proving that the area of a 2016 sided polygon is an even integer

Question

Let $P$ be a $2016$ sided polygon with all its adjacent sides perpendicular to each other, i.e., all its internal angles are either $90$°or $270$°. If the lengths of its sides are odd integers, prove that its area is an even integer.

I think visualising what this polygon might look like would more or less be the key to getting started on the problem, but I'm having quite a hard time doing so. Since the internal angles are all $90$° or $270$°, I think the shape would look like a rectangle or a square with smaller squares or rectangles protruding out from each side — but mini rectangles protruding from sides would not be possible because degrees of $180$° are not allowed. And to prove that the area is an even integer, if there are mini squares protruding from the sides there will have to be even number of those squares in total. And then I also need to make sure that at least the width or length of 'bigger' square or rectangle is even. But I seem to get stuck trying to draw a possible diagram...

Answer 1

Not a complete proof.

However, maybe you can visualize the polygon like this:

All the internal angles are either $90^\circ$ or $270^\circ$.

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Another way of visualizing:

Observe that you have a $12$-sided polygon here, comprised of $5$ squares. The next acceptable polygon would be this: Notice that $2$ sides are lost and $6$ more sides are added, thus we have a $16$-sided polygon. Notice too that there are $3$ more additional squares.

Therefore, for a polygon in this form, the number sides can be expressed as: $$12+4x \Rightarrow 12+4x=2016\iff x=501$$ Using $x$, the number of squares that compose this polygon will be: $$5+3x\Rightarrow 12+3\cdot501=1508$$ Therefore, we have an even number of squares therefore the area is an even integer.

Answer 2

Let $m = 504$ and $p_0, p_1, \ldots, p_{4m-1}$ be the $2016$ vertices of the polygon $\Delta$ ordered in counterclockwise manner. Let $p_{4m} = p_0$ and $(x_k,y_k)$ be the coordinates of $p_k$. WOLOG, we will assume all $x_k, y_k$ are integers and the edges are either horizontal or vertical.

Cyclic reordering if necessary, we will assume $y_0 = \min \{ y_k \}$ and $x_0 = \min \{ x_k : y_k = y_0 \}$.
Shift the origin so that $p_0 = (0,0)$. It is easy to see for all $k$, the parity of coordinates satisfy:

$$p_k = (x_k,y_k) \equiv \begin{cases} (0,0), & k \equiv 0 \pmod 4\\ (1,0), & k \equiv 1 \pmod 4\\ (1,1), & k \equiv 2 \pmod 4\\ (0,1), & k \equiv 3 \pmod 4 \end{cases}$$

Using Green's theorem, we can evaluate the area of $\Delta$ as a line integral over its boundary $\partial \Delta$:

$$\verb/Area/(\Delta) = \int_\Delta dx dy = \int_{\partial \Delta} x dy$$

Since $\partial\Delta$ are a bunch of horizontal or vertical segments, we can convert the last integral to a sum over corresponding edges${}^{\color{blue}{[1]}}$.

We can split the 2016 edges of the polygon into $4$ groups.

• $p_{4\ell} \to p_{4\ell+1}$ : doesn't contribute because this is a horizontal edge.
• $p_{4\ell+1} \to p_{4\ell+2}$ : $x$ is constant and an odd number. Since $y_{4\ell+2} - y_{4\ell+1}$ is also odd, this contributes an odd number to area.
• $p_{4\ell+2} \to p_{4\ell+3}$ : doesn't contribute because this is a horizontal edge.
• $p_{4\ell+3} \to p_{4\ell+4}$ : $x$ is constant and an even number. Since $y_{4\ell+4} - y_{4\ell+3}$ is an integer, this contributes an even number to the area.

This means in general, we have

$$\int_{p_{k} \to p_{k+1}} xdy \equiv \begin{cases} 1 \pmod 2, & k \equiv 1 \pmod 4\\ 0 \pmod 2, & k \not\equiv 1 \pmod 4 \end{cases}$$ Summing $k$ from $0$ to $4m-1$, we obtain

$$\verb/Area/(\Delta) = \sum_{k=0}^{4m-1} \int_{p_{k} \to p_{k+1}} xdy \equiv m \pmod 2$$

Since $m$ is an even integer, so does the area of original polygon $\Delta$.

Notes

• $\color{blue}{[1]}$ - For those who don't want to use calculus, one can replace the line integral by following specific form of shoelace formula: $$\verb/Area/(\Delta) = \left|\sum_{k=0}^{4m-1} \frac{x_k + x_{k+1}}{2}(y_{k+1} - y_k)\right|$$ The rest of arguments will be exactly the same.

Answer 3

Consider the following structure $A$:

1. Every axis-parallel rectangle whose side lengths are odd integers is in $A$.

2. Let $a \in A$ be an arbitrary element and $b \in A$ be an rectangle. The union $a \cup b$ is in $A$ if it satisfies the following properties:

   • The intersection $a \cap b$ is a subset of exactly one side of $a$ and $b$ respectively.
   • The length of the intersection $a \cap b$ is an even integer. (The intersection needs to be a line segment by the first property.)

   • Note: If created in this way $a \cup b$ has:

     • odd area
     • #(sides of $a$) + #(sides of $b$) sides
     • $4n$ sides, where $n$ is the total number of rectangles used in the process of creating $a \cup b$

It is easy to check that this structure contains only polygons with perpendicular adjacent sides where all side lengths are odd.

Now consider a arbitrary polygon with perpendicular adjacent sides where all side lengths are odd. Wlog. assume all sides are axis-parallel (rotate otherwise). We can now slice the polygon into axis-parallel rectangles the following way:

• "Sweep" the polygon with a line parallel to an axis (e.g. x-axis)
• Whenever the line contains a side of the polygon, extend that side into the interior of the polygon until you hit another side. Cut the polygon along the extended side into parts (one part is a rectangle).

One can check that after a complete sweep the resulting parts are axis-parallel rectangles whose side lengths are odd integers which can be combined following the procedure of structure $A$. Thus every (axis-parallel) polygon with perpendicular adjacent sides where all side lengths are odd is in $A$.

Thus the given polygon is (after rotating) in $A$ and can be created by $2016/4 = 504$ non-overlapping "base rectangles". Each of those rectangles has odd area. Thus the area of the polygon is the sum of 504 odd numbers which is an even number.

Answer 4

here is specific math to prove even areas are formed by certain products of sides

if we use a test case of |3n-2| which I somehow found element wise enumerates sides of the polygon drawn two different ways

The test case generalized a form for area similar to shoelace but working for Orthogonal Irregular Polygons specifically, where the sides are ordered by construction as in {8,5,2,1,4,7,10,13} that makes the 8 sided figure drawn twice in the image.

This same case can be solved for area by breaking it down into three rectangles two ways and setting those areas of 116 equal to each other.

This would be a1a2+(a7-a5)(a8-a2) = a1a2 + a4(a7-a5) + a6a7 Where {a1,a2,a3,a4,a5,a6,a7,a8} are edges of my 8 sided test case to find the general form of these parallel polygons in order to get the 2016 odd sided case.

Since a7 ≠ a5 a8 = a2 + a4 + a6, where a4 is the middle term and a6 is the n-2 term.

Now let’s plug in odd numbers for our terms we will always find even areas. It is like Gauss when he added terms of the outer bounds of a sequence times the number of terms. We find that the 2016 sided polygon has vector-product sums that are all even. If it has 8 sides and they are 270 or 90, for example—the form works out to

(a5 - a7) (a2 + a4 + a6 - a8)

Expands as

al a2 + (a7 - a5) (a8 - a2) + a5 a6 = al a2 + a4 (a7 - a5) + a6a7

a8 = a2 + a4 + a6

Is true, therefore the closed irregular orthogonal figure has a(n/2) = a4 so the other a2 and a6 are thereby the second and third to last elements. (a2 and a(n-2))

I set these two divisions of the shape into equal areas and generalize the area of any of the similar polygon shapes. Then I show that if the sides are all odd, and the side number is even, then the parity reduces to 4(odd)-4(odd) which is (even)(odd)-(even)(odd) meaning even-even —> even.

Not a complete proof either, but I hope this helps.