Let S be the set of all points inside or on a triangle. Let T be the set S with one interior point excluded. Show that one can find points Pi, Qi such that Pi and Qi are distinct and the closed segments PiQi are all disjoint and have union T.
Solution
Let the excluded interior point be O. Let the vertices be A, B, C. Take points D, E, F on BC, CA, AB. Now T is the disjoint union of T1, T2, T3, where: T1 is the quadrilateral OFAE and its interior, but excluding O, F and the segment OF; T2 is the quadrilateral ODBF and its interior, but excluding O, D and the segment OD; and T3 is the quadrilateral OECD and its interior, but excluding O, E and the segment OE. But it is now trivial to express T1 as a disjoint union of closed intervals - just take each interval to have one endpoint on AF and the other on EO. Similarly for T2 and T3. [If you wish, you can make it even more trivial by choosing D, E, F so that OD is parallel to AB, OE to BC and OF to AC.]
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002