Floor patterns with three polygons at each corner

Floor patterns with three polygons at each corner

IP 1
7 marks

A designer wants to cover a floor with regular polygon tiles so that at every corner exactly three tiles meet with no gaps: two regular \(n\)-gons and one regular \(m\)-gon.

  1. Using the interior angle \(180^\circ-\dfrac{360^\circ}{n}\), show that \(\dfrac4n+\dfrac2m=1\).[2]
  2. Find all the pairs of whole numbers \((n,m)\), with \(n,m\ge3\), that satisfy this equation.[3]
  3. One solution uses two regular pentagons and a regular decagon at a corner. Explain why this corner cannot be repeated around every corner of a pentagon, so this pattern cannot cover a whole floor.[2]

Solution:

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Answer:(a) \(2\left(180-\dfrac{360}{n}\right)+180-\dfrac{360}{m}=360\) simplifies to \(\dfrac4n+\dfrac2m=1\) (b) \((5,10),(6,6),(8,4),(12,3)\) (c) the neighbours of a pentagon must alternate between pentagon and decagon, impossible around \(5\) sides

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