Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For $n \geq 3$ proof that the number of diagonals of a polygon is $\frac{n(n-3)}{2} $ using induction.

I don't know how to start this problem, can you give me a hint?

share|improve this question
1  
How many diagonals meet at each vertex? Hm, but that way doesn't use induction. –  Daniel Fischer Feb 3 at 21:03

4 Answers 4

up vote 0 down vote accepted

Let $d_n$ be the number of diagonals of a polygon.

$d_3=0$ (polygon is triangle) Assume that the number of diagonals of a $n$-polygon is $\frac{n(n−3)}{2}.$

Сonsider $n+1-$poligon $A_1 A_2 \ldots A_{n+1}.$ Under the assumption the number of diagonals of $n-$poligon $A_1 A_2 \ldots A_{n}$ is $\frac{n(n−3)}{2}.$ Besides there are $n-1$ diagonals $A_{n+1} A_2,\,\,$ $A_{n+1} A_3,\ldots , A_{n+1} A_{n-1},$ and $A_1A_n$ for $n+1-$poligon $A_1 A_2 \ldots A_{n+1}.$ Therefore the number of diagonals of a $n+1$-polygon $A_1 A_2 \ldots A_{n+1}$ equal to $d_{n+1}=\frac{n(n−3)}{2}+n-1=\frac{n^2−3n+2n-2}{2}=\frac{(n+1)((n+1)-3)}{2}.$ Sorry for my English.

share|improve this answer

I don't know what you mean by induction. But it's simple: For each vertex (n), there is one diagonal linking for all vertex, except the same vertex and the two neighbours (n-3). Thus you have n(n-3), that you got to divide by two or else each diagonal will be counted going and coming back.

share|improve this answer
    
"Induction" stands for a basic logical way of proving something. Basically it is used, when something need to be proven $\forall n \in \mathhbb{N}. So, e.g. here, where polygon can have arbitrary number of vertices, it is good to use induction. Not enough room for details, but basically: proof by induction is done in two steps, the first base, and then the induction step. For example look at my answer, for more, search the internet;) –  quapka Apr 9 at 6:08

Suppose you have an $n$-gon with vertices $V_1, \ldots, V_n$ which by induction hypothesis has $n (n - 3)/2$ diagonals. Now let us add another vertex $V_{n+1}$. How many new diagonals have appeared? Add that to the existing diagonals and verify that you get $(n + 1) (n + 1 - 3)/2$. (Do not forget that the side $V_n V_1$ becomes a diagonal when you add $V_{n+1}$ beween $V_n$ and $V_1$.)

share|improve this answer

Hint

Well start with the first step: prove, that $\frac{n(n-3)}{2}$ is the number of diagonals for the smallest possible polygon. In this case it is $n$-polygon, where $n=3$.

In the second step assume, that $\frac{k(k-3)}{2}$ is the number of diagonals for $k$-polygon and you must show that $\frac{(k+1)(k+1-3)}{2}$ is the number of diagonals for $(k+1)$-polygon. It might be helpful to draw a simple picture with arbitrary number of vertices and add one vertex and find out, what it does with the number of diagonals.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.