Without looking up the proof, can you prove it?

We can draw the new boundary of the inner polygon in the outer polygon by using interior lattice points. We put B as B_n, I as I_n, Area as A_n, then the outermost one is can be written like this:

A_n = I_n + B_n/2 - 1

You can get the area of inner polygons with the same rules, so if the theorem is true, we assume that below is true:

A_n-1 = I_n-1 + B_n-1 - 1

When we call the innermost I as I_0 and the innermost B as B_0, we can get I_n like below:

I_n = sum(i=[0, n-1], B_i)

by this, we can pull the equation below:

A_n - A_n-1 = (B_n-1 + B_n)/2

what the above means is that the ring-shape area between two boundaries can be known by the points of which the area consists, which is true because the area of all triangular fragments that contribute the ring-shape area is 1/2 and its number is B_n-1 + B_n.

now we know that the above is true, therefore the theorem holds.

P1486
>the area of all triangular fragments that contribute the ring-shape area is 1/2
This is the critical step that remains to be proven.
The base case also needs to be proven. Then this should work by induction.

P1486
I went in a completely different direction.

Any such shape can be constructed by adding and removing right angled triangles and complete small squares. Let's first verify that the theorem holds for both classes of objects.

The theorem only holds for shapes that have no hole.

A small square has an area A=1, B=4 points on its border and I=0 inner points. As such, I+B/2-1=1=A.

Let's consider only such triangle that have no point on the hypotenuse. Such an n×m triangle verifies that n and m are relatively prime, so at least one of (n-1) and (m-1) is even. Such a triangle has an area A=n×m/2, B=n+m+1 points on its border and I=(n-1)(m-1)/2 inner points. As such, I+B/2-1=(n×m/2-n/2-m/2+1/2)+(n/2+m/2+1/2)-1=n×m/2=A.

Let's verify that adding two shapes that verify the theorem creates a shape that verifies the theorem.
Let's note a0 and a1 the areas of the two shapes, b0 and b1 the border points and i0 and i1 the inner points. Let's note A the area of the new shape, B the number of border points and I the number of inner points.
When merging the two shapes, 2n points that were on the border of one or the other shape become n points on the inside of the new shape.
There are also 4 points that were on the border of one or the other shape that become 2 points on the border of the new shape.
Therefore:
A=a0+a1
B=b0+b1-2n-2
I=i0+i1+n
As such, I+B/2-1=(i0+b0/2-1)+(i1+b1/2-1)=a0+a1=A

Let's verify that removing a square from the outside of a shape that verifies the theorem creates a new shape that verifies the theorem.
A1=A0-1
There are multiple scenarios:
The removed square was connected to the bigger shape by one side: B1=B0-2, I1=I0; or
The removed square was connected to the bigger shape by two sides: B1=B0, I1=I0-1; or
The removed square was connected to the bigger shape by three sides: B1=B0+2, I1=I0-2.
In any case, I1+B1/2-1=I0+B0=A0-1=A1

Since any lattice polygon can be constructed by merging smaller shapes and removing squares, the theorem is true for all polygons that have no holes.


If the polygons can have holes, A=I+B/2-R+H with R the number of contiguous regions and H the number of holes. Can you prove that?

P9386
Looks good. The proof that
>adding two shapes that verify the theorem creates a shape that verifies the theorem
is probably the most critical part. You can easily generalize it as [tex: f(S_1) + f(S_2) = f(S_1 \cup S_2)], where [tex: f(S) = I_S+B_S/2-1] and [tex: S_1], [tex: S_2], and [tex: S_1 \cup S_2] are connected grid polygons with no holes. Using this you can simplify the proof a lot. For example, not only is it an easy corollary that adding two shapes that verify the theorem produces a new shape that verifies the theorem, but so is subtracting one shape from another and cutting a shape into two congruent halves. Every grid polygon can be constructed by adding/removing grid triangles, every grid triangle can be constructed by removing grid right triangles from a circumscribed grid rectangle, every grid right triangle can be constructed by cutting a grid rectangle in half, and every grid rectangle can be constructed by adding unit grid squares.