## amilapsn one year ago An advanced challenge for math enthusiasts:Given any 9 reals show that there will be at least 2 distinct reals a, b such that $$\left|\frac{a-b}{1+ab}\right|\leq\sqrt{2}-1$$

1. amilapsn

@rational @amistre64 @AngusV

2. amistre64

why 9 of them? does it fail with 8 of them?

3. mathmate

pigeon-hole?

4. amistre64

sqrt(2) - 1 is less then 1/2

5. amilapsn

@mathmate si

6. amistre64

good luck

7. amilapsn

@amistre64 yes

8. amilapsn

$\Large\sf{\text{Here is the answer:}\\ \forall x \in \mathbb{R}\ \exists\ \theta\in(-\frac{\pi}{2},\frac{\pi}{2})\ such\ that\\ x=tan\theta\\ \text{For 9 reals there'll be 9 corresponding }\theta s\\ \text{Let's divide the interval }(-\frac{\pi}{2},\frac{\pi}{2})\\ \text{8 equal parts. So out of 9 }\theta s\\ \text{there'll be at least 2 such that}\\ \theta_1-\theta_2<\frac{\pi}{8}\\ \Rightarrow tan(\theta_1-\theta_2)<tan\frac{\pi}{8}\\ \Rightarrow}\left|\frac{a-b}{1+ab}\right|<\sqrt{2}-1$

9. mathmate

@amilapsn Brilliant. I went as far as splitting the circle in eight parts, but went nowhere. Didn't think of mapping to tan theta. Thank you!

10. amilapsn

That idea suddenly came to my mind with $$\Large\sf{\tan{\frac{\pi}{8}}=\sqrt{2}-1}$$