6. Edges $A B$ and $C D$ of the triangular pyramid $A B C D$ are mutually perpendicular and equal in length to 4. The lengths of edges $B C$ and $B D$ are 3 and 2, and the lengths of edges $A C$ and $A D$ are $2 \sqrt{3}$ and $\sqrt{17}$, respectively. Find the volume of the pyramid.

## Answers and solutions

Problem 1 Answer: $E_{f}=[-1 ; 9]$
Solution

The number $a \in E_{f}$, if the equation $\frac{7 x^{2}+22 x+16}{x^{2}+2 x+2}=a$ has a solution:

$x^{2}+2 x+2 \neq 0 \rightarrow 7 x^{2}+22 x+16=a x^{2}+2 a x+2 a \rightarrow(a-7) x^{2}+2(a-11) x+2 a-16=0$.

For $a \neq 7$, the quadratic equation has a solution if its discriminant is non-negative:

$D / 4=(a-11)^{2}-(a-7)(2 a-16) \geq 0 \rightarrow a^{2}-8 a-9 \leq 0 \rightarrow a \in[-1,9], a \neq 7$.

For $a=7$, the equation becomes: $-8 x-2=0$ and it also has a solution $x=-1 / 4$. Therefore, $a=7$ also belongs to the range of the function.

Problem 2 Answer: $\cos 2 x=-\frac{1}{3}$

Solution

Let $a=\operatorname{arctg} \sqrt[6]{2}$.

Solving the equation $\sin (x+3 a)=3 \sin (a-x)$:

$\sin x \cos 3 a+\cos x \sin 3 a=3(\sin a \cos x-\cos a \sin x) \rightarrow$

$\rightarrow \sin x(\cos 3 a+3 \cos a)+\cos x(\sin 3 a-3 \sin a)=0 \rightarrow$

$\rightarrow 4 \sin x \cdot \cos ^{3} a-4 \cos x \cdot \sin ^{3} a=0 \rightarrow \operatorname{tg} x=\operatorname{tg}^{3} a=\sqrt{2}$

The function $\cos 2 x=\frac{1-\operatorname{tg}^{2} x}{1+\operatorname{tg}^{2} x}=\frac{1-2}{1+2}=-\frac{1}{3}$

Problem 3 Answer: $x=6 m, m \in Z, m \geq 1$

Solution

Notation: $\stackrel{p}{=} b$, if $a-b$ is divisible by $p$. Properties: if $a^{p}=b$ and $c^{p}=d$, then $a \cdot c \stackrel{p}{=} b \cdot d$ and $a+c \stackrel{p}{=} b+d, a^{k} \stackrel{p}{=} b^{k}$ for any natural $k$.

Case $1 . x=3 m$

Then

$$
\begin{aligned}
& (x-2)=3 m-2 \stackrel{3}{=}^{1} \rightarrow(x-2)^{x+1} \stackrel{3}{=}=1 \\
& (x+2)=3 m+2 \stackrel{3}{=}_{=}-1 \rightarrow(x+2)^{x}=(-1)^{3 m}=(-1)^{m}
\end{aligned}
$$

If $m=2 k$ is even, then the expression $(x-2)^{x+1}-(x+2)^{x}=1-1=0$, i.e., it is divisible by 3 without a remainder and $x=6 k, k \in Z, k \geq 1$.

Case $2 . x=3 m+1$

Then

$(x-2)=3 m-1 \stackrel{3}{=}-1 \rightarrow(x-2)^{x+1} \stackrel{3}{=}(-1)^{3 m+2}=(-1)^{m}$ the expression $(x-2)^{x+1}-(x+2)^{x}=(-1)^{m} \neq 0$.

$(x+2)=3 m+3 \stackrel{3}{=} 0 \rightarrow(x+2)^{x}=0$

Thus, in case 2, the expression is not divisible by 3 for any $m$.

Case 3. $x=3 m-1$

Then

$(x-2)=3 m-3 \stackrel{3}{=} 0$

and the expression $(x-2)^{x+1}-(x+2)^{x} \stackrel{3}{=}(-1) \neq 0$.

$(x+2)=3 m+1=1)^{3} \rightarrow(x+2)^{x}=1$.

Thus, in case 3, the expression is not divisible by 3 for any $m$.

Problem 4 Answer: four solutions $d=1,2,3,6$

Solution

For any $n$ there exists $m$ such that $a_{n}^{2}=a_{m} \rightarrow(3+d(n-1))^{2}=3+d(m-1)$. This is possible only for $d>0$. Let $t=n-1$. Then $d(m-1)=6+6 d t+d^{2} t^{2} \rightarrow m-1=\frac{6}{d}+6 t+d t^{2}$. By the condition, the right-hand side of the equation is an integer for any integer $t \geq 0$. Substituting $t=0$ we get that $\frac{6}{d}$ is an integer. Then $d t^{2}=m-1-\frac{6}{d}-6 t$ is an integer for any $t$. Setting $t=1$ we get that $d=m-7-\frac{6}{d}$ is an integer. Therefore, the sought difference of the progression $d>0$ and is a divisor of the number 6.

Problem 5 Answer: $a=0, a=1$

Solution.

The number of integer solutions $(x ; y)$ of the first inequality in the system does not exceed 13 for any $a$ and equals 13 for integer $a$. The geometric portrait of the first inequality is a square with diagonals parallel to the coordinate axes, with a length of 4, centered at the point $(3 a-3 ; 2 a)$. For integer $a$, it contains 13 points with integer coordinates, for non-integer $a$ there are fewer such points. The points $\left\{\begin{array}{c}x=3 n-3 \\ y=2 n\end{array}, n \in Z\right.$ inside the square $|x|+|y| \leq 6$ are only two: $|3 n-3|+|2 n| \leq 6 \leftrightarrow\left[\begin{array}{c}\left\{\begin{array}{c}n \geq 1 \\ 5 n \leq 9\end{array}\right. \\ \{n=1 . \\ \left\{\begin{array}{c}n \leq 0 \\ 5 n \geq-3\end{array}\right.\end{array} \rightarrow n=0\right.$.

For $n=0$ the square $|x+2|+|y| \leq 2$ is completely inside the square $|x|+|y| \leq 6$ and therefore the system has 13 integer solutions. For $n=1$ the points with integer coordinates of the square $|x|+|y-2| \leq 2$ lie within the square $|x|+|y| \leq 6$, and the system has the maximum number of solutions 13.

Problem 6 Answer: $v=\frac{\sqrt{71}}{3}$

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_63c93fb681b93ab9d7e4g-03.jpg?height=803&width=603&top_left_y=501&top_left_x=178)

$B P=h_{1}, A P=h_{2}, K P=h_{3}$

1) $\sqrt{4-h_{1}^{2}}+\sqrt{9-h_{1}^{2}}=4 \quad u_{1}=\sqrt{4-h_{1}^{2}}, v_{1}=\sqrt{9-h_{1}^{2}} \rightarrow\left\{\begin{array}{l}u_{1}+v_{1}=4 \\ v_{1}^{2}-u_{1}^{2}=5\end{array} \rightarrow\left\{\begin{array}{l}u_{1}+v_{1}=4 \\ v_{1}-u_{1}=\frac{5}{4}\end{array}\right.\right.$

$v_{1}=\frac{21}{8} \rightarrow h_{1}^{2}=\frac{135}{64}$

2) $\sqrt{17-h_{2}^{2}}+\sqrt{12-h_{2}^{2}}=4, h_{2}^{2}=\frac{647}{64}$
3) $\sqrt{\frac{647}{64}-h_{3}^{2}}+\sqrt{\frac{135}{64}-h_{3}^{2}}=4 \rightarrow h_{3}^{2}=\frac{71}{64}$
4) $V=\frac{1}{3} S_{A B P} \cdot C D=\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{\sqrt{71}}{8} \cdot 4 \cdot 4=\frac{\sqrt{71}}{3}$