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# An Intriguing combinatorics problem

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.

Note by Soham Chanda
4 years, 3 months ago

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You 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?

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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?

- 4 years, 3 months ago