Diagonals of a Polygon

Recently one of my co-workers was teaching a Geometry class on the topic of polygons, when the students asked if there was a formula for finding the number of diagonals according to the number of sides.  They drew out different polygons on the board and counted the number of diagonals in each:

Unfortunately, nobody was able to derive an equation on the spot, but fortunately, we were able to Google it.  The formula is  

d = ½n(n – 3)

where d is the number of diagonals and n is the number of sides.

The reason the formula is so difficult to find is because the relation is quadratic (the highest exponent of the variable is a square, when the formula is re-written as d = ¹/₂n² – ³/₂n).  It is much more intuitive to find an equation in which the relation is linear instead (like when converting units).  This and the fact that the formula also requires fractions made it difficult to find.

Instead of trying to find a pattern in the numbers, a geometrical approach should be used for this problem.  Surprisingly, it is easier to see the relationship in a polynomial with more sides than with less.  Consider all the diagonals of a hexagon from a single vertex:

From that vertex, a diagonal can be made to all the other vertices in the hexagon, with the exception of itself and the two adjacent vertices.  In other words, that vertex connects to all the other vertices in the hexagon except for 3 of them, for a total of 6 – 3 = 3 diagonals.  The same argument can be made 6 times, one for each vertex, so now we have counted 6(6 – 3) = 18 diagonals.  But since each diagonal has 2 vertices each, we have actually counted each diagonal in the hexagon twice.  To correct this, we half our answer, which means there are ½(6)(6 – 3) = 9 diagonals.

The same argument can be made for all polygons.  A polygon with n sides also has n vertices.  From each of these vertices, a diagonal can be made to all the other vertices except for 3 of them (itself and the two adjacent vertices), for a total count of n(n – 3) diagonals.  But since each diagonal has 2 vertices each we have counted each diagonal in the polygon twice, so we must half our answer, for the final and generalized formula of d = ½n(n – 3).