7th Australian 1986 Problem B1 C is a circle of unit radius. For any line L define d(L) to be the distance between the two points of intersection if L meets C in two points, and zero otherwise. For any point P define f(P) to be the maximum value of d(L) + d(L') for two perpendicular lines through P. Which P have f(P) > 2?
points inside (but not on) the circle with the same center and radius √(3/2) Solution It is obvious that points inside C qualify (take L to be a diameter), and fairly obvious that points on C qualify (L and L' meet C again at the ends of a diameter). So we consider P outside C. Let O be the center of C. Obviously if PO ≥ √2, then only one of L, L' meets C, so f(P) = 2. So suppose P is close enough for L and L' to meet C. Let OP = r. Take the angle x as shown. Then d = d(L) + d(L') = 2√(1 - r 2 cos 2 x) + 2√(1 - r 2 sin 2 x). So d 2 /4 = 2 - r 2 + 2√(1 - r 2 + r 4 sin2xcos2x) = 2 - r 2 + √(4 - 4r 2 + r 4 sin2x). That is obviously maximised by taking x = π/4 giving 2 - r 2 + √(4 - 4r 2 + r 4 ) = 4 - 2r 2 , since r 2  2 iff 4 - 2r 2 > 1, or r < √(3/2). Solutions are also available in Australian Mathematical Olympiads 1979-1995 by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust. It can be ordered by email. 7th Aus 1986 © John Scholes jscholes@kalva.demon.co.uk 5 Jan 2004 Last corrected/updated 5 Jan 04