11.4. $n$ vectors in space are such that any pair of them forms an obtuse angle. What is the largest value that $n$ can take?
Answer: 4. Solution. Let $\vec{a}_{1}, \vec{a}_{2}, \ldots, \vec{a}_{n}$ be the given vectors. Direct the $z$-axis of the coordinate space along $\vec{a}_{n}$. Then the $z$-coordinate of the other vectors must be negative (this follows from the formula for the cosine of the angle between vectors through the scalar product of vectors). Let $\vec{a}_{1}=\left(x_{1}, y_{1}, z_{1}\right), \ldots, \vec{a}_{n-1}=\left(x_{n-1}, y_{n-1}, z_{n-1}\right)$ be the coordinates of the vectors, and let the two-dimensional vectors $\vec{b}_{1}=\left(x_{1}, y_{1}\right), \ldots, \vec{b}_{n-1}=\left(x_{n-1}, y_{n-1}\right)$ be the projections of these vectors onto the $x O y$ plane. Then the angle between any pair of vectors $\vec{b}_{i}$ and $\vec{b}_{j}$ is obtuse: indeed, $\cos \left(\vec{b}_{i} \wedge \vec{b}_{j}\right)=\frac{x_{i} x_{j}+y_{i} y_{j}}{\left|\vec{b}_{i}\right| \cdot\left|\vec{b}_{j}\right|}<0$, since $x_{i} x_{j}+y_{i} y_{j}<x_{i} x_{j}+y_{i} y_{j}+z_{i} z_{j}<0$ (in the first of these inequalities, it is used that $z_{i}$ and $z_{j}$ are negative, and in the second, that the angle between $\vec{a}_{i}$ and $\vec{a}_{j}$ is obtuse). Thus, on the $x O y$ plane, the vectors $\vec{b}_{1}, \ldots, \vec{b}_{n-1}$ have the same property, namely: the angle between any pair of them is obtuse. Therefore, $n-1 \leq 3$ (this is easy to show if, by placing the vectors from one point, we sum the adjacent angles around the circle, since they sum up to $360^{\circ}$). Thus, we obtain the estimate $n \leq 4$. An example for four vectors can be given as follows: $\vec{a}_{1}=(1,0,-\varepsilon), \vec{a}_{2}=\left(\cos \frac{2 \pi}{3}, \sin \frac{2 \pi}{3},-\varepsilon\right), \vec{a}_{3}=\left(\cos \frac{4 \pi}{3}, \sin \frac{4 \pi}{3},-\varepsilon\right), \vec{a}_{4}=(0,0,1)$, where $\varepsilon$ is a sufficiently small positive number, which can be taken as $\varepsilon=0.1$.

## 11th Grade