How many distinct equilateral triangles can be formed in a regular nonagon having atleast two of their vertices as the vertices of nonagon.As the question says two vertices are common so, I did 9 C 2 = 36 but after that i am not able to proceed further. how should I think?

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How many distinct equilateral triangles can be formed in a regular nonagon having atleast two of their vertices as the vertices of nonagon.As the question says two vertices are common so, I did           9        C    2    =  36 but after that i am not able to proceed further. how should I think?

America Barrera · Accepted Answer

This is simple using a simplified* form of inclusion-exclusion. There are                     (                    9      2                      )              =  36 chords in the 9-gon, each giving two distinct equilateral triangles, for 72 in all. But some triangles are counted more than once; it is not hard to see that if two side of such a triangle are chords, the third one is as well, and the vertices then are regularly spaced at distance 3 around the 9-gon. This gives 3 triangles that are over-counted, namely 3 times each; total over-counting   3  ×  (  3  −  1  )  =  6, so 72−6=66 triangles in all.By the way, I am reading &quot;having two of their vertices...&quot; as a minimal requirement; if you want to insist on exactly two vertices, then the 3 special triangles should not be counted at all, so you should correct by   3  ×  3  =  9 rather than by 6.*This is simplified because we can get a direct measure of the over-counting being done. Normally one would enumerate all pairs of initial data that produce the same result and compensate for that, then all triples that thus lead to over-compensation, correcting for them, and so forth. But here that seems unnecessarily complicated.

How many distinct equilateral triangles can be formed in a regular nonagon having atleast two of the

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