Regular Polygon inscribed in a circle

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tanmoy
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Regular Polygon inscribed in a circle

Unread post by tanmoy » Mon Oct 20, 2014 10:11 pm

Let $A_{1}A_{2}A_{3}......A_{21}$ be a $21$ sided regular polygon inscribed in a circle with center $O$.How many triangles $A_{i} A_{j} A_{k}$,$1\leq i< j< k\leq 21$,contain the point $O$ in their interior
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seemanta001
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Re: Regular Polygon inscribed in a circle

Unread post by seemanta001 » Sat Jun 27, 2015 5:06 pm

If point $O$ is on any side of a triangle, shall we count it as inscribed??
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tanmoy
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Re: Regular Polygon inscribed in a circle

Unread post by tanmoy » Sat Jun 27, 2015 7:50 pm

seemanta001 wrote:If point $O$ is on any side of a triangle, shall we count it as inscribed??
No
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seemanta001
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Re: Regular Polygon inscribed in a circle

Unread post by seemanta001 » Mon Jul 06, 2015 9:33 pm

A regular polygon inscribed in a circle with center $O$ has $n$ points(where $n$ is odd).
Let, $$N=\lfloor\dfrac{n}{2}\rfloor$$.
Then the number of the triangles having $O$ as inscribed is $N^2+(N-1)^2+(N-2)^2+........+1$.
In this case the answer is $385$.
Please let me know if there are any other types of solutions.... :)
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Tahmid
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Re: Regular Polygon inscribed in a circle

Unread post by Tahmid » Wed Jul 08, 2015 4:09 am

general solution:
first choose any of $n$ points .
now we have even numbers of rest points .
draw a line with two vertices such that the line divides the rest points into equal parts. [the line must be parallel to one of the adjacent side of the choosen point]
so one part has $\frac{n+1}{2}+1$ points [with choosen point] and another part has $\frac{n+1}{2}$ points [2 points are common in each part].
(the process has been shown it the attachment for $n=9$ , where 'A' is our first choosen point)
by angle chasing it is easy to show that $O$ is situated in that part which has the first choosen point .
so if we choose two points from another part and make a triangle with first point , it will contain $O$ in its interior .
let $\frac{n+1}{2}=m$
so for every $n$ we can choose two another points in ${m \choose 2}$
total triangle $=n{m \choose 2}$
but each triangle is counted three times .
so $\frac{n}{3}{m \choose 2}$ triangles contain $O$ in their interior.

for $n=21$ , number of triangles =$\frac{21}{3}{11 \choose 2}=7*55=385$
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Tahmid
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Re: Regular Polygon inscribed in a circle

Unread post by Tahmid » Wed Jul 08, 2015 4:20 am

i just linked up my solution with seemanta's one ,
$\frac{n}{3}{m \choose 2}$
$=\frac{(2m-1)}{3}\frac{m(m-1)}{2}$ (as$ n=2m-1$)
$=\frac{(m-1)m(2m-1)}{6}$
$={1}^{2}+{2}^{2}+......+{(m-1)}^{2}$ :)
seemanta001 wrote: $N^2+(N-1)^2+(N-2)^2+........+1$.

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