
Re: equilateral triangle (Spoiler)
Posted:
Jan 8, 2002 4:18 PM


On 23 Nov 2001, Rouben Rostamian wrote:
> Dan Hoey <haoyuep@aol.com> proved that: > > Let ABC be an equilateral triangle and let P be an arbitrary > point anywhere in the plane of ABC. Then PC <= PA + PB. > > [Proof omitted] > > Dan, that's a wonderful proof. Much neater than I could have > imagined. It ought to become a classic. Congratulations!
It suggests a number of questions. For starters, what's the set of shapes for ABC that still yield this (a): one way round, and (b): all three ways round (so that PA,PB,PC can be used as triangle edges)?
John Conway

