
Re: equilateral triangle (Spoiler)
Posted:
Nov 22, 2001 1:27 PM


Thanks to all on geometry.puzzles, and to Andrei Ismail on sci.math who identified this as the "Van Schooten theorem," though I'm not quite sure whether that name refers to the inequality as well as the equality.
Here's my simple geometric proof of the inequality (though I'm sure it's been done before). Let ABC be an equilateral triangle and P a point in the plane. Prove: AP + BP >= CP.
Proof: Erect equilateral triangle APD, with the angles <PAD and <BAC in the same rotational direction about point A. Depending on whether the angles overlap, we have either
<BAP = <BAC  <PAC = <PAD  <PAC = <DAC or <BAP = <BAC + <PAC = <PAD + <PAC = <DAC .
Also, AB=AC and AP=AD , so by sideangleside the triangles ABP and ACD are congruent. Therefore, BP=CD . So by the triangle inequality, CP >= CD + DP = BP + AP, QED.
Dan Hoey haoyuep@aol.com

