If P and Q are two points in the plane, let m(PQ) be the perpendicular bisector of PQ. S is a finite set of n > 1 points such that: (1) if P and Q belong to S, then some point of m(PQ) belongs to S, (2) if PQ, P'Q', P"Q" are three distinct segments, whose endpoints are all in S, then if there is a point in all of m(PQ), m(P'Q'), m(P"Q") it does not belong to S. What are the possible values of n?
Answer
n = 3 (equilateral triangle), 5 (regular pentagon).
Solution
There are n(n-1)/2 pairs of points. Each has a point of S on its bisector. But each point of S is on at most two bisectors, so 2n ≥ n(n-1)/2. Hence n ≤ 5.
The equilateral triangle and regular pentagon show that n = 3, 5 are possible.
Consider n = 4. There are 6 pairs of points, so at least one point of S must be on two bisectors. wlog A is on the bisectors of BC and BD. But then it is also on the bisector of CD. Contradiction.
Many thanks to Tsimerman for this.
© John Scholes
jscholes@kalva.demon.co.uk
25 Nov 2003
Last corrected/updated 25 Nov 03