A diagonal of a polygon is any segment joining two vertices other than an edge.

Triangles have no diagonals, while convex quadrilaterals have two interior diagonals, and concave quadrilaterals have one interior and one exterior diagonal.

Number of diagonals

The number of diagonals of a polygon with $n$ vertices is given by $\frac{n(n-3)}{2}$.


Proof 1: By direct counting.

Pick any vertex $P$ of the $n$-gon. There are $n-3$ diagonals with this point as an endpoint, one for every vertex except $P$ itself and the two vertices adjacent to $P$ (because the segments connecting $P$ to those two points give edges, not diagonals). There are $n$ vertices and for each vertex there are $n-3$ diagonals so there are $n(n-3)$ diagonals. However, when counting our diagonals, we counted each of them twice, once for each endpoint. So, we simply divide by 2 to get our final formula, $\frac{n(n-3)}{2}$.

Proof 2: By minor trickery.

Every two vertices determine either an edge or a diagonal. Thus, there are a total of $n \choose 2$ edges and diagonals. Since there are exactly $n$ edges, this leaves ${n \choose 2} - n = \frac{n(n-1)}2 - n = \frac{n(n-3)}2$ diagonals.


Polyhedra have two different kinds of diagonals, face diagonals and space diagonals. A face diagonal of a polyhedron is a diagonal of one of the faces of the polyhedron, while a space diagonal is any segment joining two vertices which is neither an edge nor a face diagonal.

For example, tetrahedra have no space or face diagonals. Octahedra have no face diagonals but have 3 space diagonals. Cubes have 12 face diagonals (2 on each face) and 4 space diagonals.

Following the same method as Proof 2 for polygons, we see that the number of edges plus the number of face diagonals plus the number of space diagonals of a polyhedron with $n$ vertices is equal to $n \choose 2$.

2004 AIME I problem 3 is a problem related to polyhedral diagonals.

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