[ Area and orthogonal projection ]

Find the area of the section of a regular triangular prism ABCA1B1C1 by a plane passing through vertex C and the midpoint of side B1C1 of the base A1B1C1, and parallel to the diagonal AC1 of the lateral face AA1C1C, if the distance between the line AC1 and the cutting plane is 1, and the side of the base of the prism is $\sqrt{14}$.
Let's denote the side of the base of the prism by \( a \) ( \( a = \sqrt{14} \) ). The plane of the lateral face \( AA_1C_1C \) passes through the line \( AC_1 \), which is parallel to the intersecting plane, and has a common point \( C \), so the line \( I \) of intersection of these planes is parallel to the line \( AC_1 \). Let the line \( I \) intersect the lines \( A_1C_1 \) and \( AA_1 \) at points \( P \) and \( Q \) respectively. Then \( C_1P = AC = A_1C_1 = a \) as opposite sides of the parallelogram \( ACRP \). Similarly, \( AQ = CC_1 = AA_1 \). Therefore, \( AC_1 \) is the midline of the triangle \( PA_1Q \). Let \( M \) be the midpoint of \( B_1C_1 \), and the line \( PM \) intersects the edge \( A_1B_1 \) at point \( N \). Draw a line through vertex \( B_1 \) parallel to \( A_1C_1 \), and extend \( PM \) to intersect this line at point \( T \). From the equality of triangles \( TMB_1 \) and \( PMC_1 \), it follows,

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-42.jpg?height=86&width=1949&top_left_y=783&top_left_x=0)
the edge \( AB \) at point \( L \). Then \( AL \) is the midline of the triangle \( NQA_1 \), so

\[
AL = \frac{1}{2} A_1N = \frac{1}{2} \cdot \frac{2}{3} A_1B_1 = \frac{1}{3} A_1B_1 = \frac{1}{3} AB,
\]

and the section of the prism by the given plane is a trapezoid \( MNLC \). Let's find the area of the orthogonal projection of the section on the plane of the base \( ABC \). Let \( N' \) and \( M' \) be the projections of points \( N \) and \( M \) on this plane. Then \( M' \) is the midpoint of \( BC \), and point \( N' \) is such that \( \frac{BN'}{BA} = \frac{1}{3} \), so,

\[
\begin{aligned}
& S' = S_{LN'M'C} = S_{\triangle BLC} - S_{\triangle BN'M'} = \frac{BL}{BA} S_{\Delta ABC} - \frac{BN'}{BA} \cdot \frac{BM'}{BC} S_{\triangle ABC} =
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-42.jpg?height=113&width=780&top_left_y=1344&top_left_x=639)

Let \( K \) be the foot of the perpendicular dropped from point \( C_1 \) to the line \( MP \). Then, by the theorem of three perpendiculars, \( CK \perp PM \), and the line \( CK \) lies in the intersecting plane, so \( CKC_1 \) is the linear angle of the dihedral angle between the intersecting plane and the base plane of the prism. Denote \( \angle C K C_1 = \phi \). From the triangle \( MC_1P \) by the cosine theorem, we find that

\[
MP = \sqrt{a^2 + \frac{a^2}{4} - 2 \cdot a \cdot \frac{a}{2} \cdot \cos 120^\circ} = \frac{a \sqrt{7}}{2} = \frac{7 \sqrt{2}}{2}
\]

By writing the area of the triangle \( MC_1P \) in two ways, we get that

\[
C_1K = \frac{C_1P \cdot C_1M \cdot \sin \angle MC_1P}{MP} = \frac{\frac{a^2 \cdot \sin 120^\circ}{\frac{a \sqrt{7}}{2}}}{MP} = \sqrt{\frac{3}{2}}
\]

If \( C_1H \) is the height of the right triangle \( CC_1K \), then the distance between the line \( AC_1 \) and the intersecting plane is the length of the segment \( C_1H \), i.e., \( C_1H = 1 \). From the right triangle \( CC_1K \), we find that

\[
\sin \phi = \frac{C_1H}{C_1K} = \frac{1}{\sqrt{\frac{3}{2}}} = \sqrt{\frac{2}{3}}, \cos \phi = \frac{1}{\sqrt{3}}
\]

Therefore, if \( S \) is the desired area of the section, then

\[
S = \frac{S'}{\cos \phi} = \frac{\frac{7 \sqrt{3}}{4}}{\frac{1}{\sqrt{3}}} = \frac{21}{4}
\]

Answer

## \(\frac{21}{4}\)

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