\circ\$. I have to show that \$\displaystyle m\geq n\$. So, it's straightforward when I draw it, that there are less than \$\displaystyle (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 \$\displaystyle \tfrac12n(n+1)\$ black triangles and \$\displaystyle \tfrac12n(n-1)\$ white triangles. Each rhombus must contain a white triangle, and there must be at least n left-over black triangles.
• Mar 10th 2010, 11:23 AM
atreyyu
sweet! (Bow)