[Theorems of Ceva and Menelaus]

In a regular triangular pyramid $SABC$ ($S$ is the vertex), point $P$ is the midpoint of the apothem $SD$ lying in the face $SBC$. On the edge $AB$, a point $M$ is taken such that $MB: AB = 2: 7$. A sphere with its center on the line $MP$ passes through points $A$ and $C$ and intersects the line $BC$ at point $Q$ such that $CQ = m$. Find the volume of the pyramid $SABC$ if it is known that the radius of the sphere is $\frac{\frac{m}{\sqrt{3}}}{\sqrt{3}}$.
Let $O$ be the center of the sphere, $H, O^{\prime}$, and $P^{\prime}$ be the orthogonal projections of points $S, O$, and $P$ onto the base plane of the pyramid (Fig.2). Then $P^{\prime}$ is the midpoint of segment $H D$, so

$$
D P^{\prime}=\frac{1}{2} D H=\frac{1}{2} \cdot \frac{1}{3} A D=\frac{1}{6} A D
$$

If the line $C P^{\prime}$ intersects segment $A B$ at point $N$, then by Menelaus' theorem applied to triangle $A B D$ and line $C N$,

$$
\frac{A N}{N B} \cdot \frac{B C}{C D} \cdot \frac{D I^{N}}{P^{I} A}=1, \text { or } \frac{A N}{N B} \cdot \frac{2}{1} \cdot \frac{1}{5}=\frac{5}{2}
$$

from which $\frac{A N}{N B}=\frac{5}{2}$, and since by the problem's condition $\frac{A M}{M B}=\frac{5}{2}$, point $N$ coincides with $M$. Therefore, point $O^{\prime}$ lies on line $C M$. On the other hand, $O^{\prime} A=O^{\prime} C$ as projections of equal oblique lines $O A$ and $O C$, so point $O^{\prime}$ lies on the perpendicular bisector of side $A C$ of the equilateral triangle $A B C$, i.e., on line $B H$. Thus, $O^{\prime}$ is the intersection point of lines $C M$ and $B H$. Let $A B=a$. Then

$$
D P^{\prime}=\frac{1}{6} A D=\frac{a \sqrt{3}}{12}, C D=\frac{1}{2} a
$$

Consider triangle $D H B$ and line $C M$. By Menelaus' theorem,

$$
\frac{H O^{\prime}}{O^{\prime} B} \cdot \frac{B C}{C D} \cdot \frac{D P^{N}}{I^{\prime} H}=1, \text { or } \frac{H O^{\prime}}{O^{\prime} B} \cdot \frac{2}{1} \cdot \frac{1}{1}=1
$$

from which $\frac{H O^{\prime}}{O^{\prime} B}=\frac{1}{2}$

$$
\begin{aligned}
& \text { and the projection of point } O^{\prime} \text { onto side } C Q, \text { then } \\
& \frac{\partial F^{\prime}}{F^{\prime} B}=\frac{H O^{\prime}}{O^{\prime} B}=\frac{1}{2}, \text { or } \frac{C F^{\prime}-C D}{B C-C F^{\prime}}=\frac{\frac{m}{2}-\frac{a}{2}}{a-\frac{\pi^{m}}{2}}=\frac{1}{2}
\end{aligned}
$$

from which $a=\frac{\mathbf{3}}{\mathbf{4}} m$. Consider triangle $A P^{\prime} M$ and line $B H$. By Menelaus' theorem,

$$
\frac{P^{\prime} O}{\sigma^{\prime} M} \cdot \frac{M H}{B A} \cdot \frac{B H}{H P^{\prime}}=1, \text { or } \frac{I^{\prime} O}{\sigma^{\prime} M} \cdot \frac{2}{7} \cdot \frac{4}{1}=1
$$

from which $\frac{\rho^{\prime} O}{\sigma^{\prime} O}=\frac{7}{8}$. Therefore, $\frac{p^{\prime} \nu^{\prime} M}{\sigma^{\prime} M}=\frac{15}{8}$. Let $R 1$ be the radius of the section circle. From triangle $A C Q$ by the cosine theorem, we find that

$$
A Q=\sqrt{C Q^{2}+A C^{2}-2 C Q \cdot A C \cos 60^{\circ}}=m \sqrt{1+\frac{9}{16}-\frac{3}{4}}=\frac{m \sqrt{13}}{4}
$$

Then by the sine theorem,

$$
R 1=\frac{A Q}{2 \sin 60^{9}}=\frac{\frac{m \sqrt{13}}{13}}{\sqrt{3}}=\frac{m \sqrt{13}}{4 \sqrt{3}}
$$

From the right triangle $C O^{\prime} O$ (Fig.1), we find that

$$
O O^{\prime}=\sqrt{C O^{2}-O^{\prime} C^{2}}=\sqrt{\frac{m^{2}}{3}-\frac{13 m^{2}}{4 b}}=\frac{m}{4}
$$

and from the similarity of triangles $P P^{\prime} M$ and $O O^{\prime} M$,

$$
P P^{\prime}=O O^{\prime} \cdot \frac{\eta^{\prime} M}{\sigma^{\prime} M}=\frac{m}{4} \cdot \frac{15}{8}=\frac{15}{32} m
$$

Therefore,

$$
S H=2 P P^{\prime}=\frac{15}{16} m
$$

Thus,

$$
V_{S A B C}=\frac{1}{3} S_{\triangle A B C} \cdot S H=\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4} \cdot \frac{15}{16} m=\frac{1}{3} \cdot \frac{9^{\frac{2}{6}} m^{2} \sqrt{3}}{4} \cdot \frac{15}{16} m=\frac{45}{1024} \sqrt{3} m 3 .
$$

## Answer

## [ Regular Pyramid ]  Problem 110918 Topics: [ Theorems of Ceva and Menelaus ]  $[$ Cosine Theorem $]$

In a regular quadrilateral pyramid $S A B C D$ ( $S$ is the vertex) $S A=2 A B$. The perpendicular dropped from point $B$ to edge $S D$ intersects it at point $K$. On the apothem $S F$ of face $S A B$, point $M$ is taken such that $S M: S F=4: 5$. A sphere with center on line $M K$ passes through points $B, K$ and intersects line $A B$ at point $P$, and $B P=d$. Find the length of segment $A B$.

## Answer

$\frac{\frac{8}{9}}{9} d$.