Answer
the statement that the area of a simple polygon with n sides and vertices all at lattice points is given by I (P) + B(P)/2 − 1.
- We will prove P(n) for all n ≥ 3.
We begin with an additivity lemma:
-If P is a simple polygon with all vertices at lattice points, divided into polygons P1 and P2 by a diagonal,
- then I (P) + B(P)/2 − 1 = [I (P1) + B(P1)/2 − 1] + [I (P2) + B(P2)/2 − 1].
Work Step by Step
--Let P(n) be the statement that the area of a simple polygon with n sides and vertices all at lattice points is given by I (P) + B(P)/2 − 1.
- We will prove P(n) for all n ≥ 3.
We begin with an additivity lemma:
-If P is a simple polygon with all vertices at lattice points, divided into polygons P1 and P2 by a diagonal,
- then I (P) + B(P)/2 − 1 = [I (P1) + B(P1)/2 − 1] + [I (P2) + B(P2)/2 − 1].
-- To prove
this, suppose there are k lattice points on the diagonal, not
counting its endpoints.
-Then I (P) = I (P1) + I (P2) + k and
-B(P) = B(P1) + B(P2) − 2k − 2; and
-the result follows
by simple algebra.
- What this says in particular is that if Pick’s formula gives the correct area for P1 and P2,
- then it must give the correct formula for P, whose area is the sum of the areas for P1 and P2; and similarly if Pick’s formula gives the correct
area for P and one of the Pi ’s,
-then it must give the correct formula for the other Pi .
-- Next we prove the theorem for rectangles
whose sides are parallel to the coordinate axes.
-Such a rectangle necessarily has vertices at (a, b), (a, c), (d, b), and
(d, c), where a, b, c, and d are integers with b < c and a < d.
Its area is (c − b)(d − a).
- Also,
-B = 2(c − b + d − a) and I = (c−b−1)(d−a−1) = (c−b)(d−a)−(c−b)−(d−a)+1.
Therefore,
I + B/2 − 1 = (c − b)(d − a) − (c − b) − (d −a) + 1 + (c − b + d − a) − 1 = (c − b)(d − a),
which is the desired area. Next consider a right triangle whose legs are
parallel to the coordinate axes.
-This triangle is half a rectangle of the type just considered, for which Pick’s formula holds, so by the additivity lemma, it holds for the triangle as well.
--(The values of B and I are the same for each of the two triangles,
so if Picks’s formula gave an answer that was either too small
or too large, then it would give a correspondingly wrong answer
for the rectangle.)
- For the next step, consider an arbitrary triangle with vertices at lattice points that is not of the type already
considered.
- Embed it in as small a rectangle as possible.
-There are several possible ways this can happen,
-but in any case (and adding one more edge in one case), the rectangle
will have been partitioned into the given triangle and two or
three right triangles with sides parallel to the coordinate axes.
-Again by the additivity lemma, we are guaranteed that Pick’s
formula gives the correct area for the given triangle.
-- This completes the proof of P(3), the basis step in our strong induction proof.
- For the inductive step,
-given an arbitrary polygon, use Lemma 1 in the text to split it into two polygons.
-Then by the additivity lemma above and the inductive hypothesis,
- we know that Pick’s formula gives the correct area for this polygon.
