9. From the given origin $O$, draw 4 rays so that all 6 angles between them are congruent to each other.
9. Let point $O$ be the origin of four radius vectors $\overrightarrow{O A}_{1}, \overrightarrow{O A}_{2}, \overrightarrow{O A}_{3}, \overrightarrow{O A}_{4}$, which are equal in length and form equal angles with each other. As in the previous problem, it is not difficult to prove that each of these vectors is perpendicular to the line passing through the ends of any two other radius vectors, so any of these four vectors is perpendicular to the plane passing through the ends of the other three radius vectors. Due to the equality of the lengths of the vectors and the angles between them, the triangles $A_{1} O A_{2}, A_{1} O A_{3}$, $A_{1} O A_{4}, A_{2} O A_{3}, A_{2} O A_{4}, A_{3} O A_{4}$ are congruent to each other, and thus the triangles $A_{1} A_{2} A_{3}, A_{1} A_{2} A_{4}, A_{1} A_{3} A_{4}, A_{2} A_{3} A_{4}$ are equilateral. Perpendiculars drawn from each point $A_{k}$ to the plane of the opposite equilateral triangle will pass through $O$ and through the center of this triangle (explain). If a unit mass is placed at each of the points $A_{1}, A_{2}, A_{3}, A_{4}$, then point $O$ will be the center of these masses, since the center of mass lies on each of the lines connecting one of the points with the center of mass of the opposite triangle. For the same reasons, point $O$ divides the distance from point $A_{k}$ to the opposite plane in the ratio $3:1$. From this follows the construction.
1) Through $O$ we draw a line $a_{1}$ and take a point $A_{1}$ on it.
2) On the line $O A_{1}$ from point $O$ in the direction opposite to $A_{1}$, we lay off a segment $O O^{\prime}$ of length $\left|O O^{\prime}\right|=\frac{1}{3}\left|O A_{1}\right|$.
3) Through $O^{\prime}$ we draw a plane perpendicular to the line $a_{1}$, and on this plane from point $O^{\prime}$ we draw three rays $a_{2}, a_{3}, a_{4}$, forming angles of $120^{\circ}$ with each other.
4) On the ray $a_{2}$ we find a point $A_{2}$ such that $\left|O A_{2}\right|=\left|O A_{1}\right|$.

Similarly, on the rays $a_{3}, a_{4}$ we find points $A_{3}$ and $A_{4}$. The vectors $\overrightarrow{O A}_{1}, \overrightarrow{O A}_{2}, \overrightarrow{O A}_{3}, \overrightarrow{O A}_{4}$ satisfy the conditions of the problem. This follows from the congruence of the six triangles $A_{1} O A_{2}, A_{1} O A_{3}, A_{1} O A_{4}$, $A_{2} O A_{3}, A_{2} O A_{4}, A_{3} O A_{4}$. The common angle $\varphi$ between the vectors is determined from the equality: $\cos \varphi=-\frac{1}{3}$, from which $\varphi \approx$ $\approx 109^{\circ} 28^{\prime} 16^{\prime \prime}$.