Answer: \(12\). Construction. Let \(ABCD\) be a regular tetrahedron with center \(O\). Then choose the rays \(a_0, a_1, a_2\) through \(O\) so that they make very acute angles with \(OA\). Similarly define the rays \(b_i, c_i, d_i\) for \(i\in\{0, 1, 2\}\). Then for each ray among these \(12\) rays there are exactly two which make with it acute angles. For example for the ray \(a_i\) only \(a_j\) and \(a_k\) make acute angles with it where \(\{i, j, k\} = \{0, 1, 2\}\). Estimation. Let \(S\) be the set of the given rays. For any ray \(x\in S\) define \(S_x\subset S\setminus\{x\}\) as the subset of the rays which make acute angle with \(x\). According to the problem condition \(|S_x| = 2\) for any \(x\in S\). Let \(O\) be the center of the rays. For any ray \(x\) from \(O\) draw the plane \(\pi_x\) through \(O\) perpendicular to \(x\) and denote the half-spaces containing \(x\) and not containing \(x\) by \(H_x^+\) and \(H_x^-\), respectively, so that \(\pi_x\subset H_x^-\) and \(\pi_x\cap H_x^+=\{O\}\). Suppose for the sake of contradiction that \(|S|\ge 13\). Take any ray \(a\in S\). Then exactly two other rays of \(S\) lie in \(H_a^+\). Choose another ray \(b\subset H_a^-\) of \(S\). Again exactly two other rays of \(S\) lie in \(H_b^+\). Then choose another ray \(c\subset H_a^-\cap H_b^-\) of \(S\). Exactly two other rays of \(S\) lie in \(H_c^+\). At last take another ray \(d\subset H_a^-\cap H_b^-\cap H_c^-\). Exactly two other rays of \(S\) lie in \(H_d^+\). Thus, up to this moment we have counted \(12\) rays. Let \(e\in S\) be the ray we have not yet come across. Clearly \(e\subset H_a^-\cap H_b^-\cap H_c^-\cap H_d^-\). Then one of \(a, b, c, e\) is the complement ray of \(d\). Indeed, suppose that none of \(a, b, c\) complements \(d\). It is not difficult to see that if not all the rays among \(a, b, c\) lie in \(\pi_d\) then \(H_a^-\cap H_b^-\cap H_c^- \subset H_d^+\). Thus all of \(a, b, c\) lie in \(\pi_d\). In this case the intersection \(H_a^-\cap H_b^-\cap H_c^- \) is a ray which should coincide with \(e\). Clearly \(e\perp \pi_d=\pi_e\). In the other case, i.e. when one of \(a, b, c\) is the complement of \(d\) we can simply swap its notation with \(e\). So we can safely assume from now on that the complement of \(d\) is \(e\) and \(a, b, c\subset \pi_d\). Among the rays of \(S\) three lie in \(H_e^+\) and three lie in \(H_d^+\) so the remaining at least \(7\) should lie in \(\pi_d\). Let \(T\) be the set of these at least \(7\) rays. Let \(S_d = \{d_1, d_2\}\) and \(S_e = \{e_1, e_2\}\). Note that for each \(x\in S_d\cup S_e\) exactly one ray of \(S_x\) lies in \(T\). Indeed, consider \(d_1\) as \(x\). \(d\in S_{d_1}\) thus no more than one ray of \(S_{d_1}\) may lie in \(T\). Suppose that \(S_{d_1}\cap T=\varnothing\). But then all the rays of \(T\) lie in the closed half-plane \(\pi_d\cap H_{d_1}^-\). Indeed, if we divide this half-plane into two quadrants then one of these quadrants (with its borders) will contain at least \(\lceil 7/2 \rceil =4\) rays of \(T\). Thus one of them will make acute angles with three others — contradiction. Consequently \(|S_{d_1}\cap T|=1\). It follows that \(S_{d_1}=\{d, t_1\}\) and \(S_{d_2}=\{d, t_2\}\) for some \(t_1, t_2\in T\). Particularly \(d_2\notin S_{d_1}\). So \(d_2\in H_{d_1}^-\) and it can be checked that the intersection of \(\pi_d\cap H_{d_1}^-\) and \(\pi_d\cap H_{d_2}^-\) is an infinite closed angle (with its border lines) with degree no more than \(\pi/2\) (to see this, for example, notice that the projection of \(d_2\) on \(\pi_d\) does not lie in \(\pi_d\cap H_{d_1}^+\)). But this angle contains \(|T\setminus\{t_1, t_2\}|\ge 5\) rays so one of them makes acute angles with at least three (even four) of them. This contradiction finishes the proof.

kiyoras_2001 wrote: It is not difficult to see that if not all the rays among \(a, b, c\) lie in \(\pi_d\) then \(H_a^-\cap H_b^-\cap H_c^- = \{O\}\). Thus all of \(a, b, c\) lie in \(\pi_d\). Can you please explain why this is true? If $a,b,c,d$ are $OA,OB,OC,OD$ from the beginning of your solution, then they don't lie in $\pi_d$, why this isn't the case?

Thanks, atdaotlohbh for finding the flaw! I am happy that there are people who are interested in the geometry of higher dimensions rather than just the flat plane . In fact I should have had written that "if not all the rays among \(a, b, c\) lie in \(\pi_d\) then \(H_a^-\cap H_b^-\cap H_c^- \subset H_d^+\)" but now I noted that there is a slight problem with this claim so edited this paragraph a little bit. Now the solution should be correct, I guess.

kiyoras_2001 wrote: That is, at least six rays of \(T\) lie in this half-plane. But then, if we divide this half-plane into two quadrants (for example one closed and one open) then one of these quadrants will contain a ray which makes acute angles with three rays -- contradiction. I don't see why... Why can't it be the case when we have 3 almost identical rays pointing to the left, and 3 almost identical rays pointing to the right? kiyoras_2001 wrote: Note that \(S_x\subset T\) for each \(x\in S_d\cup S_e\). But if $x \in S_d$, then $d \in S_x \subset T$, but $d \notin T$, which is a contradiction, and same with $e$, so I don't see why you need the last paragraph

atdaotlohbh wrote: kiyoras_2001 wrote: That is, at least six rays of \(T\) lie in this half-plane. But then, if we divide this half-plane into two quadrants (for example one closed and one open) then one of these quadrants will contain a ray which makes acute angles with three rays -- contradiction. I don't see why... Why can't it be the case when we have 3 almost identical rays pointing to the left, and 3 almost identical rays pointing to the right? Oops. This time the flaw you discovered was fatal for my argument, so I had to rewrite completely the last two paragraphs. atdaotlohbh wrote: kiyoras_2001 wrote: Note that \(S_x\subset T\) for each \(x\in S_d\cup S_e\). But if $x \in S_d$, then $d \in S_x \subset T$, but $d \notin T$, which is a contradiction, and same with $e$, so I don't see why you need the last paragraph Very good remark. How could I overlook this simple implication . I used this to repair the solution. Now I hope everything is correct.