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

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 Mary Krimmel Posts: 629 Registered: 12/3/04
Re: equilateral triangle
Posted: Nov 12, 2001 1:49 PM

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

Date Subject Author
11/11/01 manuel
11/12/01 Mary Krimmel
11/12/01 Virgil
11/13/01 Mary Krimmel
11/14/01 Virgil
11/14/01 Rouben Rostamian
11/17/01 Dan Hoey
11/20/01 Dan Hoey
11/20/01 Dan Ross
11/22/01 Dan Hoey
11/23/01 Rouben Rostamian
1/8/02 John Conway
11/13/01 Steve Brian
11/14/01 Virgil