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Thread: Equilateral triangle

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    Equilateral triangle

    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?
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    Quote Originally Posted by atreyyu View Post
    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.
    Attached Thumbnails Attached Thumbnails Equilateral triangle-triangles.gif  
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    sweet!
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