9. Inside the tetrahedron $ABCD$ there is a point $O$, such that the lines $AO, BO, CO, DO$ intersect the faces $BCD, ACD, ABD, ABC$ at points $A_1, B_1, C_1, D_1$ respectively, and $\frac{AO}{A_1O}=\frac{BO}{B_1O}=\frac{CO}{C_1O}=\frac{DO}{D_1O}=k$. Find all possible values of $k$.
(1968 Bulgarian Competition Problem)
9. Let $V$ be the volume of tetrahedron $ABCD$, then we have $\frac{V}{V_{OBCD}}=\frac{AA_{1}}{OA_{1}}=\frac{AO}{A_{1}O}+\frac{OA_{1}}{OA_{1}}=k+1$, similarly $\frac{V}{V_{OACD}}=\frac{V}{V_{OABD}}=\frac{V}{V_{OABC}}=$ $k+1$. Therefore, $k+1=\frac{4V}{V_{OBCD}+V_{OACD}+V_{OMBD}+V_{OABC}}=\frac{4V}{V}=4$, solving for $k$ gives $k=3$.