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

Mayra Berry

Mayra Berry

Answered question

2022-06-18

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?

Answer & Explanation

America Barrera

America Barrera

Beginner2022-06-19Added 23 answers

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 "having two of their vertices..." 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.

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