In triangle $ABC$, construct points $C_1$ on side $AB$ and $A_1$ on side $BC$ such that the lines $AA_1$ and $CC_1$ intersect at point $P$ where $AP / PA_1 = 3 / 2$ and $CP / PC_1 = 2 / 1$. In what ratio does point $C_1$ divide side $AB$?
Consider the problem as solved. Draw a line through $P$ parallel to $BC$ and $AB$, and let the intersection points of these parallel lines with the other two sides of the triangle be denoted by $R, Q, S, T$ respectively. According to the theorem of parallel intercepts,

$$
\frac{A R}{R C}=\frac{A Q}{Q B}=\frac{A P}{P A_{1}}=\frac{3}{2} \quad \text { and } \quad \frac{C T}{T A}=\frac{C S}{S B}=\frac{C P}{P C_{1}}=\frac{2}{1}
$$

Thus, the points $R$ and $Q$ divide the sides $AC$ and $AB$ in the ratio $3:2$, and the points $T$ and $S$ divide the sides $CA$ and $CB$ in the ratio $2:1$. Therefore, the points $R, Q, S, T$ can be constructed, and the intersection of the lines $RQ$ and $ST$ is precisely the sought point $P$. The problem clearly always has a solution.

The ratio $A C_{1}: C_{1} B$ can also be calculated using the theorem of parallel intercepts.

$$
\begin{gathered}
\frac{A Q}{Q B}=\frac{A P}{P A_{1}}=\frac{3}{2}, \text { so } A Q=\frac{3}{2} Q B . \\
\frac{C_{1} Q}{Q B}=\frac{C_{1} P}{P C}=\frac{1}{2}, \text { so } C_{1} Q=\frac{1}{2} Q B, \text { hence: } \\
\frac{A C_{1}}{C_{1} B}=\frac{A Q-Q C_{1}}{C_{1} Q+Q B}=\frac{\frac{3}{2} Q B-\frac{1}{2} Q B}{\frac{1}{2} Q B+Q B}=\frac{2}{3} .
\end{gathered}
$$

Thus, $C_{1}$ divides the side $AB$ in the ratio $2:3$.

László Nyúl (Kecskemét, Katona J. Gymn., II. o. t.)

Remark. The problem can also be solved in the general case. If $\frac{A P}{P A_{1}}=\frac{\alpha}{\beta}$ and $\frac{C P}{P C_{1}}=\frac{\gamma}{\delta}$, then by similar calculations, we can obtain that $\frac{A C_{1}}{C_{1} B}=\frac{\alpha \gamma-\beta \delta}{\gamma \beta+\delta \beta}$.