When I was at the gym a few days ago, a woman who works at our bank introduced me to her son, who is a ninth grader at a local high school. He's eager to go to a top math/science school, and she wanted me to talk to him a little bit about how to get there from where he is now. He had a problem with him from his geometry class - how to tile the plane with 3 different regular polygons. We worked on the problem a little bit, and he figured it out. This of course raised the question: In how many ways can we tile the plane with regular polygons? Here we'll investigate regular tilings, in which the same number of the same types of regular polygons meet at every point.

A first swing at it, inspired by the teacher's string of assignments (tile with one type of polygon, then two types, then three), might have been to group the possibilities by number of different types of polygons used.

Suppose we only use one polygon. A regular polygon with n sides has exterior angles with measure 360/n, so each interior angle has measure 180 - 360/n. Since the angles around a point must total to 360 degrees and all of our polygons have the same angle measure, if we have m polygons, we must have



Therefore, we must have:

$\begin{eqnarray*} m &=& \frac{360^\circ}{180^\circ-\frac{360^\circ}{n}}\\ &=&\frac{2n}{n-2} \end{eqnarray*}$

Clearly, we must have n > 2, and we can quickly see that n = 3, 4, 6 are the only solutions. (n=5 fails, and n=6 gives m = 3. As n gets larger past 6, 2n/(n-2) gets smaller, but it can't ever reach 2 - make sure you see why!)

That gives us these tilings:





Then we move on to two types of polygons. Immediately we're thrown into casework. Suppose one polygon is triangles. We could have one triangle, then we have to think of how to divide the remaining 300 degrees among polygons with the same number of sides. We can get there with two dodecagons, which each would contribute 150 degrees. But can we get there with any other combination?

When we go on to three types of polygons, the casework gets even more gory. Instead of this, maybe we should try a different classification.

We could instead classify each tiling by the number of polygons that meet at each point.

Clearly we have to have at least three polygons at each point. Again, we can try a little uninspired casework, or we can do a little math first to refine our investigations. Suppose the three polygons have m, n, and p sides. Then, we must have



Simplifying this gives



Hence, all three of m, n, p cannot be greater than 6. If we order the polygons such that , we need only consider the cases m = 3, 4, 5, 6.

For m = 3, we have



From here, we have a few choices how to proceed. First, we could proceed as before and note that n and p can't both be greater than 12. Then pound away with casework. Or, we could use Simon's Favorite Factoring Trick. Since we love writing 'Simon's Favorite Factoring Trick,' that's what we'll do:

$\begin{eqnarray*} 6(n+p) &=& np\\ 0&=&np-6(n+p)\\ 36&=&np-6(n+p)+36\\ 36&=&(n-6)(p-6) \end{eqnarray*}$

Note the use of Simon's Favorite Factoring trick, namely that .

From , we use the factors of 36 to read off the possible solutions in integers, thus getting the number of ways we can arrange a triangle and two other polygons about a point:

$\begin{eqnarray*} 36=1\cdot 36&\rightarrow& n=7, p = 42\\ 36=2\cdot 18&\rightarrow& n=8, p = 24\\ 36=3\cdot 12&\rightarrow& n=9, p = 18\\ 36=4\cdot 9&\rightarrow& n=10, p = 15\\ 36=6\cdot 6&\rightarrow& n=12, p = 12 \end{eqnarray*}$

There are probably a few surprises in that list! However, these can't all make tilings! Note that if XYZ is our triangle and three polygons meet at a point, then the polygon besides the triangle which has XY as a side must be the same as the one which has XZ as a side (otherwise - what will be along side YZ!) This sort of argument doesn't only for three polygon at a point set-ups in which a triangle is involved, but for any three polygon at a point set-up in which one of the polygons has an odd number of sides!

Hence, our only tiling here is 3, 12, 12:



Moving on m = 4, and working through the algebra similarly, we have

$\begin{eqnarray*} \frac{1}{n} + \frac{1}{p} &=&\frac{1}{4}\\ 4(n+p)&=&np\\ 0&=&np-4(n+p)\\ 16&=&np-4(n+p)+16\\ 16&=&(n-4)(p-4) \end{eqnarray*}$

Here, we can go through the factors of 16 to get the number of ways we can surround a single point with a square and two other polygons.

$\begin{eqnarray*} 16=1\cdot 16&\rightarrow& n=5, p = 20\\ 16=2\cdot 8&\rightarrow& n=6, p = 12\\ 16=4\cdot 4&\rightarrow& n=8, p = 8 \end{eqnarray*}$

As we've seen before, 'odd-sided polygon plus two others of different sides meet at each point' can't give a tiling, so we have only the tilings 4-6-12 and 4-8-8:





Examining m=5, we come to 1/n + 1/p = 3/10, which we'll find has only (5,10) as a solution, but this won't give us a tiling due to our odd-sided polygon plus two others of different sides' rule.

m=6 clearly leads to only n=p=6:



See if you can use the same logic we've followed above to analyze the case of 4 polygons meeting at a point! Find both the different ways in which 4 regular polygons can completely surround a point, and the number of these which are valid tilings!

Also, see if you can find some tilings in which the polygons that meet at each point are not always the same from point to point.