35*. What is the greatest number of rays in space that form pairwise obtuse angles?
35. It is easy to see that it is possible to draw four rays, each pair of which forms obtuse angles: for example, it is sufficient to connect the center $O$ of a regular tetrahedron $ABCD$ with all its vertices (it is clear, for example, that $\angle AOB > \angle AO_1B = 90^\circ$, where $O_1$ is the center of the face $BCD$ of the tetrahedron). Now let's prove that it is impossible to specify more than four rays in space, each pair of which forms obtuse angles.

Let $l_1, l_2, l_3$, and $l_4$ be four rays emanating from a point $O$ and forming obtuse angles with each other, and let $\pi_1, \pi_2, \pi_3$, and $\pi_4$ be planes passing through the point $O$ and perpendicular to the rays $l_1, l_2, l_3$, and $l_4$, respectively. All rays forming obtuse angles with the ray $l_1$ must lie on the other side of the plane $\pi_1$ from the ray $l_1$. All rays forming obtuse angles with both rays $l_1$ and $l_2$ must lie within the dihedral angle formed by the half-planes $\pi_1$ and $\pi_2$; this dihedral angle (Fig. 83a) is acute, since the angle between the rays perpendicular to its faces is obtuse. All rays forming obtuse angles with each of the three rays $l_1, l_2, l_3$ must lie within the intersection of the dihedral angles $\pi_1\pi_2$ and $\pi_1\pi_3$. This intersection is a trihedral angle $\pi_1\pi_2\pi_3$, or $OABC$ (Fig. 83b), which, as proven above, has all dihedral angles acute. The ray $l_4$ must lie within it; we will show that it forms acute angles with all other rays lying inside this angle.

Let the plane passing through $OA$ and the ray $l_4$ intersect the face $BOC$ along the ray $OA'$. One of the dihedral angles with edge $OA'$ is acute; let this be the angle $AOA'C$. Then all dihedral angles of the trihedral angle $OAA'C$ are acute; hence, all plane angles of this trihedral angle are also acute (see, for example, the solution to problem 28 in the book by D. O. Sharygin et al., cited in the footnote on page 151). But $\angle AOL_4$ is only a part of the angle $AOA'$; therefore, the angle between the rays $OA$ and $l_4$ is acute. Similarly, the ray $l_4$ forms acute angles with $OB$ and $OC$. Therefore, all

1) Consider how this reasoning changes if we are talking about points in space (although the case where all points lie in the same plane does not contradict the condition of problem b).
three rays $OA, OB, OC$ lie on the same side of the plane $\pi_4$ as the ray $l_4$, and thus all internal points of the trihedral angle $OABC$ lie on the same side of $\pi_4$ as the ray $l_4$. Therefore, any ray $l_5$ located inside the trihedral angle $\pi_1\pi_2\pi_3$,

![](https://cdn.mathpix.com/cropped/2024_05_21_d4ffc236376258137647g-161.jpg?height=403&width=445&top_left_y=256&top_left_x=265)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_d4ffc236376258137647g-161.jpg?height=329&width=340&top_left_y=318&top_left_x=714)
b)

Fig. 83.

lies on the same side of the plane $\pi_4$, and thus no such ray $l_5$ can form an obtuse angle with $l_4$. This proves the required statement.