25) On the triangle $A B C$, a pyramid with vertex $V$ and base $A B C$ is constructed. $P$ is a point on the edge $V A$ such that $B P$ and $C P$ are perpendicular to each other and are altitudes of the triangles $B A V$ and $C A V$, respectively. Knowing that $P$ divides $V A$ into two segments of length $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ and that the altitudes $B P$ and $C P$ are $3 \mathrm{~cm}$ and $4 \mathrm{~cm}$ long, respectively, determine the volume (in $\mathrm{cm}^{3}$) of the pyramid.
$\begin{array}{llll}\text { (B) } 6 & \text { (C) } 9 & \text { (D) } 12\end{array}$
(E) no such pyramid exists.
25) The answer is (B). Since the segment $C P$ is orthogonal to both the side $V A$ and the segment $B P$, it is orthogonal to the plane containing the face $V A B$ of the pyramid. Then, considering $V A B$ as the base of the pyramid, it is easy to calculate its volume: in $V A B$, the segment $B P$ is the height relative to the side $V A$, and thus the area of $V A B$ in $\mathrm{cm}^{2}$ is

$$
\frac{V A \cdot B P}{2}=\frac{(1+2) 3}{2}=\frac{9}{2}
$$

then, since $C P$ is the height of the pyramid relative to $V A B$, its volume is in $\mathrm{cm}^{3}$

$$
\frac{(\text { area } V A B) \cdot C P}{3}=\frac{9 \cdot 4}{3}=6
$$

SECOND SOLUTION.

The line $A V$, being perpendicular to $C P$ and $B P$, is perpendicular to the plane $B C P$. The pyramid $A B C V$ is thus decomposed by $B C P$ into two pyramids having $B C P$ as the base, and $P V, P A$ as their respective heights. The volume of $A B C V$ is the sum of the volumes of these two pyramids, and is

$$
\frac{1}{3}(\operatorname{area} B C P) \cdot(A P+V P)=\frac{1}{3} \cdot \frac{1}{2} \cdot 3 \cdot 4 \cdot(1+2)=6 \text {. }
$$