Problem 23. The base of the quadrilateral pyramid $M A B C D$ is the parallelogram $A B C D, \overline{D K}=$ $=\overline{K M}, \overline{B P}=0.25 \overline{B M}$; point $X$ is the intersection of line $M C$ and plane $A K P$. Find the ratio $M X: X C$.
Solution. We choose the following vectors as the basis vectors

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-133.jpg?height=408&width=442&top_left_y=2206&top_left_x=453)

Fig. 11

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-133.jpg?height=314&width=517&top_left_y=2290&top_left_x=1112)
\(\overline{A B}, \overline{A D}, \overline{A M}\). Since point \(X\) lies in the plane \(A K P\), we have

\[
\begin{gathered}
\overline{A X}=m \overline{A K}+n \overline{A P}=m(0.5 \overline{A D}+0.5 \overline{A M})+ \\
+n(0.25 \overline{A M}+0.75 \overline{A B}=0.75 n \overline{A B}+0.5 m \overline{A D}+ \\
+(0.5 m+0.25 n) \overline{A M}
\end{gathered}
\]

Let \(\overline{M X}=k \overline{M C}\). Then

\[
\begin{aligned}
& \overline{A X}=(1-k) \overline{A M}+k \overline{A C}= \\
& =(k-1) \overline{A M}+k \overline{A D}+k \overline{A B}
\end{aligned}
\]

From equations (1) and (2), we have:

\[
0.75 n=k ; 0.5 m=k ; 0.5 m+0.75 n=1-k
\]

. Solving this system of equations, we get \(k=3: 7\), i.e., \(M X: X C=3: 4\).