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Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

Congratulations Tiffany from Island School on your very clear and well explained solution.

Let the number of interior $k$-points be $I=Ak^2-Bk+C$, and the number of $k$-points in the closed rectangle be $R=Ak^2+Bk+C$.

When $k=1$ \begin{eqnarray} I=A-B+C &= 2\times 1 = 2 \quad (1) \\ R=A+B+C &= 2 + 10 =12 \quad (2) \end{eqnarray} Subtracting (2)-(1) gives $2B=10, B=5$.
Substituting into the original equations $$A+C=7 \quad (1').$$ When $k=2$ \begin{eqnarray} I=4A-2B+C &= 5\times 3 = 15 \quad (3) \\ R=4A+2B+C &= 15 + 20 = 35 \quad (4) \end{eqnarray} Subtracting (4)-(3) gives $4B=20, B=5$, (agreeing with the above).
Substituting into the original equations $$4A+C=25 \quad (3').$$ 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 $k=3$ \begin{eqnarray} I=9A-3B+C &= 8\times 5 = 40 \quad (5) \\ R=9A+3B+C &= 40 + 30 = 70 \quad (6) \end{eqnarray} Subtracting (6)-(5) gives 6B=30, B=5, (agreeing with the above). Substituting $A=6$, $B=5$, $C=1$ into equation (5): $$9\times 6 -3\times 5 + 1 = 54-15+1 = 40$$ and therefore these values of $A$, $B$ and $C$ hold: $A=6,\ B=5,\ C=1$
p by q rectangle
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 $p\times q$.

The number of interior $k$-points $I=Ak^2-Bk+C =(pk-1)(qk-1)$. 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 $$Ak^2-Bk+C = pqk^2 - (p+q)k +1.$$

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 $Ak^2-Bk+C$ interior $k$-points, is given by $$\lim_{k\to \infty} {{\rm number of interior k-points}\over k^2}$$ consider the interior of the polygon divided into a very large number of tiny squares of area ${1\over k}\times {1\over k}.$ The number of these small squares is roughly the same as the number of interior $k$-points, that is $Ak^2-Bk+C.$ (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.) $${\rm Area of polygon} = \lim_{k\to \infty}[{\rm number of interior k-points} \times {1\over k^2}]= \lim_{k\to \infty}\left({Ak^2+Bk+C\over k^2}\right) = \lim_{k\to \infty}\left(A + {B\over k} + {C\over k^2}\right) = A.$$ (This is because as $k$ approaches infinity, ${B\over k}$ and ${C\over k^2}$ both approach zero.)

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