> > manuel <firstname.lastname@example.org> wrote: > > I need an hint for this problem: > > Considerer an equilateral triangle ABC and a point P on the plane. > > Prove that |AP|+|BP|>=|CP|. > > Can we have the equality?
> What is odd about the exponents of the polynomial > (r+s+t)(-r+s+t)(r-s+t)(r+s-t) ? > Application of this hint has enabled me to prove the inequality > algebraically, but I'd certainly like a geometric proof.
Well, darn it, I knew where that polynomial came from, but I didn't see until now how it will yield a geometric proof. Try this: Erect equilateral triangle AA'P (resp BB'P) in the half-plane of AP (BP) that includes the ray PC. (If P is outside the circumcircle, the erected triangles will overlap.)
Prove that CA'PB' is a parallelogram.
I've never been able to prove these Napoleonic things geometrically, but I know they can be done in principle. I hope it's easier than phonying up axes to simulate the (easy) algebraic proof.
Does anyone know the provenance or any history of the problem?