Centroid of triangle: The
coordinates of the centroid of the triangle with vertices A (x1, y1,
z1) B (x2, y2, z2) and C (x3,
y3, z3)
[(x₁
+ x₂ + x₃)/ 3, (y₁ + y₂ + y₃)/ 3, (z₁ + z₂ + z₃)/ 3]
Proof: Let D be the mid-point of AC then coordinates
of D are
[(x₂
+ x₃)/ 2, (y₂ + y₃)/ 2, (z₂ + z₃)/ 2]
Let
G be the centroid of ΔABC
Then,
G divides AD in the ratio 2:1
So,
coordinate of D are
[{1.x₁
+ 2(x₂ + x₃/ 2)}/ 1 + 2, {1.y₁ + 2(y₂ + y₃/ 2)}/ 1 + 2, {1.z₁ + 2(z₂ + z₃/ 2)}/
1 + 2]
i.e,
[(x₁ + x₂ + x₃)/ 3, (y₁ + y₂ + y₃)/ 3, (z₁ + z₂ + z₃)/ 3]
Centroid of tetrahedron: Let
ABCD be a tetrahedron such that the coordinates of its vertices are A (x1,
y1, z1) B (x2, y2, z2) C
(x3, y3, z3) and D (x4, y4,
z4) the coordinates of the centroid of faces ABC, DAB, DBC and DCA
are respectively
G₂
= [(x₁ + x₂ + x₄)/ 3, (y₁ + y₂ + y₄)/ 3, (z₁ + z₂ + z₄)/ 3],
G₃
= [(x₂ + x₃ + x₄)/ 3, (y₂ + y₃ + y₄)/ 3, (z₂ + z₃ + z₄)/ 3],
G₄
= [(x₄ + x₃ + x₁)/ 3, (y₄ + y₃ + y₁)/ 3, (z₄ + z₃ + z₁)/ 3].
Now,
coordinates of point G dividing DG, in the ratio 3:1 are
[{1.x₄
+ 3(x₁ + x₂ + x₃/ 3)}/ 1 + 3, {1.y₄ + 3(y₁ + y₂ + y₃/ 3)}/ 1 + 3, {1.z₄ + 3(z₁
+ z₂ + z₃/ 3)}/ 1 + 3]
=
[(x₁ + x₂ + x₃ + x₄)/ 4, (y₁ + y₂ + y₃ + y₄)/ 4, (z₁ + z₂ + z₃ + z₄)/ 4]
Similarly
the point dividing CG1, CG2, AG3 and BG4
in the ratio 3:1 has the same coordinates.
Hence,
the point G [(x₁ + x₂ + x₃ + x₄)/ 4, (y₁ + y₂ + y₃ + y₄)/ 4, (z₁ + z₂ + z₃ +
z₄)/ 4]
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