## BdMO National Secondary/Higher Secondary 2018/4

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### BdMO National Secondary/Higher Secondary 2018/4

After drawing $m$ lines on a plane, Sabbir got exactly $200$ different intersection points on that plane.What is the lowest value of $m$?

Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

### Re: BdMO National Higher Secondary 2018/4

$21$ I got.

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm
Lemma: If we draw $x$ lines on plane, we will get maximum $\frac{x(x-1)}{2}$ intersection points.
Proof: $2$ lines needed for a intersection point. So, maximum number of intersection point is $xC2=\frac{x(x-1)}{2}$
Here $200$ intersection points.
So, $\frac{m(m-1)}{2} \geq 200 \Rightarrow m > 20$. We can easily construct the figure which has $21$ lines and exactly $200$ intersection points.
The answer is $21$