Centroid of triangle & Centroid of tetrahedron

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

G₁ = [(x₁ + x₂ + x₃)/ 3, (y₁ + y₂ + y₃)/ 3, (z₁ + z₂ + z₃)/ 3],

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]