Induction Help! (1 Viewer)

[denoz]

A regular polygon has n sides and n vertices, n>=3. Show that the number of diagonals of the polygon is given by D=0.5n(n-3).

any help please .... thanks

 

[airie]

Do you have to use induction? It kinda seems unnecessary here o.0 Just think about it like this: Take any vertex of the polygon. Then to make a diagonal, you need to connect it to another vertex, other than itself and the two vertices already connected to it by two sides of the polygon ie. the neighbouring two. Therefore, you have n(n-3) diagonals. But since a diagonal from vertex A to vertex B is the same as the diagonal connecting B to A, you half the total number. Therefore, n(n-3)/2. The polygon doesn't even have to be regular (just convex).

But if you must use induction, prove it for base case where n=3 first. Then if all k-gons has k(k-3)/2 diagonals, take one k-gon, and "add" a vertex to it to make a (k+1)-gon. All diagonals of the previous k-gon are still diagonals of this new (k+1)-gon, and the new vertex can make a further [(k+1)-3] = k-2 diagonals with other vertices, plus one more diagonal connecting the two vertices between which the new vertex is added, since they are no longer connected by a side of the polygon as before. Add, you get k(k-3)/2 + (k-2) + 1 = (k^2-k-2)/2 = (k+1)(k-2)/2 = (k+1)[(k+1)-3]/2, ie. the statement holds for (k+1)-gons as well, if it holds for any k-gon. Therefore, by induction, it is proven

EDIT: Realised there was a stupid mistake Fixed it