3. Find all values of the parameter $a$ for which the equation

$$
\frac{4 a \cos ^{2} t+4 a(2 \sqrt{2}-1) \cos t+4(a-1) \sin t+a+2}{\sin t+2 \sqrt{2} \cos t}=4 a
$$

has exactly two distinct solutions in the interval $(-\pi / 2 ; 0)$.

(12 points)
Solution. Let $x=\cos t, y=\sin t, x^{2}+y^{2}=1, x \in(0 ; 1), y \in(-1 ; 0)$. Then the equation will have the form

$$
\begin{gathered}
\frac{4 a x^{2}+4 a(2 \sqrt{2}-1) x+4(a-1) y+a+2}{2 \sqrt{2} x+y}=4 a \\
4 a x^{2}+4 a(2 \sqrt{2}-1) x+4(a-1) y+a+2=4 a 2 \sqrt{2} x+4 a y \\
4 a x^{2}-4 a x-4 y+a+2=0, \quad 2 \sqrt{2} x+y \neq 0
\end{gathered}
$$

In the end, we have the system:

$$
\left\{\begin{array} { c } 
{ 4 a x ^ { 2 } - 4 a x - 4 y + a + 2 = 0 , } \\
{ 2 \sqrt { 2 } x + y \neq 0 , } \\
{ x ^ { 2 } + y ^ { 2 } = 1 , x \in ( 0 ; 1 ) , y \in ( - 1 ; 0 ) . }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
4 y=4 a x^{2}-4 a x+a+2, \\
y \neq-2 \sqrt{2} x, \\
x^{2}+y^{2}=1, x \in(0 ; 1), y \in(-1 ; 0)
\end{array} \Leftrightarrow\right.\right.
$$

$$
\left\{\begin{array}{c}
y=a\left(x-\frac{1}{2}\right)^{2}+\frac{1}{2} \\
y \neq-2 \sqrt{2} x \\
x^{2}+y^{2}=1, x \in(0 ; 1), y \in(-1 ; 0)
\end{array}\right.
$$

Find the value of the parameter $a$ for which the parabola $y=a\left(x-\frac{1}{2}\right)^{2}+\frac{1}{2}$ passes through the point $(0 ;-1)$. We have $-1=\frac{a}{4}+\frac{1}{2}, a=-6$. Since the parabola is symmetric with respect to the line $x=\frac{1}{2}$, the system will have two different solutions for $a<-6$, except for the case when the parabola passes through the intersection point of the circle and the line $y=-2 \sqrt{2} x$. Solve the system $\left\{\begin{array}{c}y=-2 \sqrt{2} x, \\ x^{2}+y^{2}=1, x \in(0 ; 1), y \in(-1 ; 0) .\end{array}\right.$ We have $x=\frac{1}{3}, y=-\frac{2 \sqrt{2}}{3}$. Find the value of the parameter $a$ for which the parabola $y=a\left(x-\frac{1}{2}\right)^{2}-\frac{1}{2}$ passes through

![](https://cdn.mathpix.com/cropped/2024_05_06_b07d651c344f50283694g-11.jpg?height=806&width=762&top_left_y=428&top_left_x=1184)
the point $(1 / 3 ;-2 \sqrt{2} / 3)$. Solve the equation $-\frac{2 \sqrt{2}}{3}=a\left(\frac{1}{3}-\frac{1}{2}\right)^{2}+\frac{1}{2}, a=-18-24 \sqrt{2}$.

Answer: $a \in(-\infty ;-18-24 \sqrt{2}) \cup(-18-24 \sqrt{2} ;-6)$

4b. Find the angle between the planes $B C E$ and $A F E$ (see the condition of problem 4a). (8 points)

## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_b07d651c344f50283694g-11.jpg?height=906&width=1516&top_left_y=1620&top_left_x=361)

Introduce a coordinate system as shown in the figure. Then $A=(0 ; 0 ; 0), B=(0 ; 40 ; 0), E=$ $(20 ; 10 ; 20), F=(40 ; 40 ; 20), C=(20 ; 30 ; 0)$. We have $\quad \overrightarrow{B C}=\{20 ;-10 ; 0\}, \overrightarrow{B E}=\{20 ;-30 ; 20\}$,
$\overrightarrow{A E}=\{20 ; 10 ; 20\}, \overrightarrow{A F}=\{40 ; 40 ; 20\}$. Let $\vec{n} \perp \overrightarrow{B C}, \vec{n} \perp \overrightarrow{B E}, \vec{m} \perp \overrightarrow{A E}, \vec{m} \perp \overrightarrow{A F}, \vec{n}=\{a ; b ; c\}$, $\vec{m}=\{d ; e ; f\}$. Then we have two systems of linear algebraic equations:

$\left\{\begin{array}{c}20 a-10 b=0, \\ 20 a-30 b+20 c=0,\end{array} \quad\left\{\begin{array}{l}20 d+10 e+20 f=0, \\ 40 d+40 e+20 f=0 .\end{array}\right.\right.$ For the first system, we have $\left\{\begin{array}{l}b=2 a, \\ c=2 a,\end{array}\right.$ and we can choose the vector $\vec{n}=\{1 ; 2 ; 2\}$ as the normal vector to the plane $B C E$. For the second system, we have $\left\{\begin{array}{c}e=f \\ d=-3 f / 2\end{array}\right.$, and we can choose the vector $\vec{m}=\{-3 ; 2 ; 2\}$ as the normal vector to the plane $A F E$. The angle between the planes is equal to the angle between the normals to these planes. Therefore, the cosine of the desired angle $\alpha$ can be found using the formula

$$
\cos \alpha=\frac{|\vec{n} \cdot \vec{m}|}{|\vec{n}| \cdot|\vec{m}|}=\frac{5}{3 \sqrt{17}}=\frac{5 \sqrt{17}}{51}
$$

Answer: $\arccos \frac{5 \sqrt{17}}{51}$.

## Olympiad for Schoolchildren "Step into the Future"

Profile: Computer Modeling and Graphics; Mathematics and Engineering Graphics

## Variant: 3

Class: 11

Problem 1 (10 points). The sequence of real numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{100}$ satisfies the inequalities $a_{n}-2022 a_{n+1}+2021 a_{n+2} \geq 0$ for $n=1,2,3, \ldots, 98$, and $a_{99