Vector Journeys
James from Wilson's School told us
something he'd noticed about the second, third and fourth vectors
forming a square park:
The vector journey goes: $\pmatrix{3\cr 1} \pmatrix{-1\cr 3}
\pmatrix{-3\cr -1} \pmatrix{1\cr -3}$
After finding the first vector both top and bottom numbers should
be positive, next vector the top number is made negative then next
vector both numbers are negative and finally the last vector the
bottom number is negative. The second and last vectors the
left/right number switches with the up/down number.
Elliott, also from Wilson's, made some
observations:
Charlie walks on vectors of $\pmatrix{3\cr 1} \pmatrix{-1\cr
3} \pmatrix{-3\cr -1} \pmatrix{1\cr -3}$
Another square he could walk would have vectors of $\pmatrix{5\cr
2} \pmatrix{-2\cr 5} \pmatrix{-5\cr -2} \pmatrix{2\cr -5}$
These vectors must only consist of four numbers: $x, y, -x$ and
$-y$.
It can only be two numbers, and their negatives, so that all the
sides of the square are equal in length.
After travelling along the first vector, you can then move left or
right. From there, you must do the opposite of your first move,
then the opposite of your second, to get back to your original
position.
Niharika from Leicester High School
for Girls sent us this
solution. Well done to you
all.