Problem 924

We call $A_{1}, B_{1}$, and $C_{1}$ the feet of the medians from $A, B$, and respectively $C$. We call $h$ the homothety with center $G$ and ratio $-\frac{1}{2}$. We call $O_{-1}$ the point $h^{-1}(O)$, as shown in Figure 1, and we see that it suffices to show $\mathrm{O}_{-1}=\mathrm{H}$. One of the properties of homotheties shows that

$$h(A) h\left(O_{-1}\right) \| A O_{-1}$$

However, it is known that $G$ is located one-third of the way from $A_{1}$ and two-thirds of the way from $A$, so $A_{1}=h(A)$. Thus, we have:

$$\mathrm{A}_{1} \mathrm{O} \| \mathrm{AO}_{-1}$$

Since $A_{1} O$ is the perpendicular bisector of the segment $B C, A_{1} 0 \perp B C$ and, consequently, $A_{-1} \perp B C$. Similarly, $\mathrm{BO}_{-1} \perp A C$, so $\mathrm{O}_{-1}$ lies on two altitudes of the triangle. Thus $\mathrm{O}_{-1}=\mathrm{H}$.