# Problem 7. (4 points)

A plane intersects the edges of the tetrahedron $ABCD$ emanating from vertex $C$ and divides them in the ratio $4:1$ (not necessarily from vertex $C$). This plane also intersects the lines $AB$ and $BD$ at points $E$ and $F$. Find the ratio of the areas of triangles $BEF$ and $ABD$.
Answer: $1: 225$ or $256: 225$.

## Solution:

Let the points of intersection of the plane (which we will call $\alpha$) with the edges $C A, C B$, and $C D$ be $A^{\prime}, B^{\prime}$, and $D^{\prime}$, respectively. If $C A^{\prime}: A^{\prime} A = C B^{\prime}: B^{\prime} B$, the lines $A B$ and $A^{\prime} B^{\prime}$ are parallel, which means that the plane $\alpha$ does not intersect the line $A B$. Similar reasoning applies to the line $B D$. Therefore, $C B^{\prime}: B^{\prime} B \neq C A^{\prime}: A^{\prime} A = C D^{\prime}: D^{\prime} D$.

There are two possible cases. In the first case, $C B^{\prime}: B^{\prime} B = A^{\prime} A: A^{\prime} C = 4: 1$. By Menelaus' theorem, we have $\frac{B E}{E A} \cdot \frac{A A^{\prime}}{A^{\prime} C} \cdot \frac{C B^{\prime}}{B^{\prime} B} = 1$, from which $\frac{B E}{E A} = \frac{1}{16}$. In this case, point $E$ lies on the ray $A B$ beyond point $B$, so $\frac{B E}{A B} = \frac{1}{15}$. Similarly, $\frac{B F}{B D} = \frac{1}{15}$, and the ratio of the areas is $1: 225$ (since the angles at vertex $B$ of the triangles in question are equal).

In the second case, $C B^{\prime}: B^{\prime} B = A^{\prime} A: A^{\prime} C = 1: 4$. By Menelaus' theorem, we have $\frac{B E}{E A} \cdot \frac{A A^{\prime}}{A^{\prime} C} \cdot \frac{C B^{\prime}}{B^{\prime} B} = 1$, from which $\frac{B E}{E A} = 16$. In this case, point $E$ lies on the ray $B A$ beyond point $A$, so $\frac{B E}{A B} = \frac{16}{15}$. Similarly, $\frac{B F}{B D} = \frac{16}{15}$, and the ratio of the areas is $256: 225$ (since the angles at vertex $B$ of the triangles in question are equal).