how can we prove the equality between number of vertices and sides of a polygon?

Question

We know that a triangle is composed of 3 vertices and 3 sides,a square is composed of 4 vertices and 4 sides,a pentagon has 5 vertices and 5 sides. Can we prove by induction (or any other method) that this pattern is true for a polygon with n sides or n vertices?

Answer 1

Starting with a triangle - whenever you add a side you also have to add a vertice to "close" the geometric form again. Its part of what you understand as a polygon...

Answer 2

It's not a dumb question, and it's good to rigorously prove even the statements that most would assume as "obvious".

To prove that any simple polygon has the same number of sides as vertices, proceed by induction as follows:

Let $S_n$ and $V_n$ be the no. of sides and vertices in a general simple polygon.

First show that $S_3 = V_3$ for the triangle.

Then assume $S_k = V_k$ for some polygon with $k \geq 3$.

Now construct a polygon with one more vertex than the above polygon. We need the extra vertex to connect to two existing vertices. So we need to remove the side between those two vertices then add two more sides to connect the new vertex to the other two vertices.

So $V_{k+1} = V_k + 1$ and $S_{k+1} = S_k - 1 + 2 = S_k + 1$, giving $S_{k+1} = V_{k+1}$.

Hence $S_n = V_n, \forall n \geq 3$. This completes the proof.

Answer 3

By induction on the number of vertices, if a polygon is considered the closed path associated to the enumeration of its vertices: the sides of the polygon $A_1A_2\dots A_{n-1}A_nA_{n+1}$ are obtained from the $n$ sides of the polygon $A_1A_2\dots A_{n-1}A_n$ replacing the $n$-th side $[A_nA_1]$with two sides $[A_nA_{n+1}]$ and $[A_{n+1}A_1]$. Therefore it has $n-1+2=n+1$ sides.

Answer 4

A proof without induction: Let $V_n$ the number of vertices. Let $l_1,l_2,...,l_n$ the sides. By the Inclusion-Exclusion Principle,

$$V_n=|l_1|+...+|l_n|-|l_1\cap l_2|-|l_2\cap l_3|-...+|l_1\cap l_2\cap l_3|+...+(-1)^{n-1}|l_1\cap l_2\cap...\cap l_n|,$$

where $|l_i|$ is the number of vertices en $l_i$, $|l_i\cap l_j|$ is the number of vertices in $l_i\cap l_j$ and so on.

Now, each $l_j$ satisfies $|l_i|=2$. $|l_i\cap l_j|=1$ iff $j=i+1$, $i\in\{1,2,...,n-1\}$, or $i=n$ and $j=1$, and $|l_i\cap l_j|=0$ in other case (that is, only adyacent sides have a vertice in common), and the summands of three or more intersection are cero.

This simplifies to $V_n=|l_1|+...+|l_n|-|l_1\cap l_2|-|l_2\cap l_3|-...$. But there are $n$ summands of the form $l_i$ and $n$ of the form $|l_i\cap l_j|$ with $j=i+1$ and $i\in\{1,2,...,n-1\}$, or $i=n$ and $j=1$.

So $V_n=2n-n=n$.

Answer 5

A Polygonal Chain (or polygonal curve/path, polyline etc) with vertices $A_1,\,A_2,\,...,\,A_n$ consists of the line segments connecting consecutive vertices $A_1A_2,\,A_2A_3,\,...,\,A_{n-1}A_n$. It follows by mere inspection of indices that there are $n-1$ such line segments.

A polygon with $n$ vertices $A_1,\,A_2,\,...,\,A_n$ is a polygonal chain with $n+1$ vertices $A_1,\,A_2,\,...,\,A_n,\,A_{n+1}$ where $A_{n+1} \equiv A_1$ and the other vertices are all distinct. Since the polygonal line with $n+1$ vertices has $(n+1)-1=n$ line segments (called sides in the case of a polygon) it follows that the polygon with $n$ vertices has $n$ sides.