Problem 373

8. In the pyramid $FABC$, $AB=BC$, $FB=FK$, where $K$ is the midpoint of the segment $AC$, and the tangent of the angle between the planes $FAB$ and $ABC$ is in the ratio of 1:3 to the tangent of the angle between the planes $FBC$ and $ABC$. The plane $\pi$ is parallel to $AB$, divides the edge $FC$ in the ratio $1:4$, counting from the vertex $F$, and passes through the base $O$ of the height $FO$ of the pyramid $FABC$. Find the ratio of the volumes of the polyhedra into which this plane divides the pyramid $FABC$.

## Answer: $5: 11$ or $1: 19$.

Solution: 1) We will prove that the base of the height (point $O$) lies on the midline of $\triangle ABC$. Triangles $FOB$ and $FOK$ are equal (they are right triangles, $FO$ is common, and $FB=FK$), so $BO=OK$. In the plane of the base $ABC$, draw a line $l$ passing through point $O$ parallel to $AC$ and denote $L=l \cap BK$, $M=l \cap AB$, $N=l \cap BC$. Since $AB=BC$, the median $BK \perp AC$, and thus $OL \perp BK$. Then triangles $BOL$ and $KOL$ are equal (they are right triangles with a common $OL$ and equal $BO=OK$), so $BL=KL$, i.e., $MN$ is the midline of triangle $ABC$.

2) Find the ratio $OM:ON$. From point $O$, draw perpendiculars $OH_1$ to side $AB$ and $OH_2$ to side $BC$ and denote $\angle FH_1O=\alpha$, $\angle FH_2O=\beta$. The plane $FOH_1$ is perpendicular to the plane $ABC$ and the lateral face $FAB$, i.e., $\alpha$ is the angle between the planes $FAB$ and $ABC$. Similarly, $\beta$ is the angle between the planes $FBC$ and $ABC$. From triangles $FOH_1$ and $FOH_2$, we find $OH_1=FO \cdot \operatorname{ctg} \alpha$, $OH_2=FO \cdot \operatorname{ctg} \beta$. From the condition, it follows that $OH_1:OH_2=3:1$. Triangles $OMH_1$ and $ONH_2$ are similar (they are right triangles and $\angle OMH_1=\angle ONH_2$), so $OM:ON=3:1$.
3) First case. Suppose that point $O$ lies inside triangle $ABC$. In the plane $ABC$, draw a line $m$ through point $O$ parallel to $AB$. By the condition, this line lies in the cutting plane $\pi$. Denote $P=m \cap AC$, $Q=m \cap BC$. From Thales' theorem, $BQ=3QN$. Since $MN$ is the midline, we get $CQ:CB=CP:CA=5:8$. Denote the point of intersection of the plane $\pi$ and the edge $FC$ as $R$ - by the condition $CR:CF=4:5$. Thus, the plane $\pi$ intersects the edges $AC$, $BC$, and $FC$ at points $P$, $Q$, and $R$, respectively, so

$$\frac{V_{CPQR}}{V_{CABF}}=\frac{CP \cdot CQ \cdot CR}{CA \cdot CB \cdot CF}=\frac{5}{8} \cdot \frac{5}{8} \cdot \frac{4}{5}=\frac{5}{16}$$

Then the volumes of the polyhedra $CPQR$ and $ABQPRF$ are in the ratio $5:11$.

4) Second case. Suppose that point $O$ lies outside triangle $ABC$. Then $MN:ON=2:1$. Similarly, we get $BN:NQ=2:1$, and since $MN$ is the midline, $CQ:CB=CP:CA=1:4$. Then $V_{CPQR}:V_{CABF}=1:20$. Then the volumes of the polyhedra $CPQR$ and $ABQPRF$ are in the ratio $1:19$.