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.

#Physics #MathProblem #Math

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Comments

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?

Jorge Tipe - 7 years, 10 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

Hunter Killer - 7 years, 10 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

Ankit Sultana - 7 years, 10 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?

Tanishq Aggarwal - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$