# Problem 7. (4 points)

A plane intersects the edges of the tetrahedron $ABCD$ emanating from vertex $A$ and divides them in the ratio $3:1$ (not necessarily from vertex $A$). This plane also intersects the lines $BC$ and $CD$ at points $E$ and $F$. Find the ratio of the areas of triangles $EFC$ and $BCD$.
Answer: $1: 64$ or $81: 64$.

## Solution:

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

There are two possible cases. In the first case, $A C^{\prime}: C^{\prime} C = B^{\prime} B: B^{\prime} A = 1: 3$. By Menelaus' theorem, we have $\frac{B E}{E C} \cdot \frac{C C^{\prime}}{C^{\prime} A} \cdot \frac{A B^{\prime}}{B^{\prime} B} = 1$, from which $\frac{B E}{E C} = \frac{1}{9}$. In this case, point $E$ lies on the ray $C B$ beyond point $B$, so $\frac{B C}{C E} = \frac{8}{9}$. Similarly, $\frac{C D}{C F} = \frac{8}{9}$, and the ratio of the areas is $81: 64$ (since the angles at vertex $C$ of the sought triangles are equal).

In the second case, $A C^{\prime}: C^{\prime} C = B^{\prime} B: B^{\prime} A = 3: 1$. By Menelaus' theorem, we have $\frac{B E}{E C} \cdot \frac{C C^{\prime}}{C^{\prime} A} \cdot \frac{A B^{\prime}}{B^{\prime} B} = 1$, from which $\frac{B E}{E C} = 9$. In this case, point $E$ lies on the ray $B C$ beyond point $C$, so $\frac{B C}{C E} = 8$. Similarly, $\frac{C D}{C F} = 8$, and the ratio of the areas is $1: 64$ (since the angles at vertex $C$ of the sought triangles are equal).