Finding the Platonic planar tilings

Why are there only three? The simple answer is because that's how many there are! But we can show mathematically why there are only three. Our ground rules are that we are counting only the plain tiling shapes, not the number of different ways they can be labeled. And, by definition, Platonic tilings mean that the polygons making up the tiling must all be the same, and all regular, where a regular polygon is one with all of its angles and edges the same. Triangles, squares, and pentagons are all regular polygons. We can generalize and call these n-gons. Obviously we cannot have any 1-gons or 2-gons, so we must start with n = 3 and go up from there.

First, consider any n-gon. Some of you may know that every n-gon has 180 (n-2) degrees when adding up the degrees of angle of all of the n-gon's interior angles. If not, here is an explanation. First, we have to do this for a triangle (3-gon); then we can do it for any n-gon.

Start with a right triangle, which is a triangle with one right angle (= 90 degrees). Using the two perpendicular sides of the triangle as two sides, form a rectangle by adding two more sides that surround the triangle, creating a rectangle with a diagonal line enclosing two copies of the original triangle. From this you should be able to convince yourself that the two other angles (other than the 90 degree angle) of the triangle add up to 90 degrees. (We are making you do some of the work here. Don't be frightened! It is good brain exercise for you.) When you add this to the 90 degree angle you get 180 degrees in a right triangle. For an arbitrary triangle, first see that you can divide it into two right triangles by drawing a line from one corner to its opposite side to make two right angles. Then the two triangles will have 360 degrees in total, except that the two right angles combine to erase 180 degrees, leaving 180 degrees again for the original triangle.

Now, for any n-gon, draw lines from its center to each of the edge vertices. This divides the n-gon into triangles. There are 360 degrees in a circle and at the center of the n-gon the circle is divided into n equal sections, so each inner angle must be 360/n. That means the two other angles of each triangle must each be (180 - 360/n)/2 degrees, using the result we just found that there are 180 degrees in a triangle. When we go around the n-gon we see that there are 2n of these angles all around the interior angles of the n-gon, so our total after adding them all up must be 2n (180 - 360/n)/2 = 180n - 360 = 180 (n-2).

Now, back to our original n-gon and using the result we just found, it must be that each of the interior angles of the n-gon must be 180 (n-2) / n, since the total is 180 (n-2) and there are n interior angles. From the point of view of creating a tiling, we must fit some multiple k of them placed side by side to form a perfect joining around a vertex point without any gaps. Thus, we are looking for some k such that k 180 (n-2) / n = 360, for some n. We can rearrange this to be k = 2 n/(n-2), where our goal is to find values of k and n so that this equation is true. This isn't exactly easy! But we can make life simpler with a small observation.

We first observe that k=3 and n=6 is a solution for this equation. For n=6, the interior angles must be 180 (n-2) / n = 180 (6-2) / 6 = 120. If n is bigger than 6 then the angles will be bigger than 120, so that two of them will be bigger than 240, leaving less than 120 to fit in one more polygon with an angle that is bigger than 120! In other words, we can never have any n-gons with n bigger than 6. Then it is just a simple problem of trying all of the possibilities. When you do that you find just three:  {n, k} = {3,6}, {4,4}, and {6,3}.

And that's why there are only three Platonic tilings!

(But what happens if we relax our requirement of only using convex regular polygons? For example, connect every second vertex in a pentagon instead of every adjacent one to produce a star-shaped pentagram. Can you tile the plane with these, or any other such non-convex regular polygon?)

Return to Platonic tilings

Last updated: April 18, 2019