BAMO

Solution:

Let $O_1$ be the midpoint of the arc $ACB$ and let $R$ be the radius of $k$.

$\triangle AOO_1$ and $\triangle BOO_1$ are equilateral ($\angle ACB = 120^\circ \Rightarrow \angle AOB = 120^\circ$). The segments $AB$ and $OO_1$ intersect each other in their midpoint, $D$. If line $AO$ intersects $k$ in point $P$, then $\angle ABP = \angle ACP = 90^\circ$, i.e. $PB \parallel CH$ and $PC \parallel BH$, and thus $PBHC$ is a parallelogram. From here, $CH = PB$. Since $OD$ is a midsegment in $\triangle ABP$, then $PB = 2OD = OO_1$ and $CH = OO_1 = R$. Again, $OCHO_1$ is a parallelogram; moreover, it is a rhombus for $OC = R$. Thus, $O_1H = R$ and the points $A, B, O, H$ lie on a circle $k_1$ with center $O_1$ and radius $R$.

Look at the quadrilateral $CMHN$: it contains two right angles (at $M$ and $N$), and the angle at $C$ is $120^\circ$, so the angle at $H$ is $60^\circ$: $\angle AHB = 60^\circ$. But $\angle AOB = 120^\circ$ (as above), so $AHBO$ do lie on the same circle. Since $A, O, B$ lie on a circle with center $O_1$ (as above), $H$ is forced to lie on the same circle.

b.

In $k_1$, $\angle BO_1H$ is central, and $\angle BAH$ is inscribed, so that $\angle BO_1H = 2\angle BAH$ and $\angle BHO_1 = 90^\circ - \angle BAH = \angle AHC$. In the rhombus $OCHO_1$ the diagonal $OH$ is the bisector of $\angle CHO_1$. Hence, $\angle AHO = \angle AHC + \angle CHO = \angle BHO_1 + \angle OHO_1 = \angle BHO \Rightarrow OH$ is the angle bisector also of $\angle AHB$. This means that point $I$ lies on $OH$.

Let $OH$ intersect the median $CD$ in point $G$. We will show that $G$ is the centroid of $\triangle ABC$ by showing first that $CG = 2GD$. Indeed, if $E$ and $F$ are the midpoints of $CG$ and $HG$, then $EF$ is a midsegment in $\triangle CHG$ with $EF = CH/2 = OD$ and $EF \parallel OD$ (from (a)). Then $\triangle EFG \cong \triangle DOG$, and $EG = GD$, $CG = 2EG = 2GD$. Thus, $G$ is the centroid of $\triangle ABC$, and the points $I$ and $G$ lie on the line $OH$.

For those who know about the Euler line: points $O, G, H$ lie on the Euler line of $\triangle ABC$ (we even know the ratio $OG : GH = 1 : 2$). Thus, it remains to show that $I$ lies on the line $OH$, or equivalently, that $OH$ is the angle bisector of $\angle AHB$. Recall that $k_1$ was the circle described around $AHBO$ from part (a). Since $OA = OB$, the corresponding arcs $OA$ and $OB$ on $k_1$ are equal, and hence the inscribed angles are equal: $\angle AHO = \angle BHO$.