Students from Cowbridge Comprehensive School
in Wales took a look at this problem. They convinced themselves
that there are only 3 regular polygons that can be used to create
regular tessellations. Some used this spreadsheet to
justify their conclusion.
Catherine, from St. Michael's School, explained why it is possible
to produce a semi-regular tessellation with triangle, hexagon,
triangle, hexagon (or 3,6,3,6) meeting at each point:
In equilateral triangles angles are all $60^\circ$.
In regular hexagons angles are all $120^\circ$.
$2$ x $120^\circ$ = $240^\circ$ and $60^\circ$ x $2$ =
$120^\circ$ + $240^\circ$ = $360^\circ$!
She then offered another semi-regular
$4, 8, 8$ (square, octagon, octagon) meeting at each point.
You can use the interactivity provided with
the problem to check that this combination works.
Classes 9S and 9W from Aylesbury High
School tackled this problem.
We found that there were $7$ semi-regular tessellations. We
made a list of all the interior angles for regular polygons. We
tried all the combinations which add up to $360^\circ$.
Here are the solutions we found:
$150^\circ$ + $150^\circ$ + $60^\circ$: $12, 12, 3$
$150^\circ$ + $120^\circ$ + $90^\circ$: $12, 6, 4$
$135^\circ$ + $135^\circ$ + $90^\circ$: $8, 8, 4$
$120^\circ$ + $120^\circ$ + $60^\circ$ + $60^\circ$: $6, 3, 6,
$120^\circ$ + $90^\circ$ + $90^\circ$ + $60^\circ$: $6, 4, 3,
$120^\circ$ + $60^\circ$ + $60^\circ$ + $60^\circ$ +
$60^\circ$: $6, 3, 3, 3, 3$
$90^\circ$ + $90^\circ$ + $60^\circ$ + $60^\circ$ +
$60^\circ$: $4, 4, 3, 3, 3$
$150^\circ$ + $90^\circ$ + $60^\circ$ + $60^\circ$: doesn't
$144^\circ$ + $108^\circ$ + $108^\circ$: doesn't
If there are more than $12$ sides, then because the angles are
so big the shapes will overlap. So we decided we didn't need to
consider any shapes with more than $12$ sides. We did test some
out, but they didn't work.
Sue pointed out that actually there are $8$
semi-regular tessellations, because you can have both $4, 4, 3, 3,
3$ and $3, 4, 3, 4, 3$. You can see pictures of them all here