Hereis how
Emily from Beeston Rylands Junior School started on the
problem.
Damini from Newstead Wood School For
Girls discovered that not all diagonal lines that join two grid
points make a square:
You find the centre of your diagonal line and go from there away
from the line in a straight line (at right angles) until you find
the right point. Then join the point to one end of the diagonal
line and you have a side! How easy was that. Not all diagonal lines
that join two grid points make a square. If you take a piece of
square dotty paper and join up any two dots it will be hard to find
the square, as there might not even be a square.
I agree with the fact that a straight line can be the side of many
different rhombuses but to me I see no real way of determining how
many rhombuses go to one line.
I also agree that the diagonal of one rhombus can have the same
diagonal line as infinitely many other rhombuses as if you take one
line you can keep editing the size of the rhombus without actually
changing the length of the first diagonal.
Alexander from Wilson's School suggested a
strategy for finding the number of possible rhombuses that could be
made when he is given a side:
To find how many different rhombuses can be found with a single
line that is between two dots, you need to choose one end
of the line. You then look at "one side" of that point, and
see how many different ways the line can be "twisted". You then
multiply it by four because the same number of "twisted" lines can
be found on the "other side" of that point, and on the "two sides"
of the other end of the line. The answer to this operation
determines how many rhombuses can be made from one line on a dotted
piece of paper. (Editor's comment: I think that Alexander
has double counted here; each of the "twisted lines" found at one
end will make rhombuses with each of the "twisted lines" found at
the other end of the line.)
Niharika from Leicester High School for
Girls explained that if the vector $\pmatrix{a\cr b}$
represents the diagonal line, the square will have vertices at grid
points if a+b (and therefore a-b) is even.
She also suggested a strategy for finding
the number of rhombuses that could be drawn if one of the sides is
given. Her solution ishere.
Niharika also contributed a solution
to Vector Journeys,
which uses vector notation so solve a related problem.
We also received good responses from the
Maths Galaxy Explorers from North Walsham Junior
School.