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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


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.


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.  

     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
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Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry

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