# Midpoint Polygon Conjecture is false

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Contrary to my expectations, my previous post turned out to be like  Popular-­Lonely primes and Decimal Problem, i.e. I discovered nothing new.

My conjecture is false. Following counterexample is given on pp. 234 of this paper:

Counterexample of the conjecture, taken from: Berlekamp, E. R., E. N. Gilbert, and F. W. Sinden. “A Polygon Problem.” The American Mathematical Monthly 72, no. 3 (1965): 233-41. doi:10.2307/2313689.

As pointed out by uncombed_coconut, the correct theorem is:

Theorem (Berlekamp-Gilbert-Sinden, 1965). For almost all simple polygons there exist a smallest natural number $k$ such that after $k$ iterations of midpoint polygon procedure, we obtain a convex polygon.

The proof of this theorem is very interesting. Till now I thought that proving euclidean geometry theorems using complex numbers was an overkill. But using an N-tuple of complex numbers to represent the vertices of a closed polygon (in given order),  $\mathbf{z} = (z_1,\ldots , z_N)$,  we can restate the problem in terms of eigenvectors (referred to as eigenpolygons) and eigenvalues.  Following are the crucial facts used is the proof:

• An arbitrary N-gon (need not be simple) can be written as a sum of regular N-gons i.e. the eigenvalues are distinct.
• The coefficient of $k^{th}$ eigenvector (when N-gon is written as linear combination of eigenpolygons) is the centroid of the polygon obtained by “winding” $\mathbf{z}$ $k$ times.
• All vertices of the midpoint polygons (obtained by repeating the midpoint polygon procedure infinitely many times) converge to the centroid.
• The sum of two convex components of $\mathbf{z}$ is a polygon. This polygon is the affine image of a regular convex N-gon whose all vertices lie on an ellipse. (as pointed out by Nikhil)
• A necessary and sufficient condition for $\mathbf{z}$ to have a convex midpoint polygon (after some finite iterations of the midpoint polygon procedure) is that the ellipse circumscribing the sum of two convex components of $\mathbf{z}$ is nondegenerate. (The degenerate form of an ellipse is a point. )

For a nice outline of the proof, please refer to the comment by uncombed_coconut on previous post.

Since I didn’t know that this is a well studied problem (and that too by a well known mathematician!) I was trying to prove it on my own. Though I didn’t make much progress, but I discovered some interesting theorems which I will share in my future posts.

# Midpoint Polygon Conjecture

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This is going to be yet another ambitious post like New Diagonal Contribution Theorem, and Cross Diagonal Cover Problem.

Let’s begin with following definitions from Wikipedia:

A convex polygon is a simple polygon (not self-intersecting) in which no line segment between two points on the boundary ever goes outside the polygon.

A concave polygon is a simple polygon (not self-intersecting) which has at least one reflex interior angle – that is, an angle with a measure that is between 180 degrees and 360 degrees exclusive.

We know that 3 sided polygons (a.k.a triangles) are always convex.  So, they are not very interesting. Let’s define the following procedure:

Midpoint Polygon Procedure: Given a n-sided simple polygon, join the consecutive midpoints of the sides to generate another n-sided simple polygon.

Here are few examples:

Red polygon is the original polygon, green polygon is the one generated by joining midpoints.

We observe that if the original polygon was a concave polygon then the polygon generated by joining midpoints need not be a convex polygon. So, let’s try to observe what happens when we apply the midpoint polygon procedure iteratively on the 4th case above (i.e. when we didn’t get a convex polygon):

In fourth iteration of the described procedure, we get the convex polygon.

Based on this observation, I want to make following conjecture:

Midpoint Polygon Conjecture: For every simple polygon there exists a smallest natural number $k$ such that after $k$ iterations of midpoint polygon procedure, we obtain a convex polygon.

If the given polygon is convex, then there is nothing to prove since $k=1$, but I have no idea about how to prove it for concave polygons. I tried to Google for the answer, but couldn’t find anything similar. So, if you know the proof or counterexample of this conjecture, please let me know.

# Cross Diagonal Cover – III

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I found many counterexamples to my conjecture, like

• for Case 2 in 7 by 5 grid we have 12, 11 by 5 grid we have 19 and 15 by 5 grid we have 26 filled squares
• for Case 3 in 4 by 10 grid we have 10 filled squares

In a comment by grant93jr   following part of of my initial question, to determine the number of squares in $m\times n$ grid that can be covered by crosses (x) by following Cross Diagonal Cover Algorithm, was proved:

Given m>n, whenever $n-1$ divides $m-1$ we have $m$ filled squares.

Examples of such grids are: 3 by 5, 4 by 7, 4 by 10, ….

Also in that comment it has been suggested that the algorithm terminates after $k(m-1) = l(n-1)$ steps where $k, l$ are natural numbers. It is certainly true, but I haven’t been able to use this idea for counting number of filled squares when $n-1$ does not divide $m-1$ since in that case some squares are filled twice.

There is another unexplained observation:

Why no filled square has more than 2 crosses?

Frustrated by so many unanswered questions I started colouring squares, so that number of times I visited a square is not visible. Since I really don’t care about how many times I visited  given square while counting the number of filled squares this may help in understanding the underlying symmetry.

Replacing cross (x) by any colour and applying Cross Diagonal Cover Algorithm

After observing above diagrams I suspect that my algorithm leads to a non-deterministic cellular automata.

So, the question which still remains to be answered is:

How many filled squares are there when  $m>n$ and $n-1$ does not divide $m-1$ ?

Examples of such grids are: 7 by 5, 11 by 5, 15 by 5 …

# Cross Diagonal Cover – II

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I will continue the previous post. Now the question to  be answered is that:

What percentage (or how many) of given $m\times n$ grid can be covered by following Cross Diagonal Cover Algorithm?

I did some calculations, for example:

As per my calculations, I have following conjecture:

The following 3 cases are possible:

1. If $m = n$, then $m$ squares will be covered.
2. If $m>n$ and both $m,n$ are odd, then $m$ squares will be covered.
3. If at least one of $m,n$ is even then $\frac{mn}{2}$ squares will be covered.

As you can see that Case – 1 is quite trivial. I tried to prove other two cases for one week, but failed, so decided to post it as conjecture.