Advanced Vectors - Part 4

Place $\triangle ABC$ on a coordinate system so that $A(0, 0)$ , $B(c, 0)$ , and $C(t, u)$ .

As the centroid, $G$ has an $x$ -coordinate of $x = \frac{1}{3}(0 + c + t) = \frac{1}{3}(c + t)$ and a $y$ -coordinate of $y = \frac{1}{3}(0 + 0 + u) = \frac{1}{3}u$ .

Then

$\overrightarrow{GA} = (0 - \frac{1}{3}(c + t), 0 - \frac{1}{3}u) = -\frac{1}{3}(c + t, u)$

$\overrightarrow{GB} = (c - \frac{1}{3}(c + t), 0 - \frac{1}{3}u) = \frac{1}{3}(2c - t, -u)$

$\overrightarrow{GC} = (t - \frac{1}{3}(c + t), u - \frac{1}{3}u) = \frac{1}{3}(2t - c, 2u)$ .

Substituting these values into $a\overrightarrow{GA} + b\overrightarrow{GB} + c\overrightarrow{GC} = 0$ gives a $y$ -component equation of $-\frac{1}{3}au - \frac{1}{3}bu + \frac{1}{3}cu = 0$ , which simplifies to

$a + b - 2c = 0$

Substituting also gives an $x$ -component equation of $-\frac{1}{3}a(c + t) + \frac{1}{3}b(2c - t) + \frac{1}{3}c(2t - c) = 0$ , which simplifies to $t(a + b - 2c) + c(a + c - 2b) = 0$ , and since $a + b - 2c = 0$ , it further simplifies to

$a + c - 2b = 0$

The two equations $a + b - 2c = 0$ and $a + c - bc = 0$ solve to $a = b = c$ , which represents an equilateral triangle. Therefore, $\angle A = \boxed{60°}$