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Minimal Distances to a Point in a Triangle

Date: 01/05/2001 at 05:33:13
From: Kyunghoon Shin
Subject: Geometry


Prove that the smallest value of PA + PB + PC occurs when 
<APB = <BPC = <CPA = 120 degrees, for a triangle ABC and a point P.

I can't start this at all. Could you send me an answer?

Thank you.

Date: 01/05/2001 at 09:12:38
From: Doctor Floor
Subject: Re: Geometry

Hi, Kyunghoon,

Thanks for writing.

I was pointed to a very simple proof for this statement by Atul Dixit.

For the record, the message by Atul Dixit was sent to the Hyacinthos 
discussion group hosted by eGroups, at Website:   

Consider triangle ABC. The points X such that AX + BX = x for some 
constant x, form an ellipse with A and B as foci.

Let D be the point on this ellipse yielding minimal AD + BD + CD. Then 
AD must be the minimal distance from A to the ellipse, and thus AD 
must be perpendicular to the tangent to the ellipse through D.

Here's a figure:


We know in this figure that <ADG = <BDH (general for tangents to an 
ellipse), and thus also <ADC = <BDC. 

Now suppose that P is the point such that AP + BP + CP is minimal over 
all points P in the plane. Then by the above argument <APC = <BPC. But 
by similar reasoning with an ellipse with A and C as foci, we also 
find <APB = <BPC.

So <APB = <BPC = <CPA = 120 degrees.

By the way, this reasoning only works when all angles of ABC are 
smaller than 120 degrees, so that point P can be an interior point 
of ABC.

For a different approach, and some background information on this 
point, see "A Point in the Triangle" in the Dr. Math archives at:   

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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