# Looking for references about a tessellation of a regular polygon by rhombuses.

A regular polygon with an even number of vertices can be tessellated by rhombii (or lozenges), all with the same sidelength, with angles in arithmetic progression as can be seen on figures 1 to 3. Fig. 1 Fig. 2 Fig. 3

I had already seen this kind of tessellation, and I met it again in a recent question on this site (Tiling of regular polygon by rhombuses).

Let the polygon be $n$-sided with $n$ even. The starlike pattern of rhombii issued from the rightmost point, that we will call the source, can be seen as successive ''layers'' of similar rhombii. A first layer $R_1$ with the most acute angles (they are $m:=\dfrac{n}{2}-1$ of them), then moving away from the source, a second layer $R_2$ with $m-1$ rhombii, etc. with a grand total of $\dfrac{m(m+1)}{2}$ rhombii.

It is not difficult to show that rhombii in layer $R_p$ are characterized by angles $p\dfrac{\pi}{m+1}.$

In fact (I had no idea of it at first), the rhombii pattern described above is much less mysterious when seen into a larger structure such as shown in figure 4. The generation process is simple: a regular polygon with $m$ sides is rotated by successive rotations with angle $\dfrac{\pi}{m+1}$ around one of its vertices.

My question about this tessellation is twofold:

• where can I find some references?

• are there known properties/applications?

The different figures have been produced by Matlab programs. The program that has generated Fig. 2 is given below ; it uses complex numbers, especially apt to render angular relationships:

 hold on;axis equal
m=9;n=2*m+2;
i=complex(0,1);pri=exp(2*i*pi/n);
v=pri.^(0:(n-1));
for k=0:m-1
z=1-(pri^k)*(1-v(1:m+2-k));
plot([z,NaN,conj(z)],'color',rand(1,3),'linewidth',5);
end;


Edit : I am indebted to @Ethan Bolker for attracting my attention to zonohedra (or zomes, as some architects call them), a 3D extension of Fig. 4 (or an equivalent one with less or more circles) ; by 3D extension, we mean a polyhedron made of (planar) rhombic facets whose projection on $xOy$ plane is the initial figure, as shown on Fig. 5. The idea is simple (we refer here to the two left figures in Fig. 6): the central red "layer" (with the thinnest rhombi) is "lifted" as an umbrella whose highest point, the apex of the zonohedra, say at height $z=1$, with the bottom of the $n$ ribs of the umbrella at $z=1-a$. Let us denote by $V_k, \ k=1, \cdots n$ with components $\left(\cos(\tfrac{2 \pi k}{n}),\sin(\tfrac{2 \pi k}{n}), -a\right)$ the (3D) vectors issued from the apex. Layer $1$ rhombi have sides $V_k$ and $V_{k+1}$ ; by the very definition of a rhombus, layer $2$ (yellow) rhombi have sides $V_k$ and $V_{k+2}$, etc. Note that Fig. 6, unlike Fig. 5, displays a closed zonohedron obtained by gluing 2 identical zonohedra. The right part of Fig. 6 displays the same zonohedron colored in a spiraling way.

Let us remark that there is a degree of freedom, i.e., the way the initial "umbrella" with ribs $V_k$ is more or less open, i.e., $a$ can be chosen. Fig. 5 : The upper part of a regular zonohedron and its projection onto the horizontal plane. Fig. 6 : A typical regular zonohedron generated by Minkowski addition of vectors $(\cos(2k \pi/n), \sin(2k \pi/n),1)$ for $k=1,2,...n$ with $n=15$. Fig. 7 : A rhombic 132-hedron (image borrowed to the Wikipedia article).

• (+1) I don't know, but your diagrams are beautiful. – Jack D'Aurizio Feb 3 '17 at 14:51
• Nicely done, +1. Don't know about real applications, but the construction would make a great "proof without words" for this trig identity: $$\sum_{k=1}^m (m-k+1) \sin \frac{k \pi}{m+1} \,=\, \frac{m+1}{2} \,\cot \frac{\pi}{2m + 2}$$ (Well, almost without words - it's calculating the area of the regular polygon vs. the tessellation. Another identity follows from a similar radius calculation.) – dxiv Feb 4 '17 at 21:51
• @dxiv Thank you very much. I had not thought about this kind of application. – Jean Marie Feb 4 '17 at 21:59
• +1 Lovely suggested pictures. Not an answer, but perhaps useful: check out the extensive literature on zonohedra. Many links to get you started. – Ethan Bolker Apr 6 '17 at 14:21
• @JeanMarie Thanks, will do shortly. I actually had a hesitation to even post that comment, since the visuals were so nicely presented that relating them to an ugly trig identity felt almost inappropriate ;-) – dxiv Feb 5 '18 at 3:48

## 1 Answer

Elaborating some more on my previous comment:

Don't know about real applications, but the construction would make a great "proof without words" for this trig identity: $\;\sum_{k=1}^m(m−k+1) \sin \dfrac{k \pi}{m+1}= \dfrac{m+1}{2} \cot \dfrac{\pi}{2(m+1)}\,$.

With OP's notation where $\,n=2(m+1)\,$ is the number of sides of the regular polygon, it can be easily seen that there are $\,m\,$ "bands" of congruent rhombi in the tesselation. From right to left, the first band consists of $\,m\,$ rhombi with an angle of $\,\frac{\pi}{m+1}\,$, then the $k^{th}$ band is made of $\,m-k+1\,$ rhombi with increasing angles $\,\frac{k \pi}{m+1}\,$, all the way to $\,k=m\,$ which is the single leftmost rhombus.

A rhombus with side $\,a\,$ and an angle of $\,\alpha\,$ has an area of $\,a^2 \sin \alpha\,$, and the areas of all bands sum up to the area of the regular polygon $\,\frac{na^2}{4} \cot \frac{\pi}{n}\,$, from which the identity above follows.

Other trigonometric identities can be derived from this tesselation as well. For just one example, in the case of odd $\,m\,$ the horizontal diagonals of the odd-numbered rhombi add up to the diameter of the circumscribed circle $\,\frac{a}{\sin \pi/n}\,$, and therefore $\,\sum_{k=1}^{(m+1)/2} \cos \frac{(2k-1) \pi}{2(m+1)} = \frac{1}{2} \csc \frac{\pi}{2(m+1)}\,$.

• Thank you very much for your answer, and for the humor of your last comment... Besides, you may have seen that I have added, at the bottom of my text some references to zonohedra (following the advices of Ethan Bolker) which is a very pertinent keyword. I give some meaningful sites, some of them with stunning images. – Jean Marie Feb 5 '18 at 8:53
• @JeanMarie Thanks in turn for the latest edits and links. Never too late to learn fascinating new things. – dxiv Feb 6 '18 at 0:40
• A little remark about your first formula : if we divide LHS and RHS by $(m+1)^2$, we have on the LHS a Riemann sum for the integral $\int_0^1(1-x)\sin(\pi x)=\dfrac{1}{\pi}$ which is consistant with the limit of the RHS when $m \to \infty$, $\dfrac{1}{\pi}$ as well. – Jean Marie Feb 8 '18 at 22:04
• @JeanMarie Indeed, thanks for pointing that out. P.S. I'll check back on your question every now and then. Seems to be getting better every time I read it again ;-) – dxiv Feb 9 '18 at 0:22