Let the number of interior k-points be I=Ak²-Bk+C, and the number of k-points in the closed rectangle be R=Ak²+Bk+C.

When k=1

I=A-B+C = 2×1 = 2     (1) R=A+B+C = 2 + 10 = 12     (2)

Subtracting (2)-(1) gives 2B=10, B=5.
Substituting into the original equations

A+C=7     (1¢).

When k=2

I=4A-2B+C = 5×3 = 15     (3) R=4A+2B+C = 15 + 20 = 35     (4)

Subtracting (4)-(3) gives 4B=20, B=5, (agreeing with the above).
Substituting into the original equations

4A+C=25     (3¢).

Subtracting (3') - (1') gives 3A=18, A=6 and C = 7-A = 1.

To verify these values for A, B and C, we will look at the case for k=3.

When k=3

I=9A-3B+C = 8×5 = 40     (5) R=9A+3B+C = 40 + 30 = 70     (6)

Subtracting (6)-(5) gives 6B=30, B=5, (agreeing with the above). Substituting A=6, B=5, C=1 into equation (5):

9×6 -3×5 + 1 = 54-15+1 = 40

and therefore these values of A, B and C hold: A=6, B=5, C=1

In order to find a connection between A, B, C, the area of the rectangle and the number of k-points we can investigate for a rectangle of dimension p×q. The number of interior k-points I=Ak2-Bk+C = (pk-1)(qk-1). This is because there are pk-1 points on each row of interior k-points and qk-1 points in each column of interior k-points. Therefore the total number of interior k-points would be (pk-1)(qk-1). So Ak2-Bk+C = pqk2 - (p+q)k +1.

Comparing coefficients we find that

A=pq which is the area of the rectangle,

B=p+q which is half the perimeter of the rectangle

C=1 (a constant).

To explain why, for large k, the area of any polygon with Ak²-Bk+C interior k-points, is given by

lim k® ¥  number of interior k-points k2

consider the interior of the polygon divided into a very large number of tiny squares of area ¹/_k×¹/_k. The number of these small squares is roughly the same as the number of interior k-points, that is Ak²-Bk+C. (This will not be exact, some of these squares on the boundary of the polygon will be partly inside and partly outside the polygon but as k increases so the approximation gets better and better.)

Area of polygon = lim k® ¥  [number of interior k-points × 1 k2 ] = lim k® ¥  æ ç è Ak2+Bk+C k2 ö ÷ ø = lim k® ¥  æ ç è A + B k + C k2 ö ÷ ø = A.

(This is because as k approaches infinity, ^B/_k and

C k2

both approach zero.)

So A is equal to the area of the polygon as we found for the rectangle in this example.