Dhaka Higher Secondary 2010/11
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Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
$n$ points are taken on each side of a regular $m$ gon. What is the total number of straight lines that can be drawn using all those points?(except the sides of $m$ gon)
Re: Dhaka Higher Secondary 2010/11
The solution was posted here (it was a regular n gon with m points on each side).
(From the post)
If we choose any two points that are not on same side of the $n\ gon$ then we will find a straight line.
Lets number the sides of $n\ gon$ with $1,2,3,...,n$.
Start with side $1$. For each of the $m$ points of this side, we have $(n-1)m$ points. So in total $(n-1)m^2$ lines.
For each point of side #$2$, we have $(n-2)m$ points. So $(n-2)m^2$ lines.
Thus, total number of line is\[(n-1)m^2+(n-2)m^2+(n-3)m^2+...+1\cdot m^2\] \[=\frac{n(n-1)}{2}\cdot m^2\]
(From the post)
If we choose any two points that are not on same side of the $n\ gon$ then we will find a straight line.
Lets number the sides of $n\ gon$ with $1,2,3,...,n$.
Start with side $1$. For each of the $m$ points of this side, we have $(n-1)m$ points. So in total $(n-1)m^2$ lines.
For each point of side #$2$, we have $(n-2)m$ points. So $(n-2)m^2$ lines.
Thus, total number of line is\[(n-1)m^2+(n-2)m^2+(n-3)m^2+...+1\cdot m^2\] \[=\frac{n(n-1)}{2}\cdot m^2\]
Every logical solution to a problem has its own beauty.
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