7th APMO 1995 Problem 4 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.
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 OQ 2 = OX 2 + XQ 2 - 2·OX·XQ cos θ and OP 2 = OX 2 + XP 2 - 2·OX·XP cos(θ+π), where θ is the angle OXQ. But cos(θ+π) = -cos θ, so OQ 2 + OP 2 = 2OX 2 + 2XQ 2 . 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 XO 2 + XC 2 = OC 2 . So OQ 2 = 2OC 2 - OP 2 . But this quantity is fixed, so Q must lie on the circle center O radius √(2R 2 - OP 2 ), 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 √(2R 2 - OP 2 ) let X be the midpoint of PQ. Then take the chord AC to have X as its midpoint. 7th APMO 1995 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002