3. Specify all values of $a$ for which the system of equations $\left\{\begin{array}{c}x-1=a\left(y^{3}-1\right) \\ \frac{2 x}{\left|y^{3}\right|+y^{3}}=\sqrt{x}\end{array}\right.$ has at least one solution, and solve it for each $a$.

#
# Solution.

Domain of definition: $y>0, x \geq 0$.

In the domain of definition, the second equation of the system takes the form: $x=y^{3} \sqrt{x}$.
I. $x=0, y=\sqrt[3]{1-\frac{1}{a}}=\sqrt[3]{\frac{a-1}{a}}>0 \Leftrightarrow\left[\begin{array}{l}a1\end{array}\right.$

II. $x>0, y=\sqrt{x} ;(\sqrt{x}-1)(\sqrt{x}+1)=a(\sqrt{x}-1)$.

1) $\sqrt{x}=1, x=1, y=1, a \in R$.
2) $\sqrt{x}+1=a, \sqrt{x}=a-1>0, a>1$. The solution found $x=(a-1)^{2}, y=\sqrt[3]{a-1}$ coincides with the previous one if $1=a-1, a=2$. Thus, for $a \in(1 ; 2) \cup(2 ;+\infty) x=(a-1)^{2}, y=$ $\sqrt[3]{a-1}$

![](https://cdn.mathpix.com/cropped/2024_05_06_d45ee4ec72eb6c7fb975g-4.jpg?height=226&width=1468&top_left_y=207&top_left_x=320)

Answer:

$$
\begin{aligned}
& a \in(-\infty ; 0) \cup\{2\}, x_{1}=0, y_{1}=1-\frac{1}{a} ; x_{2}=1, y_{2}=1 \\
& a \in[0 ; 1], x=1, y=1 ; \\
& a \in(1 ; 2) \cup(2 ;+\infty), x_{1}=0, y_{1}=\sqrt[3]{1-\frac{1}{a}} ; x_{2}=1, y_{2}=1 ; x_{3}=(a-1)^{2}, y_{3}=\sqrt[3]{a-1}
\end{aligned}
$$

4b. Find the perimeter of the figure common to both pyramids (see the condition of problem 4a).

(8 points)

## Solution.

Lines $A C$ and $E F$ are parallel, $A C F E$ is a parallelogram, the figure common to both pyramids is the triangle $A G E$, where point $G$ is the intersection of the diagonals of the parallelogram $A C F E$. The figure common to both pyramids is the triangle $A G E$.

$$
\begin{gathered}
G E=C E / 2 . C E=\sqrt{C K^{2}+K E^{2}}=\sqrt{20^{2}+20^{2}}=20 \sqrt{2}, G E=10 \sqrt{2} \\
A F=\sqrt{40^{2}+40^{2}+20^{2}}=60, \quad A G=A F / 2=30 \\
A E=\sqrt{20^{2}+10^{2}+20^{2}}=30 \\
P_{A G E}=10 \sqrt{2}+60
\end{gathered}
$$

Answer: $10 \sqrt{2}+60$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d45ee4ec72eb6c7fb975g-4.jpg?height=819&width=1177&top_left_y=1789&top_left_x=408)

## Solution to Variant №3 (Mathematics - 10th Grade)