IMO 1959 - Problem 2

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  • čas přidán 8. 09. 2024
  • #olympiad #math #algebra #jee #trigonometry #geometry #gmat #mathstrick #olympiad2022
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Komentáře • 7

  • @snjyjn
    @snjyjn Před 2 lety

    At 4:45, when you replace the square root with the mod, I took a different path.
    Since there are 2 possible square roots (one positive, and one negative), this becomes 2 equations (only the LHS shown):
    2x+2(x-1), and
    2x-2(x-1)
    That still gives us the same 2 results, but without the same constraints on x.
    A^2 = 2, or
    x=(A^2 + 2)/4
    If we try out A=1, and x=3/4 - with the right choice of +ve square root, or -ve square roots, the solution holds good.

    • @SQRTime
      @SQRTime  Před 2 lety

      Interesting approach. Thanks for sharing

  • @user-gh9ik2vu1w
    @user-gh9ik2vu1w Před 4 měsíci

    back in 1960 i could be a imo participant lol, today it is much tougher

  • @evionlast
    @evionlast Před 2 lety

    Nice it's a simple problem liked it

  • @ProbusMihraban
    @ProbusMihraban Před 2 lety

    Cool

    • @SQRTime
      @SQRTime  Před 2 lety

      Thanks for checking 👍