Shady Symmetry
We received a couple of very good solutions from students who considered the number of different patterns that could be created from shading four triangles on a 4 by 4 isometric grid.
Mary sent us her work:
Think about the four little triangles in a vertical line in the middle of the triangle. We must have an even number of them in our pattern. We can count the different possibilities separately.
All four of the vertical ones: $1$ possibility (we don't need to colour in any more triangles).
Two of the vertical ones: there are $6$ possible pairs that don't include the middle ones, each of which can go with any of the $6$ vertical pairs, so there are $6 \times 6=36$ possibilities.
None of the vertical ones: there are $6$ pairs that don't include the middle ones. Once we've picked one of them, there are $5$ others left to choose from. But this counts every possibility twice (A with B and B with A), so there are $\frac{6 \times 5}{2}=15$ possibilities.
So there are $1+36+15=52$ total possible patterns.
Below are the twelve symmetrical pairs.
Neil sent in this solution. It's not quite right, but with just a small correction the reasoning can be used to arrive at the solution:
There are $12$ different "symmetrical pairs" in the big triangle (shown above), and $2$ different pairs must be used on each pattern. So pair number $1$ can go with any one of pairs $2$ through to $12$. This gives us $11$ patterns. Continuing in the same way:
$1$ goes with $2$ through to $12$ making $11$ patterns.
$2$ goes with $3$ through to $12$ making $10$ patterns.
$3$ goes with $4$ through to $12$ making $9$ patterns.
...up until...
$11$ goes with $12$ through to $12$ making $1$ pattern.
We need to add all integers $1$ to $11$, for which we can use the formula:
$\frac{n^2+n}{2}$
$\frac{11^2+11}{2}=\frac{121+11}{2} = \frac{132}{2} = 66$
So the answer is $66$ patterns.
In fact, this isn't quite right, because not all of the pairs can go together. For example, we can't combine both pairs shown in the fourth row above. So there aren't quite as many as $66$.
Taking this into account we can use Neil's approach and work out that there are $11+10+9+8+7+6+1=52$ possibilities.
Neil sent in this solution. It's not quite right, but it makes a good start.
There are $12$ different "symmetrical pairs" in the big triangle (shown below), and $2$ different pairs must be used on each pattern. So pair number $1$ can go with any one of pairs $2$ through to $12$. This gives us $11$ patterns. Continuing in the same way:
$1$ goes with $2$ through to $12$ making $11$ patterns.
$2$ goes with $3$ through to $12$ making $10$ patterns.
$3$ goes with $4$ through to $12$ making $9$ patterns.
...up until...
$11$ goes with $12$ through to $12$ making $1$ pattern.
We need to add all integers $1$ to $11$, for which we can use the formula:
$\frac{n^2+n}{2}$
$\frac{11^2+11}{2}=\frac{121+11}{2} = \frac{132}{2} = 66$
So the answer is $66$ patterns.
In fact, this isn't quite right, because not all of the pairs can go together. For example, we can't combine both pairs shown in the fourth row below. So there aren't quite as many as $66$.
Taking this into account we can use Neil's approach and work out that there are $11+10+9+8+7+6+1=52$ possibilities.
Mary sent us her work:
Think about the four little triangles in a vertical line in the middle of the triangle. We must have an even number of them in our pattern. We can count the different possibilities separately.
All four of the vertical ones: $1$ possibility (we don't need to colour in any more triangles).
Two of the vertical ones: there are $6$ possible pairs that don't include the middle ones, each of which can go with any of the $6$ vertical pairs, so there are $6 \times 6=36$ possibilities.
None of the vertical ones: there are $6$ pairs that don't include the middle ones. Once we've picked one of them, there are $5$ others left to choose from. But this counts every possibility twice (A with B and B with A), so there are $\frac{6 \times 5}{2}=15$ possibilities.
So there are $1+36+15=52$ total possible patterns.
Below are the twelve symmetrical pairs.