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The problem looks easy and it is but I want to sure I am right .So, i am giving it here......Please give the answer in Bangla..................................There is a simple polygon with 2016 sides where there is intersection among the sides except the intersections of the adjacent sides. Maximum how many diagonals can be drawn inside the polygon such that if any two diagonals intersect , then their of intersection can't be any other point except the vertex of the polygon?

If we mark our total diagonals as M,we find M=C(2016,2)-2016.Then our required diagonals will be less than the total number of diagonals.By cutting out a pattern.....we follow that for an N sided polygon,the required diagonals as per the question is 2n+(n-1)+(n-2).........+2+1...So,the required number of diagonals=2*2016+2015+2014....+2+1...which is less than C(2016,2)-2016

aritra barua wrote:If we mark our total diagonals as M,we find M=C(2016,2)-2016.Then our required diagonals will be less than the total number of diagonals.By cutting out a pattern.....we follow that for an N sided polygon,the required diagonals as per the question is 2n+(n-1)+(n-2).........+2+1...So,the required number of diagonals=2*2016+2015+2014....+2+1...which is less than C(2016,2)-2016

Total wrong solution.$2 \times 2016 + 2015 + 2014 +......... + 2 + 1 > C(2016,2) - 2016.$

To find the number of required diagonals, just show that the triangulation of the polygon satisfies the number of diagonals.

I am wrong indeed......but as I am still in class 8,I don't understand what triangulation of polygon is....

I think it should have been 2*2013+2012+2011......+3+2+1,because I did not consider the adjacent points in the 1st case