Regular Polygon inscribed in a circle
Regular Polygon inscribed in a circle
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
"Questions we can't answer are far better than answers we can't question"
 seemanta001
 Posts: 13
 Joined: Sat Jun 06, 2015 9:31 am
 Location: Chittagong
Re: Regular Polygon inscribed in a circle
If point $O$ is on any side of a triangle, shall we count it as inscribed??
"Sometimes it's the very people who no one imagines anything of who do the things no one can imagine"
Re: Regular Polygon inscribed in a circle
Noseemanta001 wrote:If point $O$ is on any side of a triangle, shall we count it as inscribed??
"Questions we can't answer are far better than answers we can't question"
 seemanta001
 Posts: 13
 Joined: Sat Jun 06, 2015 9:31 am
 Location: Chittagong
Re: Regular Polygon inscribed in a circle
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 $N^2+(N1)^2+(N2)^2+........+1$.
In this case the answer is $385$.
Please let me know if there are any other types of solutions....
Let, $$N=\lfloor\dfrac{n}{2}\rfloor$$.
Then the number of the triangles having $O$ as inscribed is $N^2+(N1)^2+(N2)^2+........+1$.
In this case the answer is $385$.
Please let me know if there are any other types of solutions....
"Sometimes it's the very people who no one imagines anything of who do the things no one can imagine"
Re: Regular Polygon inscribed in a circle
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 =$\frac{21}{3}{11 \choose 2}=7*55=385$
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 =$\frac{21}{3}{11 \choose 2}=7*55=385$
 Attachments

 process
 hu.PNG (23.64 KiB) Viewed 4346 times
Re: Regular Polygon inscribed in a circle
i just linked up my solution with seemanta's one ,
$\frac{n}{3}{m \choose 2}$
$=\frac{(2m1)}{3}\frac{m(m1)}{2}$ (as$ n=2m1$)
$=\frac{(m1)m(2m1)}{6}$
$={1}^{2}+{2}^{2}+......+{(m1)}^{2}$
$\frac{n}{3}{m \choose 2}$
$=\frac{(2m1)}{3}\frac{m(m1)}{2}$ (as$ n=2m1$)
$=\frac{(m1)m(2m1)}{6}$
$={1}^{2}+{2}^{2}+......+{(m1)}^{2}$
seemanta001 wrote: $N^2+(N1)^2+(N2)^2+........+1$.