Japanese Math Olympiad | A Very Nice Geometry Problem

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  • Опубликовано: 23 дек 2024

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  • @ДмитрийИвашкевич-я8т

    tgx/tg10°=tg60°/tg20°
    tg3α=tgα*tg(60°-α)*tg(60°+α)
    tg60°=tg20°*tg40°*tg80°
    tgx=tg60°*tg10°/tg20°
    tgx=tg40°*tg10°*tg80°=tg40°
    x=40°

    • @vcvartak7111
      @vcvartak7111 День назад

      But proof of tan(3a) in terms of tan(60-A) is needed?

    • @ДмитрийИвашкевич-я8т
      @ДмитрийИвашкевич-я8т День назад +4

      @vcvartak7111 tgα=t
      tg3α=(3t-t^3)/(1-3t^2)
      tg3α=t(√3+t) (√3-t)/[(1-t√3) (1+t√3)]=t[(√3+t)/(1-t√3)]*[(√3-t)/(1+t√3)]=tgα*tg(60°+α)*tg(60°-α)

    • @imetroangola17
      @imetroangola17 День назад

      Verifiquei isso, porém, resolver isso, não é fácil!

    • @ДмитрийИвашкевич-я8т
      @ДмитрийИвашкевич-я8т День назад +1

      ​@imetroangola17Формула красивая, запоминается легко, может еще пригодится 🙂

  • @RAG981
    @RAG981 День назад +3

    I do not see what is wrong with tanx = rt3tan10/tan20, which gives x=40.

  • @Z-eng0
    @Z-eng0 23 часа назад +1

    This was magnificent, absolutely genius.
    I'll never be able to know how people come up with such creative constructions, like, I can solve hard geometry puzzles, but I get stuck on this specific type.
    Can you tell me how your thinking process goes to know these type of problems, please

    • @MathBooster
      @MathBooster  5 минут назад

      It takes a lot of practice to get the intuition for solving geometric problems.

  • @와우-m1y
    @와우-m1y День назад +2

    asnwer=40 isit

    • @dickroadnight
      @dickroadnight День назад

      (atan(tan(10° )/tan(20°)*tan(60°))*180/π‎ = 40

  • @wasimahmad-t6c
    @wasimahmad-t6c День назад +1

    40