Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

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/   
    
Associated Topics:
High School Coordinate Plane Geometry
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/