While the article The Centroid of A Triangle explains how to find the centroid of any triangle, this article elaborates on how to determine the centroids of some special-typed triangles, that is isosceles, equilateral, and right-angled triangles.

 

The Centroid of An Isosceles Triangle

Consider the triangle ABC below. In this triangle, AC = BC. Let D and E be the midpoints of the line segments BC and AB, respectively.

Figure 1

Since ΔABC is isosceles with AC = BC, the median CE is an altitude of ΔABC as well. So, CE ⊥ AB. Let AC = b, BC = a and AB = c. Then, by the Pythagorean formula, it can be shown that the altitude CE has the length of tC satisfying:

In the article The Centroid of A Triangle, it has been proven that CT : TE = 2 : 1. As a consequence, TE = \frac{1}{3} t_C = \frac{1}{3} \cdot \frac{1}{2} \sqrt{4a^2-c^2}, and it follows that

If we define xC = AE and and yC = TE then: (See Figure 2 below)

If T’ is the foot of the perpendicular from T to the side BC, what are CT’ and TT’? (See Figure 2)

Figure 2

Assuming ∠T’CT = θ, we have CT’ = CT cos θ. In ΔCEB, \cos \theta = \frac{t_c}{a} whereas CT = \frac{2}{3} t_c. Hence CT \prime = \frac{2}{3a} {t_C}^2. By (*), it can be concluded that CT \prime = \frac{4a^2 - c^2}{6a}.

 

In the article The Centroid of A Triangle, it has been proven that the distance between the centroid of a triangle and the base of the triangle is one third of the length of the altitude drawn from the vertex opposite to the base. Therefore, TT \prime = \frac{1}{3} t_A where tA is the length of the altitude drawn from A to the side BC.

From the ratios tA : tC = AB : BC = c : a and the fact that t_C = \frac{1}{2} \sqrt{4a^2 - c^2}, as given by (*), it follows that t_A = \frac{c}{2a} \sqrt{4a^2 - c^2}. In addition, TT' = \frac{c}{6a} \sqrt{4a^2 - c^2}. If we let CT’ = xA and TT’ = yA, it can be concluded that:

The projection of T on the side AC side gives similar results. Let T” be the foot of the perpendicular from T to the side AC. Since ΔABC is symmetrical about the line segment CE, it holds that CT” = CT’ and TT” = TT’. If we let CT” = xB and TT” = yB, we have:

As a = b in the isosceles triangle ABC, the formulae xB and yB can be easily obtained from xA and yA, i.e. by replacing a with b, and vice versa.

 

The Centroid of An Equilateral Triangle

Equilateral triangles are also isosceles triangles. Consequently, all the formulae derived above apply to equilateral triangles as well. Suppose that ABC is an equilateral triangle. Therefore, a = b = c, where a = BC, b = AC, and c = AB. (See Figure 3 below.) Replace c with a in the xA and yA formulae above. This results in:

x_A = \frac{4a^2 - c^2}{6a} = \frac{4a^2 - a^2}{6a} = \frac{3a^2}{6a} = \frac{1}{2} a

y_A = \frac{c}{6a} \sqrt{4a^2 - c^2} = \frac{a}{6a} \sqrt{4a^2 - a^2} = \frac{a}{6} \sqrt{3}

Figure 3

Since a = b = c, x_A = x_B = x_C = \frac{a}{2} and y_A = y_B = y_C = \frac{a}{6} \sqrt{3}.

 

The Centroid of A Right-Angled Triangle

Consider a right-angled triangle ABC with AC ⊥ AB. Let AB = c, AC = b, and BC = a. Place the triangle on the Cartesian plane so that AB coincides with the x-axis, AC coincides with the y-axis. (See Figure 4 below.)

Figure 4

It follows that A(0,0), B(c,0), and C(0,b). Referring to the article The Centroid of A Triangle, the coordinates of the centroid of the triangle, that is T, can be determined by the formula:

Let \vec{A} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \vec{B} = \begin{pmatrix} c \\ 0 \end{pmatrix}, and \vec{C} = \begin{pmatrix} 0 \\ b \end{pmatrix}, it can be concluded that  T(\frac{c}{3},\frac{b}{3}).

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