# Problem 6. (3 points)

Given a quadrilateral pyramid $O A B C D$, at the base of which lies a parallelogram $A B C D$. Plane $\alpha$ intersects the edges $O A, O B, O C$, and $O D$ of the pyramid at points $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$, respectively. It is known that $\frac{O A^{\prime}}{O A}=\frac{1}{a}, \frac{O B^{\prime}}{O B}=\frac{1}{b}, \frac{O C^{\prime}}{O C}=\frac{1}{c}$. Find $\frac{V_{O A B C D}}{V_{O A^{\prime} B^{\prime} C^{\prime} D^{\prime}}}$.
Answer: $\frac{2 a b c(a-b+c)}{a+c}$

## Solution:

Solution:
$A B C D$ is a parallelogram, so $\overrightarrow{C D}=\overrightarrow{B A}=\overrightarrow{O A}-\overrightarrow{O B}$, and therefore, $\overrightarrow{O D}=\overrightarrow{O C}+\overrightarrow{O A}-\overrightarrow{O B}$.

If point $X$ lies in the plane $A^{\prime} B^{\prime} C^{\prime}$, and $\overrightarrow{O X}=x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C}$, the coefficients $x, y$, and $z$ satisfy the equation $a x+b y+c z=1$ (this is the equation of a plane, even if the coordinate system is not Cartesian, and points $A^{\prime}, B^{\prime}$, and $C^{\prime}$ clearly satisfy this equation).

$\overrightarrow{O D^{\prime}}=t \overrightarrow{O D}=t \overrightarrow{O C}+t \overrightarrow{O A}-t \overrightarrow{O B}$, so $a t-b t+c t=1$, from which $t=\frac{1}{a-b+c}$.

Since triangles $A B C$ and $A C D$ are equal, $V_{O A B C}=V_{O A C D}=\frac{1}{2} V_{O A B C D}$.

The volume of a pyramid formed by three vectors is one-sixth of the volume of the parallelogram spanned by these vectors, and this volume, in turn, (with the same direction) is proportional to the lengths of the vectors. Therefore, $\frac{V_{O A^{\prime} B^{\prime} C^{\prime}}}{V_{O A B C}}=\frac{1}{a b c}$, and $\frac{V_{O A^{\prime} C^{\prime} D^{\prime}}}{V_{O A C D}}=\frac{t}{a c}=\frac{1}{a c(a-b+c)}$.

Thus, $\frac{V_{O A^{\prime} B^{\prime} C^{\prime} D^{\prime}}}{V_{O A B C D}}=\frac{V_{O A^{\prime} B^{\prime} C^{\prime}}}{V_{O A B C D}}+\frac{V_{O A^{\prime} C^{\prime} D^{\prime}}}{V_{O A B C D}}=\frac{V_{O A^{\prime} B^{\prime} C^{\prime}}}{2 V_{O A B C}}+\frac{V_{O A^{\prime} C^{\prime} D^{\prime}}}{2 V_{O A C D}}=\frac{1}{2 a b c}+\frac{1}{2 a c(a-b+c)}=\frac{a-b+c+b}{2 a b c(a-b+c)}$ $=\frac{a+c}{2 a b c(a-b+c)}$. The reciprocal of this value is the answer to the problem.