3. Specify all values of $a$ for which the system of equations $\left\{\begin{array}{c}x-4=a\left(y^{3}-2\right), \\ \frac{2 x}{\left|y^{3}\right|+y^{3}}=\sqrt{x}\end{array}\right.$ has at least one solution, and solve it for each $a$. (12 points)
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. $x=0, y=\sqrt[3]{2-\frac{4}{a}}=\sqrt[3]{\frac{2 a-4}{a}}>0 \Leftrightarrow\left[\begin{array}{l}a2\end{array}\right.$

II. $x>0, y^{3}=\sqrt{x} ;(\sqrt{x}-2)(\sqrt{x}+2)=a(\sqrt{x}-2)$.

1) $\sqrt{x}=2, x=4, y=\sqrt[3]{2}, a \in R$.
2) $\sqrt{x}+2=a, \sqrt{x}=a-2>0, a>2$. The solution found $x=(a-2)^{2}, y=\sqrt[3]{a-2}$ coincides with the previous one if $2=a-2, a=4$. Thus, for $a \in(2 ; 4) \cup(4 ;+\infty) x=(a-2)^{2}, y=$ $\sqrt[3]{a-2}$

![](https://cdn.mathpix.com/cropped/2024_05_06_d45ee4ec72eb6c7fb975g-2.jpg?height=229&width=1470&top_left_y=208&top_left_x=316)

Answer: $a \in(-\infty ; 0) \cup\{4\}, x_{1}=0, y_{1}=\sqrt[3]{2-\frac{4}{a}} ; x_{2}=4, y_{2}=\sqrt[3]{2}$;

$a \in[0 ; 2], x=4, y=\sqrt[3]{2}$

$a \in(2 ; 4) \cup(4 ;+\infty), x_{1}=0, y_{1}=\sqrt[3]{2-\frac{4}{a}} ; x_{2}=4, y_{2}=\sqrt[3]{2} ; x_{3}=(a-2)^{2}, y_{3}=\sqrt[3]{a-2}$.

4b. Find the area of the figure common to both pyramids (see the condition of problem 4a). (8 points)

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_d45ee4ec72eb6c7fb975g-2.jpg?height=837&width=1242&top_left_y=1015&top_left_x=378)

Lines $A C$ and $E F$ are parallel, $A C F E$ is a parallelogram, and 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 area of triangle $A G E$ is one-fourth of the area of parallelogram $A C F E$.

We have $A C=\sqrt{20^{2}+30^{2}}=10 \sqrt{13}$. $E K$ is the perpendicular dropped from point $E$ to the plane $A B C$. $K$ is the perpendicular dropped from point $K$ to the line $A C$. According to the theorem of three perpendiculars, $E H$ is the height of the parallelogram $A C F E$.

$$
\begin{array}{r}
K H=\frac{2}{3} \cdot \frac{30 \cdot 20}{10 \sqrt{13}}=\frac{40}{\sqrt{13}}, E H=\sqrt{E K^{2}+K H^{2}}=\sqrt{20^{2}+\left(\frac{40}{\sqrt{13}}\right)^{2}}=20 \sqrt{\frac{17}{13}} \\
S_{A C F E}=A C \cdot E H=200 \sqrt{17}, \quad S_{A G E}=50 \sqrt{17} .
\end{array}
$$

Answer: $50 \sqrt{17}$.