An inequality involving vectors

Question

Let $n$ be a positive integer number. If $S$ is a finite set of vectors in the plane, let $N(S)$ denote the number of two-element subsets $\{\mathbf{v}, \mathbf{v'}\}$ of $S$ such that $$4\,(\mathbf{v} \cdot \mathbf{v'}) + (|\mathbf{v}|^2 - 1)(|\mathbf{v'}|^2 - 1) < 0. $$ Determine the maximum of $N(S)$ when $S$ runs through all $n$-element sets of vectors in the plane. Can someone help? This is from Romanian Olympiad. I have no idea about it.

Answer 1

First of all, for simplicity let us normalize the set $ S $ of $n$ vectors by setting $v_i=\frac{u_i}{\|u_i\|}$ for all $1\le i\le n $, which gives $\|v_i\|=1$. Therefore, the inequality reduces to be $$ v_i \cdot v_j <0$$

Now, let us observe that it is impossible to have two element subsets of the form $\{v_i, v_i\}$ for all $ i $. Since \begin{align} v_i\cdot v_i &=\|v_i\|\|v_i\| \cos (\theta) \,\,\,\,\, \text {when} \theta=0 \\ &=1 \end{align} which is not less than zero for all $ v_i $.

Recall that, in general if $v $ is nonzero vector then every vector $ v'$ can be written in a unique manner as the sum of a vector $ v'_{||} $ parallel $ v $ and a vector $ v'_{\perp}$ perpendicular to $ v $: $$ v'=v'_{\parallel}+v'_{\perp}.$$

If $ v_i $ is perpendicular to $ v_j $ then $ v_i\cdot v_j=0$ which does not satisfies the inequality ($0 <0$).

When both vectors are parallel : if the angle is zero (or the angle belong to $(3\pi/2,2\pi]\cup [0,\pi/2]$) then the dot product would be $> 0$ and the thus the inequality does not holds. If the vectors are parallel in opposite direction (the angle $\pi $) then the dot product is less than $0$. Moreover, if the angle $\theta\in (\pi/2,3\pi/2) $ then $ v_i \cdot v_j <0$ and the inequality holds. Therefore, since the vectors are unit and $\cos\theta <0$ then $$ \text{proj}_{v_j} v_i =(v_i \cdot v_j) v_j= -v_j<0 $$

We recall that $v_i, v_j$ are members of $ S$ that satisfies the inequality. When $S$ run over $n$-element vectors we have to consider all two element subsets of $S$ which are listed below.

The number of all possible two element subset of $ S $ (it may not satisfy the inequality): \begin{align} {S_1} &= \left\{ {{v_1},{v_2}} \right\}\,\,\,\, \text{gives one subset} \,\,\,\left\{ {{v_1},{v_2}} \right\}\\ {S_2} &= \left\{ {{v_1},{v_2},{v_3}} \right\} \,\,\,\, \text{could give two subsets of two elements} \left\{ {\left\{ {{v_1},{v_2}} \right\},\left\{ {{v_1},{v_3}} \right\},\left\{ {{v_2},{v_3}} \right\}} \right\} \\ {S_3} &= \left\{ {{v_1},{v_2},{v_3},{v_4}} \right\} \,\,\,\, \text{gives 6 subsets of two elements} \left\{ {\left\{ {{v_1},{v_2}} \right\},\left\{ {{v_1},{v_3}} \right\},\left\{ {{v_1},{v_4}} \right\},\left\{ {{v_2},{v_3}} \right\},\left\{ {{v_2},{v_4}} \right\},\left\{ {{v_3},{v_4}} \right\}} \right\} \\ {S_4} &= \left\{ {{v_1},{v_2},{v_3},{v_4},{v_5}} \right\}\,\,\,\, \text{gives} \\ &\left\{ {\left\{ {{v_1},{v_2}} \right\},\left\{ {{v_1},{v_3}} \right\},\left\{ {{v_1},{v_4}} \right\},\left\{ {{v_1},{v_5}} \right\},\left\{ {{v_2},{v_3}} \right\},\left\{ {{v_2},{v_4}} \right\},\left\{ {{v_2},{v_5}} \right\},\left\{ {{v_3},{v_4}} \right\},\left\{ {{v_3},{v_5}} \right\},\left\{ {{v_4},{v_5}} \right\}} \right\} \\ &\vdots \\ \text{and so on} \end{align} Therefore, we deduce that the number of all two element subsets of $S$, given in the sequence $$1,2,3,6,10,15,21,28,36,45,55,66,78,91,...$$ and it has the form $$a_1=1, a_2=2, a_3=3, a_{n}=n-1 + a_{n-1}, n\ge 4$$ Now the answer ($ N(S) $) depends on which two element subset $ S_j $ their vectors satisfy the inequality. The answer it may be one subset, or two, or three, or . . . and so on. Here I leave the final answer after more discussion.