Find the largest constant $m$ and the smallest constant $M$ , such that given any 13 distinct real numbers , there always exists 2 numbers $x$ and $y$ , with $x > y$ , such that

$m < \dfrac {x-y}{1+xy} < M.$

If the value of $m+M$ can be written as $a+b\sqrt{c}$ , where $a$ , $b$ and $c$ are integers , and $c$ is square-free, find $a+b+c$ .

The answer is 4.

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Relevant wiki: Sum and Difference Trigonometric Formulas - Problem SolvingNotice how the expression looks like the formula for the tangent of the difference of two angles. Indeed, if we substitute $x=\tan \alpha$ and $y=\tan \beta$ , where $\alpha, \beta \in [0, 180^{\circ}]$ , we get that the expression is just $\tan (\alpha - \beta)$ . Also, we have that $\alpha > \beta$ . Thus, since the expression must result in a non-negative value, we must have the $\tan(\alpha - \beta) > \tan 0 = 0$ .

We also have that $\alpha - \beta$ can take any value but $0$ . Thus, $m=0$ . Also, we have that the $\tan$ function has a period $180^{\circ}$ . Thus, by PHP, 2 of these 13 distinct reals must be within a $15^{\circ}$ interval bounded by 2 multiples of 15. Thus, the

minimummaximum value of $\tan(\alpha-\beta)$ must be $\tan 15^{\circ}=2-\sqrt{3}$ , which is given by the reals being equal to $\tan (15n^{\circ})$ . However, for this to be true, one of the real values must be $\tan 90^{\circ}$ , which is undefined. Thus, the supremum value must be $\tan 15^{\circ}=2-\sqrt{3}$ .Therefore, the sum of the infimum and supremum value is $0+2-\sqrt{3} = 2+(-1)\sqrt{3}$ . $2+(-1)+3=\boxed{4}$