If no three diagonals of a convex n-gon is concurrent,prove that the polygon is divided into

\({n \choose 4} + {(n-1) \choose 2}\)

when all of its diagonals are drawn.

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## Comments

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TopNewestYou can view the obtained figure as a graph, where the vertices are the vertices of the polygon and the intersection points. What is the degree of each interior vertex?

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No. Of diagonals of a convex n-gon is number of pair of points minus the number of sides of polygon \({ n \choose 2} \) - n

No. Of intersection points of diagonals is \({n \choose 4} \)

Now initially we have a complete n-gon. Observe that one part is added for each diagonal and one for each intersection point. Hence total no. Of parts in which the n-gon is divided is equal to

\({n \choose 2} \)- n + \({n \choose 4} \)+1=\({n \choose 4} \)+ \({n-1 \choose 2} \)

We added one because initially when we had no diagonal and no intersection point, the n-gon was 'divided' into one part which was equal to the n-gon itself

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Sorry for my poor English.it is a nice question and i have a lot of fun to cope with it.my proof uses some following results: 1.number of intersection when all of its diagonals are drawn is \(\binom{n}{4}\) 2.assume that when all of its diagonals are drawn,it will devide the convex n-polygon into a triangle,b quadrilateral,c pentagon... The answer of this problem is exactly \(a+b+c+...\) 2a) we calculate all the angles in 2 ways.then we have: \(360\binom{n}{4}+180(n-2)=180a+360b+540c+...\) We deduce that \(2\binom{n}{4}+n-2=a+2b+3c+...\) 2b) we calculate all segments in 2 ways.then we have: \(4\binom{n}{4}+n(n-3) +n=3a+4b+5c+...\) From 2a and 2b we have the answer

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Could this be proved by induction? The case is clearly true for \(n=4\), the base case. Assuming the theorem is true for the \(k\)th case, can new sections be counted in the \(k+1\)th case?

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