What is the largest n for which we can find n + 4 points in the plane, A, B, C, D, X1, ... , Xn, so that AB is not equal to CD, but for each i the two triangles ABXi and CDXi are congruent?
Answer
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 SA center A radius b, or on the circle SBcenter B radius b. Similarly, CX or DX = AB = a, so X lies either on the circle SC center C radius a, or on the circle SD center D radius a. Thus we have four pairs of circles, (SA, SC), (SA, SD), (SB, SC), (SB, SD) 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 (SA, SC) are included, then no points of (SB, SD) 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 (SA, SC) 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 = 180o - ∠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 < a = CX, M lies to the right of XY.
Now suppose there is a solution for the configuration (SB, SD). Thus there is a point Z such that ABZ and ZDC are congruent. Then AZ = CZ, so Z lies on the perpendicular bisector of AC and hence on the same side of XY as D. But it is also a distance a from D and a distance b from B, and a > b, so it must lie on the same side of XY as B. Contradiction. So there are no solutions for the configuration (SB, SD), as required. That completes the proof that n ≤ 4.
For an example with n = 4, take a regular hexagon ACDBX3X2. Extend the side X2X3 to X1X4, with X1, X2, X3, X4 equally spaced in that order, so that X1AX2 and X3BX4 are equilateral. Then ABX1 and CX1D are congruent, ABX2 and DX2C are congruent, ABX3 and X3CD are congruent, and ABX4 and X4DC are congruent.
© John Scholes
jscholes@kalva.demon.co.uk
8 February 2004
Last corrected/updated 8 Feb 04