Problem of calissons

Couple years back I met a nice math problem, which is mainly know as the problem of calissons. Given a regular hexagon with side of length n. You can fill it with rhombuses, where every rhombus is made by gluing together two equilateral triangles. Here is an example:   

As you can see the rhombuses have three different orientations. 

Theorem 1: The number of rhombuses with a given orientation is one third of the number of the rhombuses.  

It is not hard to see that we will always use $3n^2$ rhombuses to fill the hexagon. So the theorem says that we will always have $n^2$ rhombuses of each orientation. 

Proof No. 1 

This problem became known it has a nice solution. It is a “Proof Without Words”. Just look at the following picture.  

We colored the rhombuses according to their orientation. If you try to look at them in 3D, you will see some cubes, and it is easy to feel the truth of the theorem. This proof is not too rigorous though.

When I was researching this topic I found two really interesting notes on the topic from Dijkstra (who is known for Dijkstra’s algorithm). In these notes he shows that this proof obscures a more general theorem, where we need an other proof.   

If we have a figure that is covered by rhombuses with $a$ of them in one orientation, $b$ of them in an other orientation and $c$ of them in the third orientation, then we will say that $(a,b,c)$ is the frequency of the covering. So Theorem 1 says that no mater how we cover a hexagon, the frequency is the same.    

Theorem 2: If we can cover any kind of figure with the rhombuses, then the frequency is independent of the covering.

For example we the following figure always needs 8 rhombuses of each orientation:

Proof of Theorem 2.

I will show of the proof of the second theorem using the original hexagon. First divide the figure into triangles. Color the triangles black and white such that neighbors have different colors. Assume we have a covering. This way every rhombus covers one black and one white triangle. In each rhombus we can draw an arrow from the middle of the black triangle to the middle of the white triangle. For example this is what we get from the covering above. 

  And now we must use a bit more advanced math. Consider the arrows as vectors!  We will consider the sum of these vectors in two different way. 

We have three different vector, one for each orientation, denote them by $v_1,v_2$ and $v_3$. Let $(a,b,c)$ be the frequency of the covering. This means we have $a$ of the $v_1$ vector, $v$ of the $v_3$ vector and $c$ of the $v_3$ vector. 

We can now calculate the sum of these vectors: $av_1+bv_2+cv_3$. 

 We need an other way to consider this sum. The important thing here is to show that the sum does not depend on the covering. Each vector goes from a black triangle to a white triangle. This means that we can write each vector as a difference of two vectors, one pointing from zero to the center of the white triangle minus one pointing to the center of the black triangle. If we consider the sum of all vectors now, we can write everything as a difference. So the sum of the vector is just the sum of vectors pointing to the white triangles minus the sum of vectors pointing to the black triangles. This quantity does not depend on the covering, just on the figure. We will call this vector $v_{sum}$.

So we have  $av_1+bv_2+cv_3=v_{sum}$. Also $a+b+c$ is a fixed number since the number of all rhombuses can be calculated from the area of the figure. 

Since we are in two dimensions, the equation  $av_1+bv_2+cv_3=v_{sum}$ is actually two equations, one for the x-coordinates and one for the y-coordinates. 

So all together we have three equations for $a,b$ and $c$ that has to be true independently of the covering. From linear algebra we know that three equation uniquely determines three variables. So $(a,b,c)$ is independent of the covering. QED 

Proof  No. 2.

Once I posted the original problem in a math competition. Most of the students gave the following solution. We can draw $n$ paths from one side of the hexagon to the opposite site using two orientations. 

In each path we must have $2n$ rhombuses. All rhombuses outside the paths belong to the third orientation. So we have $3n^2-2n\cdot n=n^2$ rhombuses in the third orientation. By symmetric arguments we have $n^2$ rhombuses in each orientation.         

Oh and my logo was inspired by all of this :)    

Let me show you some of the cooler graphs I've dealt with in maths

f(x)=x cos x

And this one, it’s never negative, it’s a positive one :3 and the blue is the derivative

Here the purple is the derivative, but hey, let’s look at the graph itself alone:

It looks like it’s flying :O

And this one’s pretty cool, and the green is the derivative

Lastly, this one, i don’t even know what to say about this one, it is just so awesome! (The red is the derivative)

Now, let’s put them all together:


Did you think graphs were boring?


Grid Deformations

Because… reasons. (How my daughter answers questions.) GeoGebra on request. First couple have y-matters on vertical endpoints and x on horizontal, then I played with mixing and breaking symmetry.