Victor's Level 5 Challenges - 1

Define $a_{k}$ to be the value, in degrees, that satisfies $tan a_{k}=x_{k}, -90\leq a_{k}\leq 90$ for all integers $k, 1 \leq k \leq n$ Then the inequality becomes $\frac{100}{99}\leq\frac{1+x_{i}^{2}+x_{j}^{2}+x_{i}^{2}x_{j}^{2}}{1+2x_{i}x_{j}+x_{i}^{2}x_{j}^{2}}$

Subtracting 1 from both sides and taking the square root, we have $\frac{1}{\sqrt{99}}=\frac{x_{i}-x_{j}}{1+x_{i}x_{j}}=tan (a_{i}-a_{j})$ .

Now suppose that $x\geq 32$ . Then by Pigeonhole Principle, there exists two of $a_{k}$ such that they differ by less than $\frac{180}{32}$ . Since tangent is an increasing function on the interval $[-90, 90]$ , then the equation will not be satisfied (because $tan \frac{180}{32} < \frac {1}{\sqrt{99}}$ ). However, by taking the 31 numbers $x_{i}=tan \frac{180}{31}i$ , $i=1, 2, …, 31$ , we know that no two have inverse tangents differing by more than $\frac{1}{\sqrt{99}}$ (check using a calculator that $tan \frac{180}{31} > \frac{1}{\sqrt{99}}$ ).

Hence the answer is 31.