Find the number of triangles whose sides are formed by the sides and the diagonals of a regular heptagon. (The vertices of triangles need not be the vertices of the heptagon).

First of all, there are $7 \cdot 4 / 2 = 14$ diagonals and $7$ sides. Then I tried drawing this out — except that regular heptagons are hard to draw and there are a lot of lines. I thought you could choose 3 lines to be sides of triangles but that won't always work because sometimes they just don't form a triangle. How can I approach / solve this problem>

  • $\begingroup$ Are the sides and diagonals allowed to be extended, or are you looking for triangles within the original heptagon? $\endgroup$ – Mark Bennet May 26 '18 at 8:55
  • $\begingroup$ @MarkBennet I am not entirely sure — the problem doesn't explicitly say. I thought they meant within the heptagon. $\endgroup$ – jjhh May 26 '18 at 8:59
  • $\begingroup$ I think it is important to clarify if the sides and diagonals are taken as independent segments of the figure of the heptagon. This is different from the fact that the heptagon must be unchanged and the triangles can be drawn in said figure. $\endgroup$ – Piquito May 26 '18 at 12:48
  • $\begingroup$ Maybe of interest math.stackexchange.com/questions/2239005/… $\endgroup$ – cgiovanardi May 26 '18 at 13:15
  • $\begingroup$ @cgiovanardi I had already seen it, but I'm not sure how it helps (thanks though) $\endgroup$ – jjhh May 26 '18 at 13:31

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