Triangle Centroid in 3-SpaceDate: 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/ |
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