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Topic: equilateral triangle
Replies: 13   Last Post: Jan 8, 2002 4:18 PM

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Dan Hoey

Posts: 172
Registered: 12/6/04
Re: equilateral triangle
Posted: Nov 20, 2001 7:15 AM
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> > manuel <mnas@netcabo.pt> 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?

I hinted:

> 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

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

Dan Hoey

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