Minimal Distances to a Point in a TriangleDate: 01/05/2001 at 05:33:13 From: Kyunghoon Shin Subject: Geometry Problem: 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: http://egroup.com/ 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: http://mathforum.org/dr.math/problems/john02.12.99.html If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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