# Converse of the British Flag Theorem

Exist a theorem known as British Flag Theorem. It say that in a rectangle $ABCD$ we have $PA^2 - PB^2 + PC^2 - PD^2 = 0$, for any point $P$ in the plane.

I was thinking in a type of converse of this theorem. Given a polygon with vertices $A_1,A_2,\ldots,A_n$, consider the function

$$f(P) = \sum_{i=1}^{n} (-1)^{i} {PA_i}^2,$$

where $P$ is a point in the plane. So I conjectured that

If a polygon with vertices $A_1,A_2,\ldots,A_n$ is such that $f(P) = 0$, for all point $P$ in the plane, then this polygon is a rectangle.

I can to prove easily that if $n = 4$, then this is true. Somebody know this problem? This conjecture is true? If this question is easy, but requires a expert argument, can to give me a hint of how to solve this?

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I'm not sure what you mean by "this polygon is a rectangle". For example, with $n=8$, we can easily create a polygon that satisfies your conditions, by taking the union of any 2 rectangles. – Calvin Lin Aug 15 '13 at 0:26
Well, I assume implicitly that $A_1,\ldots,A_n$ are all distinct. So, how will this union of two rectangles? Remember that a polygon is a closed curve. – Walner Aug 15 '13 at 0:35

I think you can use linear algebra to get a necessary and sufficient condition for your $f$ to be zero.

Write points in the plane as vectors. Then the hypothesis is that $A_1, \ldots, A_n$ are such that $$\sum_{i=1}^n (-1)^i ||P-A_i||^2 = 0$$ for all $P$. You can expand this into $$\sum_{i=1}^n (-1)^i\langle P-A_i, P-A_i \rangle = 0$$ or $$(\sum_{i=1}^n (-1)^i)||P||^2 + \sum_{i=1}^n (-1)^i ||A_i||^2 -2\langle P , \sum_{i=1}^n (-1)^i A_i \rangle = 0.$$ If this is true for $P$, it must also be true for $kP$ for any scalar $k$. So for a fixed $P$ you have a polynomial of degree $2$ in $k$, which must vanish identically. It follows that a necessary and sufficient condition to have $f(P)$ identically zero is

1. $n$ is even.
2. $\sum_{i=1}^n(-1)^i ||A_i||^2 =0.$
3. $\sum_{i=1}^n (-1)^i A_i = 0.$

You can find many polygons satisfying these. For example, let all the $A_i$ have equal length and let $n$ be even. Then you just need $A_1-A_2 + \cdots - A_{n-1} + A_n =0$.

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Thanks! This indeed solve the question! – Walner Aug 15 '13 at 2:09
Just only think, that I left to pass unnoticed: "If this is true for $P$, it must also be true for $kP$ for any scalar $k$." Why? – Walner Aug 15 '13 at 2:19
Sorry, that statement is not right. It should say, if this is true for all $P$ (as we are claiming), and we fix a $P$, then it must also be true for $kP$ for every $k$. That's just because we are claiming that it's true for every $P$, and then looking at the multiples of a particular $P$. – Flounderer Aug 15 '13 at 2:22
All right! I got it. But I don't understand how the big equation above together with this fact about the polynomial in $k$ implies in the three conditions. – Walner Aug 15 '13 at 2:40
If $ak^2 + bk =c = 0$ for all $k$ then $a=b=c=0$. From the coefficient of $k^2$ we get $\sum_{i=1}^n (-1)^i =0$ which is the same as saying $n$ is even. From the constant term being zero, we get the second condition. From the coefficient of $k$ being zero, we get that the inner product of $P$ with $\sum_{i=1}^n (-1)^i A_i$ is zero. But since this is now true for any $P$, we must have $\sum_{i=1}^n (-1)^i A_i=0$ as zero is the only vector which is orthogonal to everything. – Flounderer Aug 15 '13 at 2:43

Consider the points

$A_1 = (a,b), A_2 = (a,d) A_3 = (w,x), A_4 = (y,x), A_5 = (c,d), A_6 = (c,b), A_7 = (y,z), A_8 = (w,z)$

This satisfies your conditions, since the vertices are the union of vertices of 2 rectangles.

Proof:

We know that $PA_1^2 - PA_2^2 + PA_5^2 - PA_6^2 = 0$ and $PA_3^2 - PA_4^2 + PA_7^2 - PA_8^2 = 0$. Add up both equations.

It is clearly not a rectangle. Note that this polygon can be self-intersecting.

You can easily set conditions on $a, b, c, d, w, x, y, z$ to find a convex polygon. A simple example would be the regular octagon.

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Sorry, but not yet I can to see this polygon. I think that you are drawn two rectangles disconnected. A polygon is a connect curve. – Walner Aug 15 '13 at 0:56
Just to be more clear, if $A_1,\ldots,A_n$ are the vertices of a polygon, then $A_1A_2,A_2A_3,\ldots,A_{n-1}A_n,A_nA_1$ are the edges of this polygon. – Walner Aug 15 '13 at 1:06
@Walner The exact edges of the polygon doesn't matter. Do you see why this polygon satisfies your condition? – Calvin Lin Aug 15 '13 at 1:23
No. Do you can to draw to me? I don't know if here is allowed to put a figure, but if not, do you can to send for me? My email is walner@alu.ufc.br Thanks! – Walner Aug 15 '13 at 1:28
@Walner I added a proof of the statement. This is what I meant by "union of 2 rectangles". – Calvin Lin Aug 15 '13 at 1:30