Using complex numbers to show that a triangle whose circumcenter and centroid coincide must be equilateral

Question

Given that the circumcenter and the centroid lie on the same point inside $\triangle ABC$ we have to show that the triangle is an equilateral triangle.

I am able to do this using congruence of the sub triangles but I want a proof using complex numbers . So how can we use complex numbers to derive the centroid and the circumcentre and then prove that the triangle is equilateral. Thank you!

Answer 1

Let $z_1$, $z_2$, and $z_3$ be the vertices of the triangle in the complex plane.

The centroid of the triangle, $z_g$, is given by: $ z_g = \frac{z_1 + z_2 + z_3}{3}$

Let $z_c$ be the circumcenter of the triangle, such that: $|z_c - z_1| = |z_c - z_2| = |z_c - z_3|$

Since it is given that $z_c = z_g$, we can substitute this into the equation to get: $|z_g - z_1| = |z_g - z_2| = |z_g - z_3|$

Using this result, we can prove that the three medians are equal in length. (The proof involves using the property that the centroid divides the median in the ratio 2:1, but it has been omitted for brevity.)

From this point, we can easily prove that the three sides are equal using another theorem which states that twice the length of the median is equal to the sum of the lengths of the two sides which the median does not bisect (the containing sides).

Hence the triangle is equilateral.

Answer 2

Taking the circumradius as the length unit, you can assume WLOG that the triangle is inscribed into the unit circle.

You can furthermore assume that the vertices are

$$1,e^{ib},e^{ic}\tag{1}$$

The circumcenter is therefore the center of the unit circle, i.e., origin point $0$.

Constraining the centroid $\frac13(1+e^{ib}+e^{ic})$ to coincide with the circumcenter is equivalent to write :

$$1+e^{ib}+e^{ic}=0 \tag{2}$$

Intuitively, (2) is possible iff points (1) constitute an equilateral triangle, but we need to prove it. Here is a way. Taking real and imaginary parts in (2), we get :

$$\begin{cases}\cos b +\cos c &=&-1\\ \sin c &=&-\sin b \end{cases} \tag{3}$$

Squaring the second equation in (3), we deduce that

$$\cos b =\pm \cos c. \tag{4}$$

Plugging (4) into the first equation of (3), we see that the single possibility is with the $+$ sign in (4), giving :

$$2 \cos b=-1 \iff \cos b= \cos \frac{2 \pi}{3} \iff \text{either} \ b=\frac{2 \pi}{3} \ \& \ c=\frac{4 \pi}{3}$$

or a similar result by exchange of $b$ and $c$.