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

$$
\frac{4 a \sin ^{2} t+4 a(1+2 \sqrt{2}) \cos t-4(a-1) \sin t-5 a+2}{2 \sqrt{2} \cos t-\sin t}=4 a
$$

has exactly two distinct solutions in the interval $(0 ; \pi / 2)$.
Solution. Let $x=\cos t, y=\sin t, x^{2}+y^{2}=1, x \in(0 ; 1), y \in(0 ; 1)$. Then the equation will have the form

$$
\begin{gathered}
\frac{4 a y^{2}+4 a(1+2 \sqrt{2}) x-4(a-1) y-5 a+2}{2 \sqrt{2} x-y}=4 a \\
4 a y^{2}+4 a(1+2 \sqrt{2}) x-4(a-1) y-5 a+2=4 a 2 \sqrt{2} x-4 a y \\
4 a y^{2}+4 a x+4 y-5 a+2=0, \quad 2 \sqrt{2} x-y \neq 0
\end{gathered}
$$

In the end, we have the system:

$$
\left\{\begin{array} { c } 
{ 4 a y ^ { 2 } + 4 a x + 4 y - 5 a + 2 = 0 , } \\
{ 2 \sqrt { 2 } x - y \neq 0 , } \\
{ x ^ { 2 } + y ^ { 2 } = 1 , x \in ( 0 ; 1 ) , y \in ( 0 ; 1 ) . }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
4 y=4 a x^{2}-4 a x+a-2, \\
y \neq 2 \sqrt{2} x, \\
x^{2}+y^{2}=1, x \in(0 ; 1), y \in(0 ; 1) .
\end{array} \Leftrightarrow\right.\right.
$$

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

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 distinct solutions for $a>6$, except in the case where the parabola passes through the intersection point of the circle and the line $y=2 \sqrt{2} x$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b07d651c344f50283694g-05.jpg?height=768&width=794&top_left_y=147&top_left_x=1176)
Solve the system $\left\{\begin{array}{c}y=2 \sqrt{2} x \\ x^{2}+y^{2}=1, x \in(0 ; 1), y \in(0 ; 1) \text {. }\end{array}\right.$ 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 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: $a \in(6 ; 18+24 \sqrt{2}) \cup(18+24 \sqrt{2} ;+\infty)$

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-05.jpg?height=911&width=1513&top_left_y=1506&top_left_x=360)

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 $\overrightarrow{B C}=\{20 ;-10 ; 0\}, \overrightarrow{B E}=\{20 ;-30 ; 20\}$, $\overrightarrow{A E}=\{20 ; 10 ; 20\}, \overrightarrow{A F}=\{40 ; 40 ; 20\}$. 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 as a normal vector to the plane $B C E$, we can choose the vector $\vec{n}=\{1 ; 2 ; 2\}$. For the second system, we have $\left\{\begin{array}{c}e=f \\ d=-3 f / 2\end{array}\right.$, and as a normal vector to the plane $A F E$, we can choose the vector $\vec{m}=\{-3 ; 2 ; 2\}$. 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: $\frac{5 \sqrt{17}}{51}$.

Profile: Computer modeling and graphics; mathematics and engineering graphics tour

Problem 1 (10 points). Find the millionth digit after the decimal point in the decimal representation of the fraction $1 / 41$.

Problem 2 (10 points). Given the vertices of a regular 120-gon $A_{1}$, $A_{2}, A_{3}, \ldots, A_{120}$. In how many ways can three vertices be chosen from them to form an obtuse triangle? Justify your answer.

Problem 3 (12 points). 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 $(-\pi / 2 ; 0)$.

Problem 4a (10 points). See the reverse side of the sheet.

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

Problem 5 (20 points). See sheet 2.

## Problem $4 a$.

Given are two projections of triangle $\mathrm{