Problem 840

4 Solution Many thanks to Allen Zhang for completing the proof Assume AB = a, CD = b with a > b. If ABX and CDX are congruent, then either AX or BX = CD = b, so X lies either on the circle S A center A radius b, or on the circle S B center B radius b. Similarly, CX or DX = AB = a, so X lies either on the circle S C center C radius a, or on the circle S D center D radius a. Thus we have four pairs of circles, (S A , S C ), (S A , S D ), (S B , S C ), (S B , S D ) each with at most 2 points of intersection. X must be one of these 8 points. However, we show that if two points of intersection of (S A , S C ) are included, then no points of (S B , S D ) can be included. The same applies to each pair of circles, so at most 4 points X are possible. Finally, we will give an example of n = 4, showing that the maximum of 4 is achieved. So suppose (S A , S C ) intersect at X and Y. We must have BX = DX and BY = DY, so X and Y both lie on the perpendicular bisector of BD. In other words, XY is the perpendicular bisector of BD, so D is the reflection of B in the line XY. There is no loss of generality in taking B (and D) to be on the same side of AC as X. Let A' be the reflection of A in the line XY. Since B lies on the circle center A radius a, D must lie on the circle center A' radius A. Thus the triangles A'XC and CDA' are congruent. (Note that A and C can be on the same side of XY or opposite sides.) Hence D is the same height above AC as X, so DX is perpendicular to XY. Hence X is the midpoint of BD. Also ∠A'CD = ∠CA'X = 180 o - ∠CAX, so AX and CD are parallel. They are also equal, so ACDX is a parallelogram and hence AC = DX = BD/2. In the second configuration above, both A and C are on the same side of XY as D, so the midpoint M of AC lies on the same side of XY as D. In the first configuration, since AX = b b, so it must lie on the same side of XY as B. Contradiction. So there are no solutions for the configuration (S B , S D ), as required. That completes the proof that n ≤ 4. For an example with n = 4, take a regular hexagon ACDBX 3 X 2 . Extend the side X 2 X 3 to X 1 X 4 , with X 1 , X 2 , X 3 , X 4 equally spaced in that order, so that X 1 AX 2 and X 3 BX 4 are equilateral. Then ABX 1 and CX 1 D are congruent, ABX 2 and DX 2 C are congruent, ABX 3 and X 3 CD are congruent, and ABX 4 and X 4 DC are congruent. 13th APMO 2001 © John Scholes [email protected] 8 February 2004 Last corrected/updated 8 Feb 04