In article <4a4rom9xmy62@legacy>, email@example.com (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 > >firstname.lastname@example.org > > > >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