Given $n$ sided regular polygon

$(a)$ Total number of $\triangle$ formed in which none of the sides are the sides of that polygon

$(b)\; $Total number of equilateral $\triangle$ formed in by joining the vertices of that polygon

$(c)\; $Total number of right angle $\triangle$ formed in by joining the vertices of that polygon

$\bf{My\; Try::}$ For $(a)$ one,

$\triangle$ with no side common $=$Total $\triangle-\triangle$ with one side common$-\triangle$ with $2$ side common

So we get $$\displaystyle = \binom{n}{3}-(n)\cdot (n-4)-n$$

Now, how can I solve other $2$ parts? Help required. Thanks.


Hints for (b) and (c):

The given conditions enforce conditions on $n$. Think of peripheral and central angles. In particular: If two diagonals meet at an angle of $60^\circ$ their other endpoints subtend an angle of $120^\circ$ at the center. This is only possible if $n$ satisfies a certain condition.

  • $\begingroup$ Thanks Christian Blatter, Would you like to explain me in detail $\endgroup$ – juantheron Jun 8 '16 at 9:58

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