How Many X Can You Find?

Let us rewrite the inequality , motivated by the fact that isolating variables is sometimes quite a good technique, and for reasons that will soon be apparent. However, before we do anything, I claim that we only need to consider $1+x_ix_j > 0$ . This is motivated by the absolute value as absolute values are unpredictable and we should consider positive and negative cases separately. Indeed, notice that if $1 + x_ix_j = 0$ then clearly we have $x_i \neq x_j$ so $|x_i - x_j| > 0 = \frac{1}{100}(1+x_ix_j)$ so we are trivially done. And if $1 + x_ix_j < 0$ then $|x_i - x_j| > 0 > \frac{1}{100}(1+x_ix_j)$ .

With the above observations in mind, we rewrite our inequality as:

$|\frac{x_i - x_j}{1+x_ix_j}| > \frac{1}{100} (*)$ .

For those who are familiar with the sum and difference formula for tangents , in this new form, it should definitely ring a bell in your head. Indeed, let us now use a substitution of $x_i = \tan \alpha_i$ . We can bound $\alpha_i$ by $\alpha_i \in (-\frac{\pi}{2}, \frac{\pi}{2})$ . So now, with the substitution, rewrite $(*)$ again to the following:

$|\tan(\alpha_i-\alpha_j)| = |\frac{\tan \alpha_i - \tan \alpha_j}{1 + \tan \alpha_i\tan \alpha_j}| = | \frac{x_i - x_j}{1+x_ix_j}| (**)$

Here, we see why bounding $\alpha_i$ is of such utmost importance. Because, now we can safely conclude that there will exist at least one angle $0 < \alpha_i - \alpha_j < \frac{\pi}{n-1}$ . So $(**)$ holds iff $\tan \frac{\pi}{n-1} < \frac{1}{100} \leftrightarrow \frac{\pi}{n-1} < \arctan\frac{1}{100}$ . One easily verifies that this holds for $n < 316$ .

So now we give a construction $n = 315$ . Let us choose an arbitrarily small but non-negative $\mathfrak{c}$ . We will attempt to make $\tan (\alpha_i - \alpha_j) > \tan \frac{\pi - \mathfrak{c}}{n-1}$ $= \tan \frac { \pi - \mathfrak{c}}{314} > \frac{1}{100}$ . This motivates us to write $\alpha_l = \frac{(l-1)(\pi - \mathfrak{c} )}{n-1}$ . However, we remember that we need to fit this in the correct range , and also remember that we are considering absolute value so we can also have $\tan( \alpha_i - \alpha_j ) < 0$ . Now, finally, we can our construction for $l = 0, 1, \ldots, n$ as $\alpha_l = -\frac{\pi}{2} + \frac{\mathfrak{c}}{2} + \frac{(l-1)(\pi - \mathfrak{c} )}{n-1}$ .

In conclusion, the answer is $\fbox{315}$ .

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Or, if this is too convoluted, one might attempt a more straight-forward, constructive albeit tedious process. We conclude as above that we only have to consider positive reals. The motivating principle here is to first "order" the elements. So here, let $x_i > x_j$ to remove the absolute values. For instance, we can again rewrite the inequality as $x_i > \frac{100x_j +1}{100 - x_j}$ . Now, notice first that in this other form of $(100-x_j)x_i>100x_j+1$ , we must have $x_j < 100$ or otherwise we would violate the positive reals condition.

It is quite easy to verify that over $(-∞, 100)$ , $p(x) = \frac{100k+1}{100-k}$ is increasing. For notation sake, let $y_{i+1} = f(y_i) \space \text{and} \space y_1 = x_1<100$ . To simplify the procedure just let $y_1 \rightarrow -∞$ and see how many iterations of the function is needed to exceed $100$ .

Again, I had to use the substitution of tangents to simplify the work. By letting $100 = \tan r$ and $y_i = - \tan s$ (the negative is because of the signs in our rewritten inequality) we can perform some algebraic manipulation to get:

$1+\frac{\frac{\pi}{2}+r}{\frac{\pi}{2}-r}<n<2+\frac{\frac{\pi}{2}+r}{\frac{\pi}{2}-r}$

Solving gives: $n = \fbox{315}$ as above.

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Remark : Using both methods we can generalise that suppose we need $|x_i - x_j| > \frac{1}{m}(1+x_ix_j)$ , then we have:

• Method $1$ : $\frac{\pi}{n-1} < \arctan\frac{1}{m}$

• Method $2$ : $n=\lfloor 2+\frac{\frac{\pi}{2}+\arctan m}{\frac{\pi}{2}-\arctan m}\rfloor$

For ease of reading and writing, $x_i$ and $x_j$ will be replaced by $a$ and $b$ .

First, we notice that if the condition holds for some $a,b$ , the condition also holds for $-a,-b$ . We can also easily verify that all the numbers are distinct, or else $|a-b|=0$ and $a\cdot{b}\geq0\to\frac{1}{100}(1+a\cdot{b})>0$ .

For now, let's only look at positive reals. Because the numbers are all distinct, we can assume that $a>b$ in the given inequality, and we can get rid of the absolute value. This assumption also lets us get a strict bound on the value of $b$ given $a$ . Specifically, we solve the inequality for $b$ : $|a-b|>\frac{1}{100}(1+a\cdot{b})\to$ $a-b>\frac{1}{100}(1+a\cdot{b})\to$ $100a-1>a\cdot{b}+100b\to$ $b<\frac{100a-1}{a+100}$ . Because $a>b>0$ , we don't need to worry about a zero denominator or any strange sign switches while manipulating the inequality.

We can quickly verify that this inequality is strictly decreasing. This means that if we begin with with a specific $a_0$ value, solve for $b_0$ , and use this value for $a_1$ , the bounds on $b_1$ will be tighter than those on $b_0$ . This also tells us that we don't need to check if every pair of numbers works. We just need the ones that are closest to each other to satisfy the inequality.

The next step is to see how many positive values we can have. By increasing the largest value, $x_1$ , we can bring the second highest value, $x_2$ , arbitrarily close to $100$ , $x_3$ arbitrarily close to $\frac{100(100)-1}{100+100}=49.995$ , etc. This means that the upper bound on $x_k = f(x_{k-1})$ , where $f(x)=\frac{100x-1}{x+100}$ . We can now calculate what these upper bounds are.

To simplify the process, let's assume that $f(x_1)=x_2=100$ . Clearly, no value can actually make this happen, but we can get as close as we'd like. (From this point on, we will work under the assumption that we can bring $x_i$ arbitrarily close to its upper bound and will treat $x_i$ as being equal to its upper bound.) Repeatedly applying $f$ gives us $x_3=49.995,x_4=33.3244,$ $...,x_{158}\approx.00085,x_{159}\approx-.00915$ . Because $x_{159}<0$ , we can stop here, since we're still only looking at positive reals. There seem to be a maximum of $158$ positive reals that, in pairs, satisfy the given inequality, and because all the pairings of their opposites also satisfy the inequality, the answer should be $2*158=316$ .

However, we're overlooking something: if we include $x_{158}$ and $-x_{158}=x_{316}$ , then choosing these two as $a$ and $b$ will not satisfy the inequality because $|x_{158}-x_{316}|=2\cdot{x_{158}}<\frac{1}{100}(1-x_{158}^2)$ . This is not the case for $x_{157}$ and $-x_{157}=x_{315}$ , which also implies that it's not the case for any other $x_k,1\leq{k}\leq156$ because as $x_k$ increases, $1-x_k^2$ decreases and $2\cdot{x_k}$ increases. Therefore, we can just take out $x_{158}$ and $x_{316}$ and be sure that every other pairing satisfies the inequality.

We can easily verify that any number already in the list can be paired with $0$ to form a pair that satisfies the given inequality, since all the numbers have magnitude at least $.01$ , which means $|x_k-0|>\frac{1}{100}(1+0\cdot{x_k})$ . Therefore, we can add $0$ into our list of reals.

We can now see that there are a maximum of $2\cdot157+1=315$ reals that pairwise satisfy the given inequality, and so $n=315$ .

Consider $a_i = \arctan x_i$ . Thus we have that $\tan a_i = x_i$ . Now we note that $|\tan (a_i-a_j)| > 1/100$ . Thus $|a_i-a_j| >arctan(1/100)$ . It is simple then to note that if there are more than 315 x's, then we can simply consider the intervals [karctan(1/100),(k+1)arctan(1/100)) for k = 0,1,...,314. At least one of these intervals will contain 2 a_i's (since we limit the range of arctan from 0 to pi) and thus we attain a contradiction.

In our arrangement, for all $x_i$ , define $y_i = \tan^{-1}{x_i}$ , therefore, for some $x_i > x_j$

$|x_i - x_j| > \frac{1}{100}(1+x_i x_i) \Rightarrow \tan{(y_i - y_j)} > \frac{1}{100} \Rightarrow y_i - y_j > \tan^{-1}{\frac{1}{100}}$

Since $y_i \in (- \frac{\pi}{2}, \frac{\pi}{2})$ for all $i$ , so if we divide the interval $(- \frac{\pi}{2}, \frac{\pi}{2})$ into atmost $\displaystyle \lceil \displaystyle \frac{180}{\tan^{-1}{\frac{1}{100}}} \rceil$ parts, each $y_i$ belonging to each part, we have a valid arrangement. This yields the maximum $n \geq 315$ .

If we take one more $x_i$ or equivalently, $y_i$ , by PHP, it falls into one of the already occupied partitions & the arrangement becomes invalid, hence giving the maximum $\boxed{315}$ .