Three regular polygons, all with unit sides, share a common vertex and are all coplanar. Each polygon has a different number of sides, and each polygon shares a side with the other two; there are no gaps or overlaps. Find the number of sides for each polygon. There are multiple answers.
Formula for the third polygon
Call the number of sides of each polygon: x, y, and z.
From the formula for number of degrees per angle for a regular xgon and ygon and subtracting these from 360 degrees comes this formula for the number of sides of the third polygon. Obviously, then, one must throw out any solutions which are not integers.
z(x,y) = 2xy / (xy  2(x+y))
Ignoring for the moment the requirement that all polygons have a different number of sides, and noting that 3 hexagons are one solution, it is clear that every solution OTHER than (6,6,6) must have at least one polygon which has less than 6 sides and at least one polygon that has more than 6 sides. So simply by testing "x" values of 3, 4, 5, 6 with whatever y values produce integer results in z one can prove that the solutions already found are a complete list.

Posted by Larry
on 20180215 09:23:59 