# Task 8.

Constructing a section of a cube.

![](https://cdn.mathpix.com/cropped/2024_05_06_5c23ad681c7ac8ec03deg-06.jpg?height=724&width=1316&top_left_y=283&top_left_x=384)

The line $\mathrm{KL}$, where $\mathrm{K}$ is the midpoint of $D_{1} C_{1}$ and $\mathrm{L}$ is the midpoint of $C_{1} B_{1}$, lies in the plane face $A_{1} B_{1} C_{1} D_{1}$. Therefore, it intersects the extensions of the edges $A_{1} B_{1}$ and $A_{1} D_{1}$ at points F and E, respectively. It is easy to calculate that $D_{1} E = \frac{1}{2} A_{1} D_{1} = \frac{1}{3} A_{1} E$. Similarly, $B_{1} F = \frac{1}{2} A_{1} B_{1} = \frac{1}{3} A_{1} F$. Since points E and A lie in the same face, it is not difficult to show that the line $\mathrm{AE}$ intersects the edge $D D_{1}$ at point $\mathrm{N}$ and divides this edge in the ratio 2:1. Similarly, the line $\mathrm{AF}$ intersects the edge $B_{1} B$ at point $\mathrm{M}$ and also divides this edge in the ratio 2:1. Therefore, the section will be the pentagon AMLKN. Its projection on the lower base will be the pentagon $A B L_{1} K_{1} D$. The area of this pentagon will be $\frac{7}{8}$.

If $\mathrm{T}$ is the midpoint of $\mathrm{KL}$ and $T_{1}$ is the projection of point $\mathrm{T}$ on the lower base, then the angle $T A T_{1}$ will be the angle between the section and the lower face of the cube. Then $\cos \left(T A T_{1}\right) = \frac{3}{\sqrt{17}}$. Using the formula for the area of the projection, we get that the area of the section will be: $\frac{7 \sqrt{17}}{24}$.
Answer: $\frac{7 \sqrt{17}}{24}$.

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Chairman of the Methodological Commission for the "Mathematics" profile

![](https://cdn.mathpix.com/cropped/2024_05_06_5c23ad681c7ac8ec03deg-07.jpg?height=169&width=762&top_left_y=355&top_left_x=1069)

FGAOU VO RUT (MIIT) INTERREGIONAL INDUSTRY OLYMPIAD FOR SCHOOL STUDENTS "SAILOF HOPE" IN THE "MATHEMATICS" PROFILE 2021-2022 ACADEMIC YEAR Brief solutions to the problems of the on-site round 11th grade

# Variant 3 

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