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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 \(\left|\frac{a-b}{1+ab}\right|\leq\sqrt{2}-1\)

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  1. amilapsn
    • one year ago
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    @rational @amistre64 @AngusV

  2. amistre64
    • one year ago
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    why 9 of them? does it fail with 8 of them?

  3. mathmate
    • one year ago
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    pigeon-hole?

  4. amistre64
    • one year ago
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    sqrt(2) - 1 is less then 1/2

  5. amilapsn
    • one year ago
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    @mathmate si

  6. amistre64
    • one year ago
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    good luck

  7. amilapsn
    • one year ago
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    @amistre64 yes

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

  9. mathmate
    • one year ago
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    @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
    • one year ago
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    That idea suddenly came to my mind with \(\Large\sf{\tan{\frac{\pi}{8}}=\sqrt{2}-1}\)

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