$\left.\begin{array}{ll}{[\text { Intersecting lines, angle between them }]} \\ {[\quad \text { Prism (other) }}\end{array}\right]$

On the lateral edges $A A 1, B B 1$, and $C C 1$ of the triangular prism $A B C A 1 B 1 C 1$, points $M, N$, and $P$ are located such that $A M: A A 1 = B 1 N: B B 1 = C 1 P: C C 1 = 3: 4$. On the segments $C M$ and $A 1 N$, points $E$ and $F$ are located such that $E F \| B 1 P$. Find the ratio $E F: B 1 P$.
Note that one can draw no more than one line parallel to the line $B 1 P$ and intersecting the skew lines $C M$ and $A 1 N$. Indeed, if points $E 1$ and $F 1$, distinct from points $E$ and $F$, lie on the lines $C M$ and $A 1 N$ respectively, such that $E 1 F 1 \| B 1 P$, then the lines $E F$ and $E 1 F 1$ are parallel, which means they lie in the same plane. Then the lines $C M$ and $A 1 N$ also lie in this plane, which is impossible since they are skew. Let $E$ and $F$ be points on the segments $C M$ and $A 1 N$ respectively, such that $E F \| B 1 P$. Let the plane passing through the parallel lines $E F$ and $B 1 P$ intersect the line $A A 1$ at point $K$. Denote $A 1 K=x, A A 1=t$. The triangle $K M E$ is similar to the triangle $P C E$, the triangle $A 1 K F$ is similar to the triangle $N B 1 F$, and the triangle $E K F$ is similar to the triangle $P K B 1$, therefore

$$
\frac{K E}{P^{\prime} E^{\prime}}=\frac{M K}{P C C}=\frac{x-\frac{1}{P}}{\frac{1}{1} t}=\frac{4 x-1}{l}, \frac{K F^{\prime}}{F^{\prime} B_{1}}=\frac{A_{1} K}{B_{1} N}=\frac{x}{\lambda_{1}}=\frac{4 k}{3 t}, \frac{K E}{P^{\prime} E^{\prime}}=\frac{K F^{\prime}}{F^{\prime} B_{1}},
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_786127613a7b2f99b094g-14.jpg?height=126&width=662&top_left_y=1772&top_left_x=698)

## Answer

$\frac{1}{3}$