Rajshahi MO 2013, Secondary 1

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Fahim Shahriar
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Rajshahi MO 2013, Secondary 1

Unread post by Fahim Shahriar » Tue Jan 29, 2013 8:47 pm

Here is a regular polygon having $2013$ sides.
Find the number of intersections of its diagonals.


I tried for about 15 minutes to solve it there. But couldn't...still not. Is there any formula for finding number of intersections of diagonals for $n$ -sided polygon ? That was the first question and I didn't expect one of such type.
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Re: Rajshahi MO 2013, Secondary 1

Unread post by *Mahi* » Tue Jan 29, 2013 9:59 pm

Fahim Shahriar wrote:Here is a regular polygon having $2013$ sides.
Find the number of intersections of its diagonals.


I tried for about 15 minutes to solve it there. But couldn't...still not. Is there any formula for finding number of intersections of diagonals for $n$ -sided polygon ? That was the first question and I didn't expect one of such type.
Try this:
For any $4$ points among the $2013$ vertices of the polygon, there is one and exactly one intersection of diagonals.
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Re: Rajshahi MO 2013, Secondary 1

Unread post by Fahim Shahriar » Tue Jan 29, 2013 11:23 pm

But Mahi vai, it does not work at all. If we do according to that we get 6 sided polygon having 15 intersections. But actually it is 13.
3 intersections are same point here. It is not working for the polygons having even number(greater than 4) of sides.
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Re: Rajshahi MO 2013, Secondary 1

Unread post by Phlembac Adib Hasan » Tue Jan 29, 2013 11:34 pm

Fahim Shahriar wrote:But Mahi vai, it does not work at all. If we do according to that we get 6 sided polygon having 15 intersections. But actually it is 13.
3 intersections are same point here. It is not working for the polygons having even number(greater than 4) of sides.
২০১৩ তো জোড় না। এখানে তো কাজ করার কথা। তাও একবার চেক করে নেওয়া ভালো যে এখানে এরকম কোন coincidence আছে কিনা। ৭ ভুজ, ৯ ভুজ দেখেন আগে। যদি না থাকে তাহলে ২০১৩-র জন্য প্রুফ করার চেষ্টা করতে হবে। আমার মনে হচ্ছে থাকবে না। আর coincidence থাকলেও সমস্যা না, কোন পয়েন্টে কয়টা রেখা ছেদ করেছে তা জানতে পারলে ঐরকম বাকি সব পয়েন্ট সিমেট্রি দিয়া বের করা যাবে। এরপর ওভার কাউন্টিং বাদ দিলেই হবে।
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Fahim Shahriar
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Re: Rajshahi MO 2013, Secondary 1

Unread post by Fahim Shahriar » Wed Jan 30, 2013 12:45 am

I have just drawn a regular nonagon perfectly & also 27 diagonals and observed that there is not a single same point of intersection. :D It works for the odds. $^{2013}C_4$ is the answer. Thanks..
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Re: Rajshahi MO 2013, Secondary 1

Unread post by zadid xcalibured » Wed Jan 30, 2013 2:30 pm

Oi Mahi,that makes use of such a big assumption as for those $4$ points ,the diagonals intersect strictly inside the hull.But we can get $3$ intersection points.Namely for points $A$,$B$,$C$,$D$ ,$AB \cap CD$ ,$AD \cap BC$ and $AC \cap BD$.

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Re: Rajshahi MO 2013, Secondary 1

Unread post by *Mahi* » Wed Jan 30, 2013 3:09 pm

zadid xcalibured wrote:Oi Mahi,that makes use of such a big assumption as for those $4$ points ,the diagonals intersect strictly inside the hull.But we can get $3$ intersection points.Namely for points $A$,$B$,$C$,$D$ ,$AB \cap CD$ ,$AD \cap BC$ and $AC \cap BD$.
That's why I used $\binom {n}{4}$, as for any four points (on a regular n-gon), one of the three intersection points will be inside the convex hull created by those $4$ points.
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Re: Rajshahi MO 2013, Secondary 1

Unread post by sakibtanvir » Wed Feb 06, 2013 7:08 pm

And what is the answer for a regular n-gon where 'n' is even? I think that will be $\displaystyle \binom{n}{4}-\binom{\frac{n}{2}}{2}$ .
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Re: Rajshahi MO 2013, Secondary 1

Unread post by Mehfuj Zahir » Wed Feb 06, 2013 9:50 pm

Please make sure that you have written the question in correct from.Only one mission word can chage the statement of the question.@Fahim Shahriar.
It was the first problem of Rajshahi diviisional math olympiad 2013 (S-1,HS-2)
2013 বাহু বিশিষ্ট বহুভুজের কর্ণগুলো যোগ করলে সর্বোচ্চ কয়টা ছেদবিন্দু পাওয়া যাবে?
At most how many points of intersection can be found by joining all the diagonals of a polygon of 2013 sides?

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Re: Rajshahi MO 2013, Secondary 1

Unread post by SMMamun » Thu Feb 07, 2013 4:51 pm

As we can read the actual question in Mehfuj Zahir's comment, the answer is now quite obvious and simple because of the word "at most" in the question. And the answer is $\binom {2013} {4}$, as explained by Mahi and Adib. The formula could be quite complicated if the polygon were regular with even number of sides.

However, the wording of our BDMO question still leaves the room for improvement in language. The part "joining all the diagonals" might confuse the student, because a diagonal is, by definition, a joined line segment. So, you cannot further join the diagonals to find their points of intersection. The question could better be written as
"At most how many points of intersection can be made by all the diagonals of a polygon of 2013 sides?"

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