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

 Messages: [ Previous | Next ]
 Dan Hoey Posts: 172 Registered: 12/6/04
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 side-angle-side 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

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