Given a sphere of radius 1 with center at point $O$. From a point $A$ outside the sphere, four rays are drawn. The first ray intersects the sphere's surface sequentially at points $B 1$ and $C 1$, the second at points $B 2$ and $C 2$, the third at points $B 3$ and $C 3$, and the fourth at points $B 4$ and $C 4$. The lines $B 1 B 2$ and $C 1 C 2$ intersect at point $E$, and the lines $B 3 B 4$ and $C 3 C 4$ intersect at point $F$. Find the volume of the pyramid $O A E F$, if $A O=2$, $E O=F O=3$, and the angle between the faces $A O E$ and $A O F$ is $30^{\circ}$.
Consider the section of the sphere by a plane passing through the intersecting lines $B 1 C 1$ and $B 2 C 2$ (Fig.2). We obtain the inscribed quadrilateral $B 1 B 2 C 2 C 1$. Circles are described around triangles $A B 1 B 2$ and $E C 2 B 2$. Let $M$ be the second point of intersection of these circles. We will prove that point $M$ lies on the segment $A E$. Denote $\angle E M B 2=\alpha$. The sums of opposite angles in an inscribed quadrilateral are $180^{\circ}$, therefore

$$
\begin{aligned}
& \angle C 1 C 2 B 2=180^{\circ}-\angle E C 2 B 2=\alpha \\
& \angle A B 1 B 2=180^{\circ}-\angle C 1 B 1 B 2=\alpha \\
& \angle A M B 2=180^{\circ}-\angle A B 1 B 2=180^{\circ}-\alpha
\end{aligned}
$$

Thus, $\angle E M B 2+\angle A M B 2=180^{\circ}$. Therefore, point $M$ lies on the segment $A E \cdot$ In this case, $A E \cdot A M=A C 2 \cdot A B 2$, since if secants are drawn from a point outside a circle to the circle, the product of the entire secant and its external part is constant. Let $R$ be the radius of the sphere. Consider also the section of the sphere by a plane passing through its center $O$ and the line $B 2 C 2$. Let the ray $A O$ intersect the sphere at points $P$ and $Q$ (Fig.3). Then

$$
A E \cdot A M=A C 2 \cdot A B 2=A Q \cdot A P=(A O+R)(A O-R)=A O 2-R 2=4-1=3 .
$$

Similarly,

$$
A E \cdot E M=E B 2 \cdot E B 1=(E O+R)(E O-R)=E O 2-R 2=8
$$

Then

$$
A E 2=A E(A M+E M)=A E \cdot A M+A E \cdot E M=3+8=11 .
$$

Similarly, we obtain that $A F 2=11$. By Heron's formula

$$
S_{\triangle A O E}=S_{\triangle A O F}=\frac{1}{4} \sqrt{(2+3+\sqrt{11})(2+3-\sqrt{11})(2+\sqrt{11}-3)(3+\sqrt{11}-2)}=
$$

$$
=\frac{1}{4} \sqrt{(25-11)(11-1)}=\frac{\sqrt{140}}{4}
$$

If $S 1$ and $S 2$ are the areas of the faces of the tetrahedron, $\phi$ is the angle between these faces, and $a$ is their common edge, then the volume $V$ of the tetrahedron can be calculated by the formula $V=\frac{2}{\overline{3}}$. $\frac{S_{1} \cdot \Sigma_{2} \sin 4}{\mathbf{4}}$ (Fig.4). Therefore,

$$
V_{O A E F}=\frac{2}{3} \cdot \frac{S_{\triangle A C E} \cdot S_{\triangle A C F} \cdot \sin 30^{\circ}}{A O}=\frac{2}{3} \cdot \frac{\sqrt{\frac{\sqrt{110}}{1}} \cdot \frac{\sqrt{140}}{1}}{2 \cdot 2}=\frac{140}{3 \cdot 32}=\frac{35}{24}
$$

## [ Volume of the tetrahedron and pyramid $\quad]$  Problem 111144 Topics: [Product of segment lengths of chords and secants ] [ Auxiliary circle $\quad]$

A sphere of radius 2 with center at point $O$ is given. From a point $K$ outside the sphere, four rays are drawn. The first ray intersects the surface of the sphere successively at points $L 1$ and $M 1$, the second at points $L 2$ and $M 2$, the third at points $L 3$ and $M 3$, and the fourth at points $L 4$ and $M 4$. The lines $L 1 L 2$ and $M 1 M 2$ intersect at point $A$, and the lines $L 3 L 4$ and $M 3 M 4$ intersect at point $B$. Find the volume of the pyramid $K O A B$, if $K O=3, A O=B O=4$, and the angle between the faces $K O A$ and $K O B$ is $60^{\circ}$.

## Answer