Tag Archives: polygons

Midpoint Polygon Conjecture is false

Standard

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:

poly-counter

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

Standard

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:

concave-convex

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):

big-gon

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.

Interesting Diagonal Discovery

Standard

Yesterday (while doodling in my Number Theory class) I was just playing with diagonals of polygons and something interesting appeared on my notebook. For the first time in my life I have discovered a theorem and its proof on my own (it’s original!). In this blog-post I will share it with you all!

Consider an n-sided polygon, and start drawing diagonals from each vertex one-by-one. While doing so count the number of new diagonals contributed by each vertex. Here is the “Experiment” done for n=4,5,6,7,8.

Number written near each vertex indicate the number of new diagonals contributed by that vertex (following anti-clockwise order and starting with the red one)

Number written near each vertex indicate the number of new diagonals contributed by that vertex (following anti-clockwise order and starting with the red one)

Based on above experiment, I observed:

The number of new diagonals contributed by each vertex of a n-sided polygon follows the sequence: (n-3),(n-3),(n-4),\ldots, 1,0,0

Now let’s try to  prove this observation:

Since we can’t draw diagonals to the adjacent vertices, each vertex will have (n-1)-2 = (n-3) diagonals.

Now, let’s count the new contribution of each vertex by considering restrictions on the maximum possible contribution (which is (n-3)).

For first vertex, we have no restriction, so it will contribute (n-3) diagonals.

Also, since second vertex in adjacent to first one and both can’t affect each-other’s contribution, it will also contribute (n-3) diagonals.

But, starting with third vertex we observe that first vertex has already taken one of the diagonals from its maximum contribution (second vertex can’t affect its contribution count since it’s adjacent vertex), thus it contributes (new) (n-3)-1 = (n-4) diagonals.

Continuing same way, for k^{th} vertex, consider the restriction to contribution caused by 1^{st} to (k-2)^{th} vertices. Thus for 1<k<n we get the number of new diagonals contributed by k^{th} vertex equal to (n-3)-(k-2) = (n-k-1).

Since new contribution can’t be negative and for (n-1)^{th} vertex we end up with zero (new) contribution, n^{th} vertex will also contribute zero diagonals.

Combining all of the above arguments I complete the proof of my observation and call it “New Diagonal Contribution Theorem” (NDCT).