Pick’s Theorem 97

Proof. For any such triangle, we may adjoin one, two, or three right triangles so as to form

a rectangle, as shown previously. Since Pick’s theorem holds for the rectangle and for each

of the triangles, it follows from the key lemma 66 that Pick’s theorem holds after removing

these right triangles one at a time, and so in this way, Pick’s theorem holds for the original

triangle.

8.5 Triangulations

A final ingredient we shall need for a full proof of Pick’s theorem in the general case is the

fact that every simple polygon in the integer lattice can be triangulated by triangles in the

integer lattice.

Lemma 68. Every polygon admits of a triangulation. That is, the polygon can be par-

titioned into triangles using the vertices of the polygon, in such a way that the triangles

overlap only on their edges and the union of the triangles is the original polygon.

a

b

c

d

e

f

g

h

i

→

Proof. Since the sum of the interior angles of an n-gon is 180(n − 2)

◦

, it cannot occur that

all n vertices have interior angles exceeding 180

◦

, and so there must be a convex vertex, a

vertex whose interior angle is less than 180

◦

. Fix such a convex vertex, such as vertex d in

the figure above, and consider the triangle formed by the adjacent sides.

If this triangle is totally contained in the interior of the polygon, as it is in the figure with

cde, then we may fix this triangle as part of our triangulation, and by clipping it off, we

reduce to a polygon with one fewer side, which by induction has a triangulation, and so in

this case we are done.

It remains to consider the case of a convex vertex, such as vertex b in the figure, such

that the triangle abc formed by the adjacent sides is not contained in the interior of the

polygon. It follows by convexity that there is at least one vertex of the polygon in the

interior of this triangle, as with vertices h and f in the figure. One of these vertices interior

to abc, let us call it h, forms the smallest possible angle ∠hab. It follows that abh is

contained in the interior of the polygon, because otherwise there would be a vertex forming

a smaller angle. It follows that the edge bh lies in the interior of the polygon, and since b