BdMO National Secondary/Higher Secondary 2018/4
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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:64
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- 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.
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.