We know that square tiles can be put together to cover a floor without gaps (ignore boundary conditions). This coverability holds for triangular and hexagonal tiles as well. Are there any other symmetric tiles that can cover a floor?
Triangles, squares and hexagons have 3,4 and 6 sides respectively. Some observations:
(1) Triangle: All internal angles are 60. Six triangles meet at a corner-point
(2) Square: All internal angles are 90. Four squares meet at a corner-point
(3) Hexagon: All internal angles are 120. Three hexagons meet a corner-point
A pictorial representation is shown below (excuse me for the poor quality - it is a hand-drawn image)
Note that, for a polygon of k sides, the sum of internal angles is (k-2)*180.
In the general case, let k be the number of sides of a symmetric shape that can be tiled to cover a floor. The following must hold true:
(1)The sum of internal angles in the tile is (k-2)*180. Since the tiles are symmetric, every internal angle is the same. Hence the internal angle is (k-2)*180/k.
(2) Adjacent tiles that share a corner point, must subtend a total angle of 360 degrees at the corner point. This holds for all corner points.
If m tiles share a corner point, then the conditions above imply: m*(k-2)*180/k = 360, which simplifies to:
m=2k/(k-2). This also implies k = 2m/(m-2).
Let us try various values of k and see which result in m being in integer. Obviously k > 2, and m > 2.
k=3 implies that m=6 (Six triangular tiles meet at a corner-point)
k=4 implies that m=4 (Four square tiles meet at corner-point)
k=5 implies that m=10/3, which is an impossibility, since 10/3 is not an integer
k=6 implies that m=3 (Three hexagonal tiles meet at a corner-point)
We can keep trying for further values of k=7, 8 ,9 and so on. But let us simplify the analysis. Since we want to try for k=7, 8, 9, etc., let k = 6 + k', where k' >1. Then we get:
m = (12 + 2k')/(4 + k')
= ((4 + k') + (4 + k') + 4)/(4+k') =
= 2 + (4/(4+k'))
We observe:
(a) Since k'>1, we have (4/(4+k')) < 1. This implies that, m < 3.
(b) Since k = 2m/(m-2), we have m > 2.
No integer value of m can simultaneously satisfy the above two conditions. So, there is no valid value of m, for k=7, 8, 9 and so on. The only valid values of m are 3, 4 and 6.
Thus there are only three symmetric shapes that can cover a floor: triangles, squares and hexagons.
Triangles, squares and hexagons have 3,4 and 6 sides respectively. Some observations:
(1) Triangle: All internal angles are 60. Six triangles meet at a corner-point
(2) Square: All internal angles are 90. Four squares meet at a corner-point
(3) Hexagon: All internal angles are 120. Three hexagons meet a corner-point
A pictorial representation is shown below (excuse me for the poor quality - it is a hand-drawn image)
In the general case, let k be the number of sides of a symmetric shape that can be tiled to cover a floor. The following must hold true:
(1)The sum of internal angles in the tile is (k-2)*180. Since the tiles are symmetric, every internal angle is the same. Hence the internal angle is (k-2)*180/k.
(2) Adjacent tiles that share a corner point, must subtend a total angle of 360 degrees at the corner point. This holds for all corner points.
m=2k/(k-2). This also implies k = 2m/(m-2).
Let us try various values of k and see which result in m being in integer. Obviously k > 2, and m > 2.
k=3 implies that m=6 (Six triangular tiles meet at a corner-point)
k=4 implies that m=4 (Four square tiles meet at corner-point)
k=5 implies that m=10/3, which is an impossibility, since 10/3 is not an integer
k=6 implies that m=3 (Three hexagonal tiles meet at a corner-point)
We can keep trying for further values of k=7, 8 ,9 and so on. But let us simplify the analysis. Since we want to try for k=7, 8, 9, etc., let k = 6 + k', where k' >1. Then we get:
m = (12 + 2k')/(4 + k')
= ((4 + k') + (4 + k') + 4)/(4+k') =
= 2 + (4/(4+k'))
We observe:
(a) Since k'>1, we have (4/(4+k')) < 1. This implies that, m < 3.
(b) Since k = 2m/(m-2), we have m > 2.
No integer value of m can simultaneously satisfy the above two conditions. So, there is no valid value of m, for k=7, 8, 9 and so on. The only valid values of m are 3, 4 and 6.
Thus there are only three symmetric shapes that can cover a floor: triangles, squares and hexagons.
