5. Find all values of the parameter $a$ for which the equation $(\cos 2 x+14 \cos x-14 a)^{7}-\left(6 a \cos x-4 a^{2}-1\right)^{7}=(6 a-14) \cos x+2 \sin ^{2} x-4 a^{2}+14 a-2 \quad$ has $\quad$ two distinct solutions on the interval $\left[-\frac{2 \pi}{3} ; \pi\right]$. Specify these solutions for each found $a$.

Solution: $\quad(\cos 2 x+14 \cos x-14 a)^{7}-\left(6 a \cos x-4 a^{2}-1\right)^{7}=(6 a-14) \cos x+2 \sin ^{2} x-4 a^{2}+14 a-2 \Leftrightarrow$ $(\cos 2 x+14 \cos x-14 a)^{7}+\cos 2 x+14 \cos x-14 a=\left(6 a \cos x-4 a^{2}-1\right)^{7}+6 a \cos x-4 a^{2}-1 \Leftrightarrow$ Consider the function $f(t)=t^{7}+t$. The function $f(t)$ is increasing on the entire number line. Let $u=\cos 2 x+14 \cos x-14 a, \quad v=6 a \cos x-4 a^{2}-1$. Then we have the equation $f(u)=f(v)$, and due to the strict monotonicity of the function $f$, we arrive at the equation $u=v$, i.e., $\cos 2 x+14 \cos x-14 a=6 a \cos x-4 a^{2}-1$. The last equation is equivalent to the following $\cos ^{2} x-(3 a-7) \cos x+2 a^{2}-7 a=0$. It is necessary to find all values of the parameter $a$ for which this equation has two distinct solutions on the interval $\left[-\frac{2 \pi}{3} ; \pi\right]$. Let's make the substitution: $y=\cos x$. We arrive at the equation $y^{2}-(3 a-7) y+2 a^{2}-7 a=0$. For the condition of the problem to be satisfied, one root of this equation must belong to the interval $[-0.5 ; 1)$, and the second, if it exists, must not belong to the interval $[-1 ; 1]$, or this equation must have two distinct solutions belonging to the set $[-1 ;-0.5) \cup\{1\}$. The discriminant of the equation $D=(3 a-7)^{2}-8 a^{2}+28 a=(a-7)^{2}$. The only solution will be at $a=7$, which is the solution $y=7$. In all other cases, we have two distinct solutions $y_{1}=2 a-7$ and $y_{2}=a$.

1) $-0.5 \leq 2 a-71 \Rightarrow a \in[3.25 ; 4), \quad \cos x=2 a-7, x_{1}=\arccos (2 a-7)$, $x_{2}=-\arccos (2 a-7)$.
2) $-0.5 \leq a1 \Rightarrow a \in[-0.5 ; 1), \cos x=a, x_{1}=\arccos (a), x_{2}=-\arccos (a)$. The equation cannot have two distinct solutions belonging to the set $[-1 ;-0.5) \cup\{1\}$.
Answer: $a \in[3.25 ; 4), x_{1}=\arccos (2 a-7), x_{2}=-\arccos (2 a-7)$;

$$
a \in[-0.5 ; 1), x_{1}=\arccos (a), x_{2}=-\arccos (a)
$$

6 Find the area of the section of a regular triangular pyramid TABC by a plane passing through the center of the sphere circumscribed around the pyramid, and through the midpoint of the lateral edge $T A$ and parallel to the median $A D$ of the lateral face $A T C$, if the sides of the base are equal to 4, and the center of the sphere divides the height of the pyramid in the ratio $3: 1$, counting from the vertex.

Solution: The center of the sphere $O$ lies on the perpendicular to the base plane, drawn through the center of the base $H ; M$ - the midpoint of $T A, F$ - the midpoint of $B C$. Points $M, O, F$ belong to the plane $A T F$. We will prove that they lie on the same line. The height $A F=h$ of the base $A B C$ is divided by point $H$ in the ratio $2: 1$, counting from the vertex $A$, and $A F=\frac{a \sqrt{3}}{2}, \quad A H=\frac{a \sqrt{3}}{3}$.

Draw in the plane $A T F$ the line $T G \| A F$, where $G$ lies on the line $M F$. Triangles $G T M$ and FAM are equal, $G T=A F=h$. Let point $O_{1}$ be the intersection of $G F$ and $T H$. Triangles $G T O_{1}$ and $F H O_{1}$ are similar, and $\frac{T O_{1}}{O_{1} H}=\frac{3}{1}$, and $O_{1}=O$.

Let $R$ be the radius of the sphere circumscribed around the pyramid.
![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-22.jpg?height=1048&width=532&top_left_y=1398&top_left_x=1411)

$\frac{T O}{O H}=\frac{3}{1}, T O=R, O H=\frac{R}{3}$. By the Pythagorean theorem for triangle

AOH we have $R^{2}=R^{2} / 9+a^{2} / 3, a^{2}=8 R^{2} / 3, a=2, R=\sqrt{6}$.

Since the section plane is parallel to the median $A D$ of the lateral face $A T C$, draw a line $M N \| A D$ through point $M$, $N \in T C, T N=N D=T C / 4$.

The line $M N$ belongs to the section plane. Point $Q$ -

![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-23.jpg?height=357&width=509&top_left_y=130&top_left_x=1416)
is the intersection of lines $M N$ and $A C$. In the plane of the lateral face ATC, draw the line $T P \| A C$, where $P$ lies on the line $M N$. Triangles $T P N$ and $C Q N$ are similar, and $\frac{T P}{Q C}=\frac{T N}{N C}=\frac{1}{3}$, triangles $Q M A$ and $P M T$ are equal, $T P=Q A=a / 2$.

In the base plane $A B C$, connect points $Q$ and $F$, point

![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-23.jpg?height=420&width=611&top_left_y=521&top_left_x=1408)
K - the intersection of lines $Q D$ and $A B$. In triangle $A B C$, draw the midline $F S$. Triangles $Q A K$ and $F S K$ are equal, $S K=K A=a / 4$.

Quadrilateral $M N F K$ is the desired section

Find the angle of inclination of the section plane to the base plane. These planes intersect along the line $Q F$. From points $H$ and $A$, draw perpendiculars $H L$ and $A E$ to the line $Q F \cdot \frac{H D}{A D}=\frac{H L}{A E}=\frac{1}{3}$. Denote $H L=x$. Then $A E=3 x$.

Consider triangle $Q K A$. We have

![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-23.jpg?height=420&width=620&top_left_y=1389&top_left_x=1363)
$Q A=a / 2, A K=a / 4, \angle Q A K=120^{\circ} . \quad$ By the cosine theorem, we get $Q K^{2}=\frac{a^{2}}{4}+\frac{a^{2}}{16}+\frac{a^{2}}{8}=\frac{7 a^{2}}{16}, Q K=\frac{a \sqrt{7}}{4}$. Calculating the area of triangle $Q K A$, we have $S_{Q K A}=\frac{1}{2} Q K \cdot 3 x, S_{Q K A}=\frac{1}{2} Q A \cdot A K \sin 60^{\circ}, Q K \cdot 3 x=\frac{\sqrt{3}}{2} Q A \cdot A K, \frac{3 \sqrt{7}}{4} x=\frac{a \sqrt{3}}{16}, x=\frac{a}{4 \sqrt{21}}$.

Let $\varphi$ be the angle between the section plane and the base plane. Then from triangle $O H L$ we have $\operatorname{tg} \varphi=\frac{O H}{H L}=\frac{R}{3 x}=\frac{4 R \sqrt{21}}{3 a}=\sqrt{14}, \cos \varphi=\frac{1}{\sqrt{15}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-23.jpg?height=339&width=396&top_left_y=2212&top_left_x=1481)

Find the area of the projection of the section on the base plane.

$$
\begin{aligned}
& S_{n p}=S_{A B C}-S_{K B F}-S_{A K M_{1}}-S_{D C N_{1}}-S_{A M_{1} N_{1} C}= \\
& =S_{A B C}\left(1-\frac{3}{4} \cdot \frac{1}{2}-\frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{3}-\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{1}{3}-\frac{1}{3} \cdot \frac{7}{8}\right)=\frac{a^{2} \sqrt{3}}{24} . \\
& S_{c e 4}=\frac{S_{n p}}{\cos \varphi}=\frac{a^{2} \sqrt{3}}{24 \cos \varphi}=\frac{16 \sqrt{5}}{8}=2 \sqrt{5} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-24.jpg?height=369&width=477&top_left_y=201&top_left_x=1415)

Answer: $2 \sqrt{5}$.