Inequality in triangle

  1. Prove that in any triangle ABC with a sharp angle at the peak C apply inequality:(a^2+b^2)cos(α-β)<=2ab
    Determine when equality occurs.

    I tried to solve this problem... I proved that (a^2+b^2+c^2)^2/3 >= (4S(ABC))^2,
    S(ABC) - area
    but I dont know prove that (a^2+b^2)cos(α-β)<=2ab :(
    thanks for your help
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,043
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    welcome to pf!

    hi harry654! welcome to pf! :smile:

    (try using the X2 icon just above the Reply box :wink:)

    by "sharp angle", i assume you mean an acute angle, less than 90°?

    hint: where in that triangle can you find (or construct) an angle α-β ? :wink:
     
  4. by "sharp angle", i assume you mean an acute angle, less than 90°? yes :)
    hint: where in that triangle can you find (or construct) an angle α-β ?
    oh. I know that apply π-(α+β)=γ. But i cant see in that triangle an angle α-β. I have tried to solve this problem already 3 days, but I always proved other inequality.
     
  5. tiny-tim

    tiny-tim 26,043
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    then make one! :biggrin:

    you know that α-β is in the answer, so you know there must be an α-β somewhere

    where could you draw an extra line to make an angle α-β ? :wink:
     
  6. I can draw paralell straight line with BC. Then i draw paralell straight line with AC. And then I depict triangle ABC in axial symmetry according AB. And at the peak B(under) I have an angle α-β. Is it OK?
     
  7. tiny-tim

    tiny-tim 26,043
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    sorry, i don't understand :confused:

    can you supply some extra letters (D, E, …), to make it clearer?
     
  8. OK.
    We have triangle ABC.
    I draw paralell line p with BC and paralell line l with AC. p intersects l in the point D.
    Then angle DBA is equivalent with α. I depict triangle ABC in axial symmetry according AB. this triangle mark AEB.Then angle EBA is equivalent with β. And angle DBE is α-β
     
  9. tiny-tim

    tiny-tim 26,043
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    oh i see!

    yes, but a lot simpler would be to draw just one line

    draw CD equal to CA with D on BC … then triangle DBC has two sides the same as ABC, and angle DCB is α-β :smile:

    (btw, "depict triangle ABC in axial symmetry according AB" is weird … write "reflect triangle ABC in AB" :wink:)
     
  10. (btw, "depict triangle ABC in axial symmetry according AB" is weird … write "reflect triangle ABC in AB" :wink:)[/QUOTE]

    sorry my english is not enough good I know :(

    draw CD equal to CA with D on BC ... When I draw D on BC so B,C,D are on one line... and it isnt triangle DBC so I am confused... sorry :(
     
  11. tiny-tim

    tiny-tim 26,043
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    oops!

    sorry, i meant with D on AB :smile:
     
  12. Re: oops!

    oh yes.
    Now, Should I compare areas of triangles ABC and DBC, shouldnt I?
     
  13. tiny-tim

    tiny-tim 26,043
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    no!!

    (why are you so keen on areas anyway? :confused: you'll hardly ever need them, and certainly not here :redface:)

    forget about triangle ABC now

    just use triangle DBC :smile:
     
  14. Ok... I use cosinus law and I have equality. But I think I must something compare because I need prove inequality
     
  15. tiny-tim

    tiny-tim 26,043
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    what equality? :confused:

    (btw, we say "sine" and "cosine" :wink:)
     
  16. Ok.Maybe dont understand me. When I use cosine law I get a^2+b^2-2abcos(α-β)=h^2
    when h is DB. But I need (a^2+b^2)cos(α-β)<=2ab I now see that equality occurs when triangle is isosceles. But I try from a^2+b^2-2abcos(α-β)=h^2 get (a^2+b^2)cos(α-β)<=2ab.
     
  17. tiny-tim

    tiny-tim 26,043
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    (try using the X2 icon just above the Reply box :wink:)

    hint: eliminate ab :wink:
     
  18. hint: eliminate ab

    I found that I am stupid...
    when I eliminate ab cosine law will not apply or no?
     
    Last edited: Nov 27, 2010
  19. apply (a²+b²)cos(α-β)>= a²+b²-h²

    and then I dont know aaaaa :(
     
    Last edited: Nov 27, 2010
  20. tiny-tim

    tiny-tim 26,043
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    no, you've missed out a cos(α-β) :redface:
     
  21. 2ab=(a²+b²-h²)/cos(α-β) Is that OK?

    then a²+b²>=2ab
    a²+b²>=(a²+b²-h²)/cos(α-β) /*cos(α-β)
    (a²+b²)cos(α-β) ? (a²+b²-h²)
    so? I am lost...
     
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