Let the point where line $DC$ intersects side $AB$ of the triangle be denoted by $K$. Applying Menelaus' theorem, we get:

!

$$\frac{OH}{OA} \cdot \frac{KA}{KB} \cdot \frac{CB}{CH} = 1$$

But

$$\frac{OH}{OA} = \frac{AH - OA}{OA}, \quad \frac{KA}{KB} = \frac{AC}{BD} = \frac{AC}{AB}, \quad \frac{CB}{CH} = \frac{CB}{\overline{AC}^2} = \frac{\overline{CB}^2}{\overline{AC}^2}$$

Substituting these values into (1):

$$\frac{AH - OA}{OA} \cdot \frac{AC}{AB} \cdot \frac{\overline{CB}^2}{AC^2} = 1$$

or

$$\frac{AH - OA}{OA} = \frac{AB \cdot AC}{\overline{CB}^2}$$

But

$$\frac{AB \cdot AC}{BC} = AH$$

so from (2):

$$\frac{AH - OA}{OA} = \frac{AH}{BC}$$

Removing the denominators:

$$\begin{gathered} AH \cdot BC - OA \cdot BC = AH \cdot OA \\ (AH + BC) \cdot OA = AH \cdot BC \end{gathered}$$

from which

$$\frac{1}{AO} = \frac{AH + BC}{AH \cdot BC}$$

and finally

$$\frac{1}{AO} = \frac{1}{BC} + \frac{1}{AH}$$

(Kornis Ödön, secondary school, 6th grade, Pécs.)

The problem was also solved by: Friedmann Bernát, Grünhut Béla, Hofbauer Ervin, Riesz Frigyes.