Lemma. If an inclined line forms an angle $\phi$ with a plane, a line lying in the plane forms an angle $\alpha$ with the inclined line, and an angle $\beta$ with the orthogonal projection of this inclined line, then $\cos \phi \cos \alpha = \cos \beta$. Proof. Let $l$ be the inclined line to the plane $\Pi$ (Fig.1), $l'$ be its orthogonal projection on this plane, and $m$ be a line lying in the plane. We can assume that all three lines pass through a point $X$ in the plane $\Pi$. From point $Y$ on line $l$, we draw perpendiculars $YZ$ and $YT$ to the plane $\Pi$ and line $m$ respectively. Then point $Z$ lies on line $l'$. By the theorem of three perpendiculars, $ZT \perp XT$. From the right triangles $XYZ$, $XYT$, and $XZT$, we find that

$$
XZ = XY \cos \phi = \cos \phi, \quad XT = XY \cos \beta = \cos \beta, \quad XT = XZ \cos \alpha = \cos \phi \cos \alpha.
$$

Therefore, $\cos \phi \cos \alpha = \cos \beta$, which is what we needed to prove. Solution to the problem. Let point $O$, lying on line $SA$, be the center of the base of the cone (Fig.2). Then $OF$ is the radius of the base of the cone, so the generatrix $EF$ is seen from point $O$ at a right angle, meaning it lies on a sphere with diameter $EF$ and center $M$ at the midpoint of $EF$. Since such a cone is unique, the sphere has only one common point with line $SA$, i.e., it is tangent to line $SA$. Therefore, the radius of the sphere, drawn to the point of tangency, is perpendicular to line $SA$, i.e., $MO \perp SA$. By the Pythagorean theorem,

$$
\begin{gathered}
EF = \sqrt{CF^2 + CE^2} = \sqrt{81 + 36} = 3\sqrt{13}, \quad AF = \sqrt{AB^2 + BF^2} = \sqrt{100 + 1} = \sqrt{101} \\
AE = \sqrt{AD^2 + DE^2} = \sqrt{100 + 16} = \sqrt{116}
\end{gathered}
$$

By the formula for the median of a triangle,

$$
AM = \frac{1}{2} \sqrt{2AF^2 + 2AE^2 - EF^2} = \frac{1}{2} \sqrt{2 \cdot 101 + 2 \cdot 116 - 9 \cdot 13} = \frac{\sqrt{317}}{2}.
$$

Let $\angle MAC = \alpha$, $\angle OAM = \beta$, and $\angle SAC = \phi$. Since $OM = ME = MF = \frac{1}{2}EF = \frac{3\sqrt{13}}{2}$, from the right triangle $AOM$ we find that

$$
OA = \sqrt{AM^2 - OM^2} = \sqrt{\left(\frac{\sqrt{317}}{2}\right)^2 - \left(\frac{3\sqrt{13}}{2}\right)^2} = 5\sqrt{2}
$$

$$
\cos \beta = \cos \angle OAM = \frac{OA}{AM} = \frac{\frac{5\sqrt{2}}{\sqrt{317}}}{2} = \frac{10\sqrt{2}}{\sqrt{317}}.
$$

By the cosine rule,

![](https://cdn.mathpix.com/cropped/2024_05_06_4bea0374a35e8bb7d318g-10.jpg?height=142&width=1074&top_left_y=352&top_left_x=492)

Then, by the proven lemma,

![](https://cdn.mathpix.com/cropped/2024_05_06_4bea0374a35e8bb7d318g-10.jpg?height=149&width=683&top_left_y=560&top_left_x=693)

Let $\angle CAE = \alpha'$, $\angle SAE = \beta'$. By the cosine rule,

$$
\cos \alpha' = \cos \angle CAE = \frac{AE^2 + AC^2 - CE^2}{2AE \cdot AC} = \frac{116 + 200 - 36}{2\sqrt{116} \cdot 10\sqrt{2}} = \frac{14}{\sqrt{2} \cdot \sqrt{116}}
$$

Then, by the lemma,

$$
\cos \beta' = \cos \phi \cos \alpha' = \frac{4}{5} \cdot \frac{14}{\sqrt{2} \sqrt{116}}
$$

Let $h$ be the height of the cone, and $r$ be the radius of the base. From triangle $AOE$ by the cosine rule, we find that

$$
h = OE = \sqrt{AO^2 + AE^2 - 2AO \cdot AE \cos B^t} =
$$

$$
= \sqrt{50 + 116 - 2 \cdot 5\sqrt{2} \cdot \sqrt{116} \cdot \frac{4}{5} \cdot \frac{14}{\sqrt{2} \cdot \sqrt{116}}} = \sqrt{166 - 112} = \sqrt{54} = 3\sqrt{6}.
$$

Finally, from the right triangle $EOF$, we find that

$$
r^2 = OF^2 = EF^2 - OE^2 = 117 - 54 = 63
$$

Therefore, if $V$ is the volume of the cone, then

$$
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \cdot 63 \cdot 3\sqrt{6} = 63\pi\sqrt{6}.
$$

## Answer

$63\pi\sqrt{6}$.
The side of the base of a regular quadrilateral pyramid $S A B C D$ ($S$ - vertex) is 10. Points $E$ and $F$ are located on the edges $D C$ and $B C$ respectively, with $C E=6, C F=9$. It is known that for this pyramid, there exists a unique cone whose vertex coincides with point $E$, the center of the base lies on the line $S A$, and the segment $E F$ is one of the generators. Find the volume of this cone.