To prove this, we need only consider how multiple copies of this pentagon could possibly be arranged at a vertex. The pentagon above admits no monohedral, edge-to-edge tiling of the plane. Under what circumstances could such polygons tile the plane?įor triangles and quadrilaterals, the answer is, remarkably, always! We can rotate any triangle 180 degrees about the midpoint of one of its sides to make a parallelogram, which tiles easily. But we won’t assume the side lengths and interior angles are all the same. For example, what if we don’t restrict ourselves to regular polygonal tiles? We’ll stick with “convex” polygons, those whose interior angles are each less than 180 degrees, and we’ll allow ourselves to move them around, rotate them and flip them over. Once a specific problem is solved, we start to relax the conditions. Of course, that’s never enough for mathematicians. And with that, the regular, monohedral, edge-to-edge tilings of the plane are completely understood. Similarly for squares: Four squares around a single point at 90 degrees each gives us 4 × 90 = 360.Ī similar argument will show that after the hexagon - whose 120-degree angles neatly fill 360 degrees - no other regular polygon will work: The angles at each vertex simply won’t add up to 360 as required. This works out perfectly: The measure of each internal angle of an equilateral triangle is 60 degrees, and 6 × 60 = 360, which is exactly what we need around a single point. This chart raises all sorts of interesting mathematical questions, but for now we just want to know what happens when we try to put a bunch of the same n-gons together at a point.įor the equilateral-triangle tiling, we see six triangles coming together at each vertex. Here they are up to n = 8, the regular octagon. We can make a chart for the measure of an interior angle in regular n-gons.
What do these two facts have to do with the tiling of regular polygons? By definition, the interior angles of a regular polygon are all equal, and since we know the number of angles ( n) and their sum (180( n − 2)), we can just divide to compute the measure of each individual angle.
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