, 2023

Since $ABC$ is a scalene triangle, we may assume without loss of generality that $\angle A > \angle B > \angle C$. In fact, we can prove that $\angle OIH > \angle A$.

Claim. $A, H$ lie on the same side of $OI$.

Proof. Let $M$ be the second intersection point of $AI$ with the circumcircle $\odot(ABC)$, let $J$ be the $A$-excenter of $\triangle ABC$, and let $S$ be the intersection point of $OI$ and $AH$. The Incenter/Excenter Lemma (Fact 5) tells us that $M$ is the midpoint of $IJ$. Considering the projections $F_I, F_M, F_J$ of $I, M, I_a$ onto $AB$, we have
$$\frac{MI}{IA} = \frac{F_M F_I}{F_I A} = \frac{1}{2} \cdot \frac{F_J F_I}{F_I A} = \frac{1}{2} \cdot \frac{a}{(b+c-a)/2} = \frac{a}{b+c-a},$$
where $a = BC, b = CA, c = AB$ are the three side lengths of triangle $ABC$. Therefore
$$\overrightarrow{AH} = 2 \cos \angle BAC \cdot \overrightarrow{OM} = \frac{b^2 + c^2 - a^2}{bc} \cdot \left( \frac{MI}{IA} \cdot \overrightarrow{AS} \right) = \frac{a(b^2 + c^2 - a^2)}{bc(b+c-a)} \cdot \overrightarrow{AS}.$$
Therefore, $A, H$ lie on the same side of $OI$ if and only if
$$\begin{aligned} 1 > \frac{a(b^2 + c^2 - a^2)}{bc(b+c-a)} & \iff bc(b+c-a) > a(b^2 + c^2 - a) \\ & \iff (a-b)(a-c)(a+b+c) > 0, \end{aligned}$$
and this is clearly true (since $a > b > c$). $\square$

Returning to the original problem. Since $O, M$ lie on the same side of $AH$, $O, I$ also lie on the same side of $AH$, therefore
$$\angle OIH = \angle OSH + \angle SHI > \angle OSH = \angle IOM. \qquad (\spadesuit)$$
Combining the fact that $I, O$ lie on the same side of $BM$ together with
$$\angle MBO = 90^\circ - \frac{1}{2}\angle A = \frac{1}{2}(\angle B + \angle C) < \frac{1}{2}(\angle A + \angle B) = \angle BMI,$$
we obtain that $I, M$ lie on opposite sides of $BO$. Combining this further with the fact that $B, I$ lie on the same side of $MO$ and with $(\spadesuit)$, we obtain
$$\angle OIH > \angle IOM > \angle BOM = \angle A.$$
Therefore, there does not exist a scalene triangle $ABC$ such that triangle $ABC$ is similar to triangle $IOH$. $\square$