In space, four rays originating from a point enclose the same non-zero angle with each other. What is this angle?
Solution. Let's denote the starting point of the half-lines as $O$. From $O$, measure a distance $d$ on each of the half-lines, and mark the points obtained as $A, B, C, D$ (Figure 1). The isosceles triangles $OAB, OBC, OCD, ODA, OAC, OBD$ are congruent, since their legs are $d$ and their vertex angles are $\varphi$. Our task is to determine the size of the angle $\varphi$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9fc113e960752958ea2eg-1.jpg?height=404&width=439&top_left_y=324&top_left_x=824)

Figure 1

From the previous congruence, it follows that $AB = BC = CD = DA = AC = BD$, that is, $ABC, BCD, CDA, DAB$ are congruent equilateral triangles. The four points cannot lie in the same plane, because then the angles at the same labeled $D$ would not be $60^{\circ}$ (Figure 2). Therefore, the four points are the four vertices of a regular tetrahedron.

![](https://cdn.mathpix.com/cropped/2024_05_02_9fc113e960752958ea2eg-1.jpg?height=339&width=328&top_left_y=953&top_left_x=882)

Figure 2

The point $O$ is equidistant from all four vertices, at a distance $d$, so $O$ is the center of the circumscribed sphere of the tetrahedron, and $d$ is the radius of the sphere (Figure 3). It is known (and easily calculable) that the radius of the circumscribed sphere is $d = \frac{a \sqrt{6}}{4}$, where $a$ is the edge length of the tetrahedron.

![](https://cdn.mathpix.com/cropped/2024_05_02_9fc113e960752958ea2eg-1.jpg?height=431&width=418&top_left_y=1519&top_left_x=840)

Figure 3

Let's draw the isosceles triangle $AOB$ (Figure 4). We know that $AB = a$ and $\angle AOB = \varphi$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9fc113e960752958ea2eg-1.jpg?height=273&width=250&top_left_y=2097&top_left_x=924)

Figure 4

Let $F$ be the foot of the angle bisector of $\varphi$ on the segment $AB$, then

$$
\sin \frac{\varphi}{2} = \frac{FB}{OB} = \frac{\frac{a}{2}}{\frac{a \sqrt{6}}{4}} = \frac{\sqrt{6}}{3}
$$

From this, $\frac{\varphi}{2} \approx 54.74^{\circ}$ and the angle between the half-lines $\varphi \approx 109.47^{\circ}$.