5. Find the largest natural number $n$ for which there exist $n$ lines in space passing through the same point and the angles between any two of them are equal. (The angle between two intersecting lines is defined to be the smaller of the two angles formed by the lines).
Solution. If you are parallel to the vectors

$$
\begin{aligned}
& \left(\frac{1}{\sqrt{5}}, 0, \frac{2}{\sqrt{5}}\right),\left(\frac{1}{\sqrt{5}}, \sqrt{\frac{5+\sqrt{5}}{10}}, \frac{5-\sqrt{5}}{10}\right),\left(\frac{1}{\sqrt{5}},-\sqrt{\frac{5+\sqrt{5}}{10}}, \frac{5-\sqrt{5}}{10}\right) \\
& \left(\frac{1}{\sqrt{5}}, \sqrt{\frac{5-\sqrt{5}}{10}},-\frac{5+\sqrt{5}}{10}\right),\left(\frac{1}{\sqrt{5}},-\sqrt{\frac{5-\sqrt{5}}{10}},-\frac{5+\sqrt{5}}{10}\right),(1,0,0)
\end{aligned}
$$

then they satisfy the conditions of the problem.

We will prove that there do not exist seven lines that satisfy the conditions of the problem. Suppose there are lines $l_{i}, i=1,2,3,4,5,6,7$, that satisfy the condition of the problem and let $\vec{r}_{i}, i=1,2,3,4,5,6,7$ be unit vectors parallel to the given lines. Without loss of generality, we can take

$$
\vec{r}_{1}=(1,0,0) ; \vec{r}_{2}=(k, 0, c), c \neq 0 ; \quad \vec{r}_{i}=\left(a_{i}, b_{i}, c_{i}\right), \quad i=3,4,5,6,7
$$

Since the vectors $\vec{r}_{i}, i=1,2,3,4,5,6,7$ are unit vectors, from the condition of the problem it follows that the pairwise dot products are equal. Thus,

$$
\vec{r}_{1} \vec{r}_{2}=(1,0,0)(k, 0, c)=k ; \quad \vec{r}_{1} \vec{r}_{i}=(1,0,0)\left(a_{i}, b_{i}, c_{i}\right)=a_{i}, i=3,4,5,6,7
$$

so $a_{i}= \pm k, i=3,4,5,6,7$, i.e., at least for three of the vectors $\vec{r}_{i}, i=3,4,5,6,7$ the first coordinate is equal to $k$. Let these be the vectors $\vec{r}_{i}, i=3,4,5$, i.e.

$$
\vec{r}_{i}=\left(k, b_{i}, c_{i}\right), \quad i=3,4,5
$$

Now, $\vec{r}_{2} \vec{r}_{i}=(k, 0, c)\left(k, b_{i}, c_{i}\right)=k^{2}+c c_{i}, i=3,4,5$ and since $\vec{r}_{2} \vec{r}_{3}=\vec{r}_{2} \vec{r}_{4}=\vec{r}_{2} \vec{r}_{5}$ we get

$$
k^{2}+c c_{3}=k^{2}+c c_{4}=k^{2}+c c_{5}
$$

i.e.

$$
\begin{aligned}
& c\left(c_{3}-c_{4}\right)=0 \\
& c\left(c_{3}-c_{5}\right)=0
\end{aligned}
$$

and since $c \neq 0$ we get $c_{3}=c_{4}=c_{5}$. However, the vectors $\vec{r}_{i}, i=3,4,5$, are unit vectors, so $k^{2}+b_{3}^{2}+c_{3}^{2}=k^{2}+b_{4}^{2}+c_{3}^{2}=k^{2}+b_{5}^{2}+c_{3}^{2}=1$ i.e. $b_{3}^{2}=b_{4}^{2}=b_{5}^{2}$ which means $b_{3}, b_{4}, b_{5} \in\left\{b_{3},-b_{3}\right\}$. According to this, $b_{3}=b_{4}$ or $b_{3}=b_{5}$ or $b_{4}=b_{5}$, which means $\vec{r}_{3}=\vec{r}_{4}$ or $\vec{r}_{3}=\vec{r}_{5}$ or $\vec{r}_{5}=\vec{r}_{4}$, which is a contradiction.