Lol I really think this problem is funny. Give it a try!

I guess the triangle doesn't exist but I don't have a proof.

LoloChen wrote: I guess the triangle doesn't exist but I don't have a proof. Is this problem a counterexample to LoloChen's theorem?

No

If $\triangle ABC$ is acute, assume $\beta > \alpha > \gamma$ and let $E = BH \cap CO$, and $F= CH \cap BO$. We'll show $\angle OIH$ is obtuse, which means no similarity. $\angle HEO = \pi - \beta > \frac{\pi}{2}$ and $\angle HFO = \pi - \gamma > \frac{\pi}{2}$. Since by isogonality of $H$ and $O$ the point $I$ has to lie strictly inside the convex quadrilateral $HEOF$, the angle $\angle OIH$ has to be obtuse. For the right angle case a similar argument works, suppose $\alpha = \frac{\pi}{2}$, let $D$ be the foot of $A$ on $BC$, then $I$ has to lie strictly inside $\triangle AOD = \triangle HOD$, which is right at $D$, so $\angle OIH$ is again obtuse.

If $\triangle ABC$ is obtuse, assume $\alpha > \frac{\pi}{2} > \beta >\gamma$, and let $E = AH \cap BO$. Another easy angle chase shows $\angle AEO = \alpha$, and since $I$ is strictly contained inside $\triangle AEO$ (again by isogonality) we have $\angle HIO > \angle HAO = \alpha$, which means it's larger than all the angles of the triangle, so no similarity. A consequence of this should be the fact that $\angle OIH$ is always strictly larger than all the angles of $\triangle ABC$.

MatteD wrote: No

If $\triangle ABC$ is acute, assume $\beta > \alpha > \gamma$ and let $E = BH \cap CO$, and $F= CH \cap BO$. We'll show $\angle OIH$ is obtuse, which means no similarity. $\angle HEO = \pi - \beta > \frac{\pi}{2}$ and $\angle HFO = \pi - \gamma > \frac{\pi}{2}$. Since by isogonality of $H$ and $O$ the point $I$ has to lie strictly inside the convex quadrilateral $HEOF$, the angle $\angle OIH$ has to be obtuse. For the right angle case a similar argument works, suppose $\alpha = \frac{\pi}{2}$, let $D$ be the foot of $A$ on $BC$, then $I$ has to lie strictly inside $\triangle AOD = \triangle HOD$, which is right at $D$, so $\angle OIH$ is again obtuse.

If $\triangle ABC$ is obtuse, assume $\alpha > \frac{\pi}{2} > \beta >\gamma$, and let $E = AH \cap BO$. Another easy angle chase shows $\angle AEO = \alpha$, and since $I$ is strictly contained inside $\triangle AEO$ (again by isogonality) we have $\angle HIO > \angle HAO = \alpha$, which means it's larger than all the angles of the triangle, so no similarity. A consequence of this should be the fact that $\angle OIH$ is always strictly larger than all the angles of $\triangle ABC$. The diagrams seem broken, but yes the real inequality is that $\angle OIH$ is always larger than the three angles.

This problem from Morocco TST is somewhat related to this solution. In fact, it is the part 1 of the solution.

There is a similar question that asks whether there exists and triangle whose incenter, orthocenter and centroid form a triangle that is similar to the original triangle about 5 years ago. Surprisingly the answer is TRUE. You can see the details in the link (in simplified Chinese).

The answer is no. We prove that $\angle HIO$ is larger than the largest angle of any scalene $\Delta ABC$. If $\Delta ABC$ is acute, then consider the six regions determined by lines $AI,BI,CI$. As $H$ and $O$ are isogonal conjugates and lie inside the triangle, they must be in opposite regions. It follows that $\angle HIO > 90^\circ$ (as $\angle AIB, \angle BIC, \angle CIA$ are all obtuse). If WLoG $\angle A \geq 90^\circ$, WLoG assume $AB < AC$: note that $I$ lies on the opposite side of the $A$-altitude from $B$, as well as the opposite side of the perpendicular bisector of $AB$ from $B$, as shown. Hence $\angle HIO > 180^\circ-\angle B > \angle A$ and we are done.