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Math Help - Really need help on this hard problem IMO

  1. #1
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    Really need help on this hard problem IMO

    Let ABC be a triangle such that angle ACB = 135. Prove that:

    AB^2=AC^2+BC^2+\sqrt{2}\times AC\times BC

    I really have no Idea how to solve this, any help is really appreciated. Thanks
    Last edited by CaptainBlack; July 20th 2010 at 01:23 AM.
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  2. #2
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    Quote Originally Posted by GGGGGGGGGGGGGGGGGG View Post
    Let ABC be a triangle such that angle ACB = 135. Prove that:

    AB^2=AC^2+BC^2+\sqrt{2}\times AC\times BC

    I really have no Idea how to solve this, any help is really appreciated. Thanks
    That looks very similar to the law of cosines.. by the way you have to add spaces after \times when what follows are letters, otherwise it won't render. I fixed it while quoting you.
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  3. #3
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    Quote Originally Posted by GGGGGGGGGGGGGGGGGG View Post
    Let ABC be a triangle such that angle ACB = 135. Prove that:

    AB^2=AC^2+BC^2+\sqrt{2}\times AC\times BC

    I really have no Idea how to solve this, any help is really appreciated. Thanks
    Are you sure that's it? It is the cosine rule applied to the given triangle (or have I missed something?), in which case it looks a bit easy to be an IMO question
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  4. #4
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    Quote Originally Posted by GGGGGGGGGGGGGGGGGG View Post
    Let ABC be a triangle such that angle ACB = 135. Prove that:

    AB^2=AC^2+BC^2+\sqrt{2}\times AC\times BC

    I really have no Idea how to solve this, any help is really appreciated. Thanks
    You can also form a right-angled triangle by standing the triangle on side CB (or side AC).
    Drop a vertical to O from A and join O to C such that OCB is a straight line.
    This requires only Pythagoras' Theorem.

    Then..

    |OA|^2+|OC|^2=|AC|^2

    |OA|^2+\left(|OC|+|CB|\right)^2=|AB|^2

    |OA|=|OC| since |\angle OCA|=45^o

    |OA|^2+|OA|^2+2|OA||CB|+|CB|^2=|AB|^2

    Finish with

    2|OA|^2=|AC|^2\ \Rightarrow\ |OA|=\frac{|AC|}{\sqrt{2}}\ \Rightarrow\ 2|OA|=\sqrt{2}|AC|
    Last edited by Archie Meade; July 20th 2010 at 01:33 PM.
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