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I am trying to calculate how many triangles that can be found in an equilateral triangle with $2n$ lines starting at the bottom angles and ending at the opposite side, such that equally many lines start/end of either side.

This is rather hard to explain, so I drew that first 5 terms: Triangle animation

The 1st term has, of course, 1 triangle.

The 2nd term has 8 triangles.

The 3rd term has 27 triangles.


This problem is really killing me, so any help would be greatly appreciated.

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Well, the way I see it; break down the lines into the two groups of $n$ lines on the left and right, and add the "left" and "right" side of the original triangle to make two groups of $n+1$ lines. There are then two cases I see:

  1. One side of the triangle is from a line on the "left", one from the "right", and the bottom side of the original triangle. There are $(n+1)^2$ such triangles.

  2. Two of the sides are from lines in one group, while the third side is from a line in the other. This gets us $2(n+1)\binom{n+1}{2}$ more.

That gets me $(n+1)^2 + 2(n+1)\binom{n+1}{2} = (n+1)^3$.

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  • $\begingroup$ (And just to confirm, I did count 27 triangles for $n=2$.) $\endgroup$ – Dennis Meng Oct 24 '13 at 17:21

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