+654 find all groups of three regular polygons with side length one that can surround one point such that each polygon shares a side with the other two.
+981 Let there be three regular polygons with \(x,y,\) and \(z\) as their number of sides. Their internal angles must sum to \(360º\) and individually be supplement to their respective external angle. Therefore, we can express the sum of the three angles as:
\((180º-\frac{360º}{x})+(180º-\frac{360º}{y})+(180º-\frac{360º}{z})=360º\\ 540º-\frac{360º}{x}-\frac{360º}{y}-\frac{360º}{z}=360º\\ 180º=\frac{360º}{x}+\frac{360º}{y}+\frac{360º}{z}\\ \frac2x+\frac2y+\frac2z=1\)
\(3\le x \le 6\), because three sides is the least number of sides a regular polygon can have, and if \(x,y,\) or \(z\) is greater than 6, the sum will be less than one.
Casework:
\(x=6\Rightarrow y=6, z=6\\ x=5\Rightarrow y=5, z=10\\ x=4\Rightarrow (8,8);(6,12);(5,20)\\ x=3\Rightarrow (10,15);(9,18);(8,24);(7,42)\\\)
Therefore, there are 9 groups of regular polygons that can surrond a point.
I hope this helped,
Gavin.
+981 Let there be three regular polygons with \(x,y,\) and \(z\) as their number of sides. Their internal angles must sum to \(360º\) and individually be supplement to their respective external angle. Therefore, we can express the sum of the three angles as:
\((180º-\frac{360º}{x})+(180º-\frac{360º}{y})+(180º-\frac{360º}{z})=360º\\ 540º-\frac{360º}{x}-\frac{360º}{y}-\frac{360º}{z}=360º\\ 180º=\frac{360º}{x}+\frac{360º}{y}+\frac{360º}{z}\\ \frac2x+\frac2y+\frac2z=1\)
\(3\le x \le 6\), because three sides is the least number of sides a regular polygon can have, and if \(x,y,\) or \(z\) is greater than 6, the sum will be less than one.
Casework:
\(x=6\Rightarrow y=6, z=6\\ x=5\Rightarrow y=5, z=10\\ x=4\Rightarrow (8,8);(6,12);(5,20)\\ x=3\Rightarrow (10,15);(9,18);(8,24);(7,42)\\\)
Therefore, there are 9 groups of regular polygons that can surrond a point.
I hope this helped,
Gavin.
