A and B are two fixed points on a circle and RS is a variable diamater. What is the locus of the intersection P of AR and BS?
Two semicircle sit on the diameter of a semicircle centre O of
twice their radius. Lines through O divide the perimeter into two
parts. What can you say about the lengths of these two parts?
Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?
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 degrees 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
and PQ is
Hence SR and PQ are equal in length and parallel.
and RQ is
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.