IMC 2001 · Day 2 · P9

The unit sphere in $\mathbb{R}^n$ is defined by $$S_{n-1} = \left\{ (x_1, \dots, x_n) \in \mathbb{R}^n \;\Bigl|\; \sum_{k=1}^{n} x_k^2 = 1 \right\}.$$ The distance between the points $X = (x_1, \dots, x_n)$ and $Y = (y_1, \dots, y_n)$ is: $$d^2(X, Y) = \sum_{k=1}^{n} (x_k - y_k)^2.$$ We have $$\begin{align*} d(X, Y) > \sqrt{2} &\Leftrightarrow d^2(X, Y) > 2 \\ &\Leftrightarrow \sum_{k=1}^{n} x_k^2 + \sum_{k=1}^{n} y_k^2 + 2 \sum_{k=1}^{n} x_k y_k > 2 \\ &\Leftrightarrow \sum_{k=1}^{n} x_k y_k < 0 \end{align*}$$ Taking account of the symmetry of the sphere, we can suppose that $$A_1 = (-1, 0, \dots, 0).$$ For $X = A_1$, $\sum\limits_{k=1}^{n} x_k y_k < 0$ implies $y_1 > 0$, $\forall Y \in M_n$.

Let $X = (x_1, \overline{X})$, $Y = (y_1, \overline{Y}) \in M_n \setminus \{A_1\}$, $\overline{X}, \overline{Y} \in \mathbb{R}^{n-1}$.

We have $$\sum_{k=1}^{n} x_k y_k < 0 \Rightarrow x_1 y_1 + \sum_{k=1}^{n-1} \overline{x}_k \overline{y}_k < 0 \Leftrightarrow \sum_{k=1}^{n-1} x_k' y_k' < 0,$$ where $$x_k' = \frac{\overline{x}_k}{\sqrt{\sum \overline{x}_k^2}}, \qquad y_k' = \frac{\overline{y}_k}{\sqrt{\sum \overline{y}_k^2}}.$$ therefore $$(x_1', \dots, x_{n-1}'), (y_1', \dots, y_{n-1}') \in S_{n-2}$$ and verifies $\sum\limits_{k=1}^{n} x_k' y_k' < 0$.

If $a_n$ is the search number of points in $\mathbb{R}^n$ we obtain $a_n \le 1 + a_{n-1}$ and $a_1 = 2$ implies that $a_n \le n+1$.

We show that $a_n = n+1$, giving an example of a set $M_n$ with $(n+1)$ elements satisfying the conditions of the problem. $$\begin{align*} A_1 &= (-1, 0, 0, 0, \dots, 0, 0) \\ A_2 &= \left( \tfrac{1}{n}, -c_1, 0, 0, \dots, 0, 0 \right) \\ A_3 &= \left( \tfrac{1}{n}, \tfrac{1}{n-1} \cdot c_1, -c_2, 0, \dots, 0, 0 \right) \\ A_4 &= \left( \tfrac{1}{n}, \tfrac{1}{n-1} \cdot c_1, \tfrac{1}{n-2} \cdot c_2, -c_3, \dots, 0, 0 \right) \\ &\vdots \\ A_{n-1} &= \left( \tfrac{1}{n}, \tfrac{1}{n-1} \cdot c_1, \tfrac{1}{n-2} \cdot c_2, \tfrac{1}{n-3} \cdot c_3, \dots, -c_{n-2}, 0 \right) \\ A_n &= \left( \tfrac{1}{n}, \tfrac{1}{n-1} \cdot c_1, \tfrac{1}{n-2} \cdot c_2, \tfrac{1}{n-3} \cdot c_3, \dots, \tfrac{1}{2} \cdot c_{n-2}, -c_{n-1} \right) \\ A_{n+1} &= \left( \tfrac{1}{n}, \tfrac{1}{n-1} \cdot c_1, \tfrac{1}{n-2} \cdot c_2, \tfrac{1}{n-3} \cdot c_3, \dots, \tfrac{1}{2} \cdot c_{n-2}, c_{n-1} \right) \end{align*}$$ where $$c_k = \sqrt{\left( 1 + \frac{1}{n} \right) \left( 1 - \frac{1}{n - k + 1} \right)}, \quad k = \overline{1, n-1}.$$ We have $\sum\limits_{k=1}^{n} x_k y_k = -\frac{1}{n} < 0$ and $\sum\limits_{k=1}^{n} x_k^2 = 1$, $\forall X, Y \in \{A_1, \dots, A_{n+1}\}$.

These points are on the unit sphere in $\mathbb{R}^n$ and the distance between any two points is equal to $$d = \sqrt{2} \sqrt{1 + \frac{1}{n}} > \sqrt{2}.$$

Remark. For $n = 2$ the points form an equilateral triangle in the unit circle; for $n = 3$ the four points from a regular tetrahedron and in $\mathbb{R}^n$ the points from an $n$ dimensional regular simplex.