Triangle Centroid in 3-Space

Date: 12/30/96 at 17:31:44
From: Nathan D Chute
Subject: Centroid of a triangle

Lets say we have a triangle in 3D given by:

P1 = (x1,y1,z1) P2 = (x2,y2,z2) P3 = (x3,y3,z3)

The centroid of a triangle is the intersection of the medians. I can 
find the medians since they are half of the distance of each 
leg. How do I find the intersections to come up with the centroid?

Date: 12/30/96 at 21:05:06
From: Doctor Pete
Subject: Re: centroid of a triangle

Hi,

I claim that the centroid of a triangle with vertices {P1,P2,P3} in 
3-space is:

   C = (P1+P2+P3)/3 = ((x1+x2+x3)/3,(y1+y2+y3)/3,(z1+z2+z3)/3)

The goal, then, is to prove this using the fact that the centroid is 
the intersection point of the medians of the triangle.

Proof:

Note that the median from P1 bisects the line P2P3.  The midpoint M1 
of the line P2P3 is:

     M1 = (P2+P3)/2 = ((x2+x3)/2,(y2+y3)/2,(z2+z3)/2)

Now, we may either use the fact that the intersection of the medians 
occurs at 1/3 the distance from M1 to P1 (from which the result 
follows immediately), or if we do not wish to use this, we can find 
an expression for a 1-parameterization of the line that passes through 
P1 and M1.  In the latter case, let t be the parameter, where t=0 
gives P1, and t=1 gives M1.  Then we wish to find functions:

     (x,y,z) = (X1[t],Y1[t],Z1[t])

such that P1 = (X1[0],Y1[0],Z1[0]), and M1 = (X1[1],Y1[1],Z1[1]). 
 
Consider the x-coordinate; we see that X1[t], for some A, has the form 
A*t + x1.  Then X1[1] = A+x1, so A = (x2+x3)/2 - x1 = (-2*x1+x2+x3)/2.  
Hence:

     X1[t] = (-2*x1+x2+x3)t/2 + x1
     Y1[t] = (-2*y1+y2+y3)t/2 + y1
     Z1[t] = (-2*z1+z2+z3)t/2 + z1

This describes a line in 3-space which passes through P1 at t=0, and 
through M1 at t=1.  Similarly, by symmetry, we write down the 
parameterization for another median, say P2M2:

     X2[s] = (x1-2*x2+x3)s/2 + x2
     Y2[s] = (y1-2*y2+y3)s/2 + y2
     Z2[s] = (z1-2*z2+z3)s/2 + z2

Then the intersection of these medians occurs at some value of t,s 
where X1[t] = X2[s], Y1[t] = Y2[s], and Z1[t] = Z2[s].  Solving in the 
first coordinate:

     (-2*x1+x2+x3)t/2 + x1 = (x1-2*x2+x3)s/2 + x2
     (-2*x1+x2+x3)t + 2*x1 = (x1-2*x2+x3)s + 2*x2
     (2-2*t-s)*x1 + (t+2*s-2)*x2 + (t-s)*x3 = 0

and if this last equation is true for all values of x1, x2, x3, it 
follows that:

     2-2*t-s = 0 and t-s = 0

(why did I omit the condition for x2?) whereupon we have t=s, 
2-3*t = 0, or t = s = 2/3.  Hence the point of intersection occurs 
along the line P1M1 at t = 2/3, or:

     x = X1[2/3] = (-2*x1+x2+x3)(2/3)/2 + x1 = (x1+x2+x3)/3
     y = Y1[2/3] = (-2*y1+y2+y3)(2/3)/2 + y1 = (y1+y2+y3)/3
     z = Z1[2/3] = (-2*z1+z2+z3)(2/3)/2 + z1 = (z1+z2+z3)/3

and we are done.

I hope that made sense -- I admit that my proof involved using some 
strange tools (e.g., using parameterized lines in 3-space), but I 
think it was necessary since we didn't want to assume certain nice 
properties of medians.  Certainly the entire proof can be done in the 
plane, where things are much easier.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/