Orthocenter, Centroid and Circumcenter Coincide iff Triangle is Equilateral

Theorem

Let $\triangle ABC$ be a triangle.

Let $O$ be the circumcenter of $\triangle ABC$.

Let $G$ be the centroid of $\triangle ABC$.

Let $H$ be the orthocenter of $\triangle ABC$.

Then $O$, $G$ and $H$ are the same points if and only if $\triangle ABC$ is equilateral.

If $\triangle ABC$ is not equilateral, then $O$, $G$ and $H$ are all distinct.

Proof

Necessary Condition

Let $\triangle ABC$ be an equilateral triangle.

By definition, each side of $\triangle ABC$ is the base of an isosceles triangle.

Let $AE$, $BF$ and $CD$ be the altitudes of $\triangle ABC$ through $A$, $B$ and $C$ respectively.

From Altitudes of Triangle Meet at Point, let them intersect at $G$.

From Altitude, Median and Perpendicular Bisector Coincide iff Triangle is Isosceles:

$AE$, $BF$ and $CD$ are the medians of $\triangle ABC$

and:

$AE$, $BF$ and $CD$ are the perpendicular bisectors of $\triangle ABC$.

They all meet at a single point, that is $G$.

Hence by definition, the circumcenter, centroid and orthocenter of $\triangle ABC$ are the same point.

$\Box$

Converse Condition

Let $\triangle ABC$ not be an equilateral triangle.

Thus at least one pair of sides of $\triangle ABC$ is not equal.

Without loss of generality, suppose $AC \ne BC$.

Then $AB$ is not the base of an isosceles triangle.

Thus from Altitude, Median and Perpendicular Bisector Coincide iff Triangle is Isosceles, the altitude from $AB$, the midpoint of $AB$ and the perpendicular bisector of $AB$ are all different.

So the circumcenter, centroid and orthocenter of $\triangle ABC$ likewise cannot be the same point.

$\blacksquare$