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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?

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

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    $\begingroup$ Oh, I must have been confusing the terms. What do you call the intersection of the perpendicular bisectors? Is that the incenter? $\endgroup$ – Riley Mar 31 '18 at 2:11
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    $\begingroup$ @Riley It is circumcentre $\endgroup$ – CY Aries Mar 31 '18 at 2:13
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    $\begingroup$ 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? $\endgroup$ – Riley Mar 31 '18 at 2:15
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    $\begingroup$ @Riley: The orthocenter is the intersection of the altitudes. $\endgroup$ – hmakholm left over Monica Mar 31 '18 at 2:16
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    $\begingroup$ @HenningMakholm Yes, I got that now. I feel really stupid now :) $\endgroup$ – Riley Mar 31 '18 at 2:17
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In your example, the circumcentre is $(1,0)$, the centroid is $\left(\frac13\left(2-\sqrt3\right),0\right)$ and the orthocenter is $\left(-\sqrt3,0\right)$.

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