Take a fixed point P inside a fixed circle. Take a pair of perpendicular chords AC, BD through P. Take Q to be one of the four points such that ASCQ, ASDQ, BSCQ or BSDQ is a rectangle. Find the locus of all possible Q for all possible such chords.
Solution
Let O be the center of the fixed circle and let X be the center of the rectangle ASCQ. By the cosine rule we have OQ2 = OX2 + XQ2 - 2·OX·XQ cos θ and OP2 = OX2 + XP2 - 2·OX·XP cos(θ+π), where θ is the angle OXQ. But cos(θ+π) = -cos θ, so OQ2 + OP2= 2OX2 + 2XQ2. But since X is the center of the rectangle XQ = XC and since X is the midpoint of AC, OX is perpendicular to AC and hence XO2 + XC2 = OC2. So OQ2 = 2OC2 - OP2. But this quantity is fixed, so Q must lie on the circle center O radius √(2R2 - OP2), where R is the radius of the circle.
Conversely, it is easy to see that all points on this circle can be reached. For given a point Q on the circle radius √(2R2 - OP2) let X be the midpoint of PQ. Then take the chord AC to have X as its midpoint.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002