BdMO National Secondary/Higher Secondary 2018/4

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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samiul_samin
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BdMO National Secondary/Higher Secondary 2018/4

Unread post by samiul_samin » Fri Mar 09, 2018 6:14 pm

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

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Tasnood
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Re: BdMO National Higher Secondary 2018/4

Unread post by Tasnood » Sat Mar 10, 2018 12:13 am

$21$ I got.

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Atonu Roy Chowdhury
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Re: BdMO National Higher Secondary 2018/4

Unread post by Atonu Roy Chowdhury » Sun Mar 11, 2018 12:31 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 $21$

Remark: In contest, I proved the lemma by induction.
This was freedom. Losing all hope was freedom.

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