## 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 $m$?

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

$21$ I got.

- Atonu Roy Chowdhury
**Posts:**63**Joined:**Fri Aug 05, 2016 7:57 pm**Location:**Chittagong, Bangladesh

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

**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 $21$

Remark: In contest, I proved the lemma by induction.

This was freedom. Losing all hope was freedom.