Baltic Way 2023 Shortlist

The answer is No. Let $△ABC$ be a triangle with orthocenter $H$, circumcenter $O$ and incenter $I$. Assume $I$ lies on $OH$. We'll show that $△ABC$ must be isococles, contradicting the scalene property of the statement.

First a well-known claim: Claim. $O$ and $H$ are isogonal conjugates in $△ABC$. Proof. Let $AH$ intersect $BC$ in $D$. Then $$\angle BAH=\angle BAD=90^{\circ }-\angle CBA=90^{\circ }-\frac{1}{2}\angle AOC=\angle OAC,$$ so $AH$ and $AO$ are isogonal in $\angle BAC$. Similarly, $BH$ and $AO$ are isogonal in $\angle CBA$, so the conclusion follows. □

Note that at most one of the vertices can lie on the Euler line because the centroid lies in the interior of $△ABC$. Therefore we may assume that $OH$ passes through neither $B$ or $C$.

Now from the claim, $BI$ bisects $\angle HBO$ and $CI$ bisects $\angle HCO$. And since $I$ lies on $OH$, we get from the Angle Bisector Theorem, that $$\frac{|BH|}{|BO|}=\frac{|IH|}{|IO|}=\frac{|CH|}{|CO|}$$ so $|BH|=|CH|$ since $|BO|=|CO|$. Now this implies that $OH$ is the perpendicular bisector of $BC$. In particular, $OH\perp BC$ so $A$ lies on $OH$. That is, $A$ lies on the perpendicular bisector of $BC$ so $|AB|=|AC|$.