Deriving Trilinear Coordinates
Date: 05/18/99 at 14:59:16 From: Alejandra Smith Subject: Trilinear coordinates How do you derive the trilinear coordinates of, for example, the orthocenter of a triangle? (cos B, cos C)
Date: 05/18/99 at 15:44:33 From: Doctor Floor Subject: Re: Trilinear coordinates Hi, Alejandra, Thanks for your question! Trilinear coordinates are coordinates w.r.t. a triangle: Let P be a point in the plane of a triangle ABC. Drop perpendicular altitudes from P to the triangle sides, to find A', B', and C': C / \ / \ B' A' / P \ / \ A-----C'----B Trilinear coordinates (trilinears) are now the distances (PA', PB', PC'). In fact, the coordinates are not necessarily the distances themselves, but the ratios of the distances. These are called "homogeneous," which means that if you take coordinates (x,y,z) and (p*x,p*y,p*z) for some real number p, then these two sets of coordinates represent the same point. You can read a short introduction on trilinear coordinates on University of Evansville Professor Clark Kimberling's webpages, from which you can find a lot of triangle centers listed: http://cedar.evansville.edu/~ck6/tcenters/trilin.html To find trilinears when P is the orthocenter, note that: angle BAP = angle BAA' = 90 degrees - angle B (angle BA'A is right) angle CAP = 90 degrees - angle C So: cos(B) = sin(90-B) = sin(BAP) = PC'/AP cos(C) = sin(90-C) = sin(CAP) = PB'/AP We can conclude that: 1 1 PB':PC' = cos(C) : cos(B) = ------ : ------ cos(B) cos(C) In the same way we will find that: 1 1 PA':PB' = cos(B) : cos(A) = ------ : ------ cos(A) cos(B) This yields trilinear coordinates that are slightly different than the ones you suggested: 1 1 1 ( ------ , ------ , ------ ), or (sec(A), sec(B), sec(C)). cos(A) cos(B) cos(C) As an alternative for trilinear coordinates, you can read in the Dr. Math archives on the related topic of "Barycentric calculus": http://mathforum.org/dr.math/problems/noble1.6.99.html I hope this helps! Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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