10,11 At the base of the pyramid $SABC$ lies an isosceles trapezoid $ABCD$, in which $AD=2, BC=1$, the height of the trapezoid is 3. The height of the pyramid passes through the point $O$ of intersection of the diagonals of the trapezoid, $SO=\frac{\sqrt[4]{\sqrt{5}}}{}$. Point $F$ lies on the segment $SO$, such that $SF: FO=1: 3$. A cylinder, whose axis is parallel to the height $SM$ of the face $SAD$, is positioned such that point $F$ is the center of its upper base, and point $O$ lies on the circumference of the lower base.

Find the area of the part of the upper base of the cylinder that lies inside the pyramid.
Let $N$ be the midpoint of the base $BC$ of trapezoid $ABCD$. Since the trapezoid is isosceles, points $O, M$, and $N$ lie on the perpendicular bisector of the bases, and $OM = \frac{2}{3} MN = 3, ON = \frac{1}{3} MN = 1$, and the given pyramid is symmetric with respect to the plane $SMN$. We will cut the pyramid and the cylinder with the plane $SMN$. We will get a triangle $SMN$ with height $SO$ and a rectangle $OKLG$, one vertex of which is point $O$, and the segment $FT$, connecting the midpoints $F$ and $P$ of the opposite sides $KL$ and $OG$ respectively, is parallel to $SM$.

Let $\angle OMS = \alpha, \angle MSN = \beta$. Then

$$
\begin{aligned}
& SN = \sqrt{SO^2 + ON^2} = \sqrt{\frac{16}{5} + 1} = \sqrt{\frac{21}{5}}, \quad SM = \sqrt{SO^2 + OM^2} = \sqrt{\frac{16}{5} + 4} = \frac{6}{\sqrt{5}} \\
& \cos \alpha = \frac{SO}{SM} = \frac{\frac{1}{\sqrt{3}}}{\frac{6}{5}} = \frac{2}{3}, \quad \cos \beta = \frac{SM^2 + SN^2 - MN^2}{2 \cdot SM \cdot SN} = \frac{\frac{3(3) + 21}{5} - 9}{2 \cdot \frac{6}{\sqrt{3}} \cdot \sqrt{\frac{21}{5}}} = \frac{1}{\sqrt{21}}.
\end{aligned}
$$

Let $R$ be the radius of the base of the cylinder, and $h$ be the height of the cylinder. From the right triangle $OPF$, we find that

$$
\begin{aligned}
& h = FP = OF \cos \alpha = \frac{3}{4} SO \cdot \frac{2}{3} = \frac{3}{\sqrt{5}} \cdot \frac{2}{3} = \frac{2}{\sqrt{5}}, \\
& R = FK = PO = \sqrt{FO^2 - FP^2} = \sqrt{\frac{9}{5} - \frac{4}{5}} = 1.
\end{aligned}
$$

If $K'$ and $E$ are the points of intersection of the line $FL$ with the sides $SN$ and $SM$ respectively, then

$$
\begin{aligned}
& SE = SF \cos \alpha = \frac{1}{4} SO \cos \alpha = \frac{1}{\sqrt{5}} \cdot \frac{2}{3} = \frac{2}{3 \sqrt{5}}, \\
& FK' = EK' - EF = SE \tan \beta - SE \tan \alpha = \frac{2}{3 \sqrt{5}} \left(2 \sqrt{5} - \frac{\sqrt{5}}{2}\right) = 1,
\end{aligned}
$$

which means that point $K$ coincides with point $K'$, i.e., point $K$ lies on side $SN$. Therefore, the cylinder has only one common point $K$ with the face $BSC$. We will cut the pyramid with a plane containing the base of the cylinder passing through point $F$. We will get a circle of radius $R = 1$ with center at point $F$ and an isosceles trapezoid $K_1K_2L_2L_1$, where the line $K_1K_2$ is tangent to the circle at point $K$ - the midpoint of the base $K_1K_2$, and the vertices $L_1$ and $L_2$ are the points of intersection of the plane with the edges $SD$ and $SA$, and $D$ is the midpoint of the base $L_1L_2$. From the similarity of triangles $SK_1K_2$ and $SBC$, we find that

$$
K_1K_2 = BC \cdot \frac{SK}{SN} = 1 \cdot \frac{\frac{2 \sqrt{21}}{3 \sqrt{2}}}{\frac{\sqrt{21}}{\sqrt{5}}} = \frac{2}{3}
$$

and from the similarity of triangles $SL_1L_2$ and $SDA$,

$$
L_1L_2 = AD \cdot \frac{SE}{SM} = 2 \cdot \frac{\frac{2}{\sqrt{5}}}{\frac{5}{5}} = 2 \cdot \frac{1}{\xi} = \frac{2}{9}
$$

Let $G$ be the projection of point $L_1$ onto $K_1K_2$, and $\angle K K_1 L_1 = \gamma$. Then

$$
K_1G = \frac{1}{2}(K_1K_2 - L_1L_2) = \frac{1}{2}\left(\frac{2}{3} - \frac{2}{7}\right) = \frac{2}{9}
$$

$$
\tan \gamma = \frac{L_1G}{K_1G} = \frac{E_1K}{K_1G} = \frac{\frac{1}{2}}{\frac{2}{9}} = 6
$$

Let the lines $K_1L_1$ and $KE$ (and therefore, the lines $K_2L_2$ and $KE$) intersect at point $Q$. Then

$$
KQ = KK_1 \tan \gamma = \frac{1}{3} \cdot 6 = 2 = 2R
$$

which means that point $Q$ lies on the circle of the section. Let $T_1$ and $T_2$ be the points of intersection of the segments $QK_1$ and $QK_2$ with the circle, respectively, $S$ be the desired area, $S_1$ be the area of the sector $T_1FK$, $S_2$ be the area of the triangle $QFT_1$, and $S_3$ be the area of the triangle $QL_1L_2$. Then

$$
\begin{gathered}
\angle F T_1 Q = \angle K Q K_1 = 90^\circ - \gamma, \quad \angle Q F T_1 = 180^\circ - 2(90^\circ - \gamma) = 2 \gamma \\
\sin 2 \gamma = \frac{2 \tan \gamma}{1 + \tan^2 \gamma} = \frac{12}{1 + 36} = \frac{12}{37} \\
\cos (180^\circ - 2 \gamma) = -\cos 2 \gamma = -\frac{1 - \tan^2 \gamma}{1 + \tan^2 \gamma} = \frac{1 - 36}{1 + 36} = \frac{35}{37}, \\
S_1 = \frac{1}{2} \cdot 1 \cdot 1 \cdot \arccos \frac{35}{37} = \frac{1}{2} \arccos \frac{35}{37}, \\
S_2 = \frac{1}{2} F T_1 \cdot F Q \sin 2 \gamma = \frac{1}{2} \cdot 1 \cdot 1 \cdot \frac{12}{37} = \frac{6}{37}, \\
S_3 = \frac{1}{2} L_1 L_2 \cdot E Q = \frac{1}{2} \cdot L_1 L_2 \cdot \frac{1}{2} L_1 L_2 \tan \gamma = \frac{1}{2} L_1 L_2^2 \tan \gamma = \frac{1}{2} \cdot \frac{4}{81} \cdot 6 = \frac{2}{27}. \\
\text{Therefore,} \\
S = 2 S_1 + 2 S_2 - S_3 = \arccos \frac{35}{37} + \frac{12}{37} - \frac{2}{27}.
\end{gathered}
$$

## Answer

$$
\arccos \frac{35}{37} + \frac{12}{37} - \frac{2}{27}
$$

Submit a comment