# If a triangle is not equilateral, must its orthocenter and circumcenter be distinct?

According to Proof Wiki, if a triangle is not equilateral, then its orthocenter and circumcenter must be distinct. The exact quote is

Let △ABC be a triangle.
Let O be the circumcenter of △ABC.
Let G be the centroid of △ABC.
Let H be the orthocenter of △ABC.

Then O, G and H are the same points if and only if △ABC is equilateral.
If $$\triangle ABC$$ is not equilateral, then $$O, G$$ and $$H$$ are all distinct.

Well, it looks like I've found a counterexample: $$A=(0,0),\quad B=\left(1-\frac{\sqrt 3}{2},\frac12\right),\quad C=\left(1-\frac{\sqrt 3}{2},-\frac12\right)$$

and the orthocenter and circumcenter both coincide at $$(1,0)$$, right?

So is my counterexample valid, or did I screw something up?

## 2 Answers

In your example, the orthocentre is $(-\sqrt{3},0)$ and the circumcentre is $(1,0)$.

• Oh, I must have been confusing the terms. What do you call the intersection of the perpendicular bisectors? Is that the incenter? – Riley Mar 31 '18 at 2:11
• @Riley It is circumcentre – CY Aries Mar 31 '18 at 2:13
• Oh, I see where I was getting confused now. I noticed that the intersection of the perpendicular bisectors coincided with the point equidistant to the three vertices. I was falsely assuming one of them was the orthocenter when they are actually just different properties of the circumcenter? – Riley Mar 31 '18 at 2:15
• @Riley: The orthocenter is the intersection of the altitudes. – hmakholm left over Monica Mar 31 '18 at 2:16
• @HenningMakholm Yes, I got that now. I feel really stupid now :) – Riley Mar 31 '18 at 2:17

In your example, the circumcentre is $$(1,0)$$, the centroid is $$\left(\frac13\left(2-\sqrt3\right),0\right)$$ and the orthocenter is $$\left(-\sqrt3,0\right)$$.