making it a square.Four squares are drawn of the edges of a parallelogram ABCD as
shown in the diagram. However you change the parallelogram ABCD the
quadrilateral formed by joining the centres of the squares is
always itself a square. Iain Mitchell of Madras College used
vectors to prove that this quadrilateral is a rhombus and you might
like to try to do this yourself. He went on to prove that the
angles are all 90 making it a square.
The following proof uses complex numbers (which are, when all is
said and done, the same as vectors in two dimensions). The
advantage of this method is that it is shorter and there is not so
much to write down. You do need to use the result however that the
complex number iz represents a vector at right angles to
z. The reason for this is that when you multiply any
complex number z by the complex number i you get
iz which has the same modulus as z but an
argument 90
more than the argument of z.
Take the origin O at the centre of the parallelogram and represent the midpoints of the edges of the parallelogram by complex numbers u, - u, v and - v as shown in the diagram.
The midpoint of AB is represented by the complex number - u, and, if we call this point M, then the line segment MB corresponds to the complex number v and the line segment MP corresponds to the complex number iv. Hence P is represented by the complex number - u + iv.
Similarly Q is v + iu, R is u - iv and S is - v - iu. Hence SR is
| z | = | ( u - iv) - (- v - iu) |
| = | (1 + i) u + (1 - i) v |
and PQ is
| z 1 | = | ( v + iu) - (- u + iv) |
| = | (1 + i) u + (1 - i) v | |
| = | z. |
Hence SR and PQ are equal in length and parallel.
SP is
| w | = | (- u + iv) - (- v - iu) |
| = | ( i - 1) u + (1 + i) v |
and RQ is
| w 1 | = | ( v + iu) - ( u - iv) |
| = | ( i - 1) u + (1 + i) v = w. |
Hence SP and RQ are equal in length and parallel. It follows that iz = w which shows that SP is perpendicular to, and the same length as, SR and hence that SPQR is a square.