# Minimum distance point proof for triangle with one angle >= 120

• Jul 13th 2009, 06:32 PM
hercules
Minimum distance point proof for triangle with one angle >= 120
Hello,

I am having a little trouble developing a proof for the minimum distance point in the following scenario.

if i have a triangle with vertices A, B, and C.
measure of angle ABC is >=120 degrees
prove that the vertex B is the minimum distance point?

if there was another point call it P, then the distance from P to the vertices of the triangle which is
PA + PB + PC is greater than the distance from vertex B which is AB + BB+ BC = AB + BC.
That is saying that vertex B of triangle is the minimum distance point.

You might want to look at Fermat point.

Thank You
• Jul 19th 2009, 10:15 AM
Jhevon
Quote:

Originally Posted by hercules
Hello,

I am having a little trouble developing a proof for the minimum distance point in the following scenario.

if i have a triangle with vertices A, B, and C.
measure of angle ABC is >=120 degrees
prove that the vertex B is the minimum distance point?

if there was another point call it P, then the distance from P to the vertices of the triangle which is
PA + PB + PC is greater than the distance from vertex B which is AB + BB+ BC = AB + BC.
That is saying that vertex B of triangle is the minimum distance point.

You might want to look at Fermat point.

Thank You

did the solution i suggested work out for this?
• Jul 28th 2009, 07:53 PM
hercules
Quote:

Originally Posted by Jhevon
did the solution i suggested work out for this?

Yes, it did. We didn't even need to calculate the length using sine values
Since our triangle PBP' was isosceles by construction. We had only to realize that as the vertex angle decreased (which we found to be restricted between 0 and 60 degrees) the length of the base PP' also decreased.

Sorry, for the delayed response. I found math forum mail filtered.