Virgil
Posts:
1,119
Registered:
12/6/04


Re: "Re: equilateral triangle"
Posted:
Nov 14, 2001 10:37 AM


In article <4a4rom9xmy62@legacy>, sbrian@sbcglobal.net (Steve Brian) wrote:
> On Mon, 12 Nov 2001, Mary Krimmel wrote > > >The second part looks easy. If P coincides with A then AP = 0 and > >AP + > >BP = BP = CP. > >Similarly we have equality if P coincides with B. > > >I think otherwise we cannot have equality. > > >Would it help to consider separate cases, one when P is on same side > >as C > >of a line through AB and one when P is on the other side of the line > >and > >one when P is on the line? (I don't know whether it would help. I > >think > >that I would first look at the third case  it looks simpler.) > > >Mary Krimmel > >mary@krimmel.net > > > >At 05:34 PM 11/11/2001 0500, you wrote: > >>I need an hint for this problem: > >> > >>Considerer an equilateral triangle ABC and a point P un the plane, > >>prove that: > >>AP+BP>=CP. > >> > >>Can we have the equality? > >> > >>manuel > > Consider the circle centered at C with radius r. Then consider the > ellipse consisting of all points Q such that AQ + BQ = r. If the > circle and ellipse intersect then the points of intersection satisfy > the equality. >
For a triangle of edge length a, r >= a, since AQ + BQ >= a.
The case of r = a is trivial, and has been considered above.
The relevant range of r is a <= r <= 2*sqrt(3)/3*a ~ 1.1547*a

