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

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Virgil

Posts: 1,119
Registered: 12/6/04
Re: "Re: equilateral triangle"
Posted: Nov 14, 2001 10:37 AM
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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





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