16.15. (England, 68). Find the maximum number of points that can be placed on a sphere of radius 1 so that the distance between any two of them is: a) not less than $\sqrt{2} ;$ b) greater than $\sqrt{2}$.
16.15. a) Let's prove that if points $A_{1}, A_{2}, \ldots, A_{n}$ are located on a sphere with center $O$ and radius 1 such that the distance between any points $A_{i}, A_{j} (i \neq j)$ is at least $\sqrt{2}$, then $n \leqslant 6$. Indeed, let $n>6$. By the cosine theorem, we have

$$
A_{i} A_{j}^{2}=2-2 \cos \angle A_{i} O A_{j} \geq 2
$$

from which $\angle A_{i} O A_{j} \geqslant 90^{\circ}$ and $\overrightarrow{O A_{i}} \cdot \overrightarrow{O A_{j}} \leq 0$. Let's choose a rectangular coordinate system in space with the origin at point $O$ as follows. First, note that among the vectors $\overrightarrow{O A}_{i}$, there must be three non-coplanar vectors (otherwise, all $n>4$ vectors would lie in the same plane, and the angle between some two of them would be acute). Without loss of generality, we can assume that the vectors $\overrightarrow{O A}_{1}, \overrightarrow{O A}_{2}, \overrightarrow{O A}_{3}$ are non-coplanar. The axis $O X$ is directed along the line $O A_{i}$ so that the abscissa of point $A_{1}$ is 1. Next, the axis $O Y$ is directed so that point $A_{2}$ lies in the plane $X O Y$ and has a positive ordinate. Finally, the axis $O Z$ is directed so that the applicate of point $A_{3}$ is positive. Let $x_{i}, y_{i}, z_{i}$ be the coordinates of point $A_{i}$. Then

$$
\overrightarrow{O A}_{1}=(1 ; 0 ; 0), \overrightarrow{O A}_{2}=\left(x_{2} ; y_{2} ; 0\right), \overrightarrow{O A}_{3}=\left(x_{8} ; y_{8} ; z_{3}\right),
$$

where $y_{2}>0, z_{3}>0$, so for all values of $i>1$ we have

$$
\overrightarrow{O A}_{i} \cdot \overrightarrow{O A}_{1}=x_{i} \leqslant 0
$$

further, for all values of $i>2$ we have

$$
\overrightarrow{O A}_{i} \cdot \overrightarrow{O A_{2}}=x_{i} x_{2}+y_{i} y_{2} \leqslant 0
$$

from which $y_{i} \leqslant 0$ (since $x_{2} \leqslant 0, x_{i} \leqslant 0, y_{2}>0$); finally, for all values of $i>3$ we have

$$
\overrightarrow{O A}_{i} \cdot \overrightarrow{O A_{3}}=x_{i} x_{3}+y_{i} y_{3}+z_{i} z_{3} \leqslant 0
$$

from which $z_{i} \leqslant 0$ (since $x_{3} \leqslant 0, x_{i} \leqslant 0, y_{3} \leqslant 0, y_{i} \leqslant 0, z_{3}>0$). Among the four vectors $\overrightarrow{O A}_{4}, \overrightarrow{O A}_{5}, \overrightarrow{O A}_{8}, \overrightarrow{O A}_{7}$, at least two have the same negative coordinate, and therefore their scalar product is positive. The obtained contradiction proves that $n \leqslant 6$. Since 6 points with coordinates

$$
( \pm 1 ; 0 ; 0),(0 ; \pm 1 ; 0),(0 ; 0 ; \pm 1)
$$

satisfy the condition of part a), the maximum number of points is 6. 276
b) Let's prove that if points $A_{1}, A_{2}, \ldots, A_{n}$ are located on the same sphere such that the distance between any points $A_{i}, A_{j} (i \neq j)$ is greater than $\sqrt{2}$, then $n \leq 4$. Then $\overrightarrow{O A}_{i} \cdot \overrightarrow{O A}_{j} < 0$ for all $i \neq j$. For $n \geq 3$ we will have $x_{i} < 1 / 2$ and $n < 6$ 

(the last inequality also follows from the fact that the dihedral angle $\angle A_{1} H A_{2} = 360^{\circ} / n$ at the lateral edge $O A_{0}$ is greater than the plane angle $\angle A_{1} A_{0} A_{2} = 60^{\circ}$ of the pyramid's base), and for the other values of $n=2,3,4,5$ we have

$$
r_{n}=\sqrt{1-\frac{1}{4 \sin ^{2}\left(180^{\circ} / n\right)}}
$$