6. As part of the renovation program, it was decided to demolish an old house and build a new one in its place. During the demolition of the old house, two problems arose.
1) The accumulation of construction debris blocked access to certain points on the construction site, between which it was necessary to measure the distance. The foreman measured the distance from point C to point A and from point C to point B (see figure). It turned out that $\mathrm{AC}=4$ m, $\mathrm{BC}=10$ m. In addition, he managed to determine the distances $\mathrm{CR}=1 \mathrm{m}, \mathrm{VR}=3 \mathrm{m}, \mathrm{CV}=2.5$ m for points $\mathrm{R}$ and $\mathrm{V}$, located on the extensions of the paths connecting objects A, C and C, B, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-08.jpg?height=459&width=720&top_left_y=336&top_left_x=1245)

Determine the distance between points A and B (it was not possible to measure this distance on the site due to the debris) and the area of the part of the construction site $\mathrm{CADB}$, given that the distances CD and $\mathrm{BD}$ are twice the distance CB.

2) A rigid piece of rebar - a stake - was sticking out of the ground, with a height of 0.5 m above the ground. Attempts to pull it out of the ground were unsuccessful. Then the workers moved it with a bulldozer so that the end of the stake was level with the ground and was 1.5 m from the initial point where the stake entered the ground. Find the length of the part of the stake hidden in the ground, assuming that its lower end did not move. (20 points)
# Solution.

1) Using the similarity of triangles ACB and RCV:

$$
\frac{AC}{CR}=\frac{AB}{VR} \quad \Rightarrow \quad AB=\frac{AC}{CR} \cdot VR=\frac{4}{1} \cdot 3=12
$$

The area of quadrilateral CADB will be found as the sum of the areas of triangles CAB and ABD: the area of the first triangle will be found as half the product of the lengths of the sides by the sine of the angle between them, the angle will be found using the cosine theorem.

$\cos \beta=\frac{AB^2 + CB^2 - AC^2}{2 AB \cdot CB}=\frac{144+100-16}{2 \cdot 12 \cdot 10}=\frac{36+25-4}{2 \cdot 3 \cdot 10}=\frac{57}{6 \cdot 10}=\frac{19}{20} \Rightarrow \sin \beta=\frac{\sqrt{39}}{20}$, $S_{CAB}=\frac{1}{2} \cdot CB \cdot AB \cdot \sin \beta=\frac{1}{2} \cdot 10 \cdot 12 \cdot \frac{\sqrt{39}}{20}=3 \sqrt{39}$, since triangle CDB is isosceles, the cosine of the angle at the base is $\cos \alpha=\frac{CB}{2 CD}=\frac{CB}{2 \cdot 2 \cdot CB}=\frac{1}{4} \Rightarrow \sin \alpha=\frac{\sqrt{15}}{4}$ $S_{ABD}=\frac{1}{2} \cdot DB \cdot AB \cdot \sin (\alpha-\beta)=\frac{1}{2} \cdot 20 \cdot 12\left(\frac{19}{20} \cdot \frac{\sqrt{15}}{4}-\frac{1}{4} \cdot \frac{\sqrt{39}}{20}\right)=\frac{57 \sqrt{15}}{2}-\frac{3 \sqrt{39}}{2}$,

![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-08.jpg?height=331&width=291&top_left_y=2290&top_left_x=223)

$S_{CADB}=\frac{57 \sqrt{15}}{2}+\frac{3 \sqrt{39}}{2}=\frac{57 \sqrt{15}+3 \sqrt{39}}{2} \approx 119.74$

b) $(AE + AC)^2 = EB^2 = AE^2 + AB^2 \Rightarrow AE = \frac{AB^2 - AC^2}{2 AC} = \frac{1.5^2 - 0.5^2}{2 \cdot 0.5} = 2$

Answer: a) $12, \frac{57 \sqrt{15} + 3 \sqrt{39}}{2} \approx 119.74, \quad$ b) 2.

## Olympiad for Schoolchildren "Step into the Future"

Profile: Mathematics

Variant: 3

Class: 10

Problem 1 (12 points). The fraction $\frac{1}{5}$ is written as an infinite binary fraction (in the binary number system). How many ones are among the first 2022 digits after the decimal point in such a representation?

Problem 2 (16 points). There is a cube fixed on legs and six different colors.

In how many ways can all the faces of the cube be painted (each in one color, not all colors need to be used) so that adjacent faces (having a common edge) are of different colors?

Problem 3 (16 points). Point $M$ lies on the leg $AC$ of a right triangle $ABC$ with a right angle at $C$, and $AM=2, \quad MC=16$. Segment $MH$ is the altitude of triangle $AMB$. Point $D$ is located on line $MH$ such that angle $ADB$ is $90^\circ$, and points $C$ and $D$ lie on the same side of line $AB$. Find the length of segment $BL$, where $L$ is the intersection of $BD$ and $AC$, and the tangent of angle $ACH$ is $1/18$.

Problem 4 (16 points). Find all values of the parameter $a$ for which the system

$$
\left\{\begin{array}{c}
\left(a\left|y^3\right| + a\left|x^3\right| - 8\right)\left(x^6 + y^6 - 3a^2\right) = 0 \\
\sqrt{x^6 y^6} = a
\end{array}\right.
$$

has exactly eight distinct solutions.

Problem 5 (20 points). Inside a regular quadrilateral pyramid $SABCD$ with base $ABCD$, there is a regular quadrilateral prism $KLMNK_1L_1M_1N_1$, the base $KLMN$ of which lies in the plane $ABCD$. The center of the base $KLMN$ of the prism is located on segment $AC$, $KL \parallel AC$, $KN \parallel BD$ (points $K$ and $B$ lie on the same side of $AC$), the side of the base of the prism is 2, and the lateral edge $KK_1$ of the prism is 3. The vertices $L_1$ and $M_1$ of the upper base of the prism $KLMNK_1L_1M_1N_1$ lie on the lateral faces $SBC$ and $SCD$ of the pyramid $SABCD$ respectively. Plane $\gamma$ passes through line $BD$ and point $L_1$. Find the area of the section of the pyramid $SABCD$ by plane $\gamma$, if the side of the base of the pyramid is $8\sqrt{2}$, and its height is 12.

Problem 6 (20 points). To improve living conditions in some cities, a renovation program is being implemented - replacing non-salvageable housing stock through demolition and new construction on the freed-up land. Typically, the demolition of old buildings and the clearing of the site for new construction occur on a limited area. Therefore, equipment (rotary excavators) is needed that can clear the largest area with the fewest movements.

The diagram shows a schematic working area and a schematic movement of the excavator. Standing at the initial point $E$, the excavator first clears the nearest area to itself (mainly containing metal scrap). This area is a sector of a circle bounded by the large arc $FG$ of a circle with center at point $E$ and radius $2\sqrt{2}$ m, and by the segments of its radii $EF$ and $EG$. Then the excavator moves to point $O$ and clears the farthest area from itself (mainly containing brick scrap). This area is bounded externally by the large arc $CD$ of a circle with center at point $O$ and radius 6 m, by segments $CG$ and $FD$ (points $F$ and $G$ are on the corresponding radii $OC$ and $OD$), and internally by the large arc $FG$ of the first circle with center at point $E$.

Determine the area under the brick scrap (without metal residues), if point $O$ is 2 meters from point $E$ in the direction of the center of the metal scrap, and angle $EOD$ is $45^\circ$.

## Solution to Variant 3 (Mathematics - 10th Grade)