# Really need help on this hard problem IMO

• July 19th 2010, 11:21 PM
GGGGGGGGGGGGGGGGGG
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
• July 19th 2010, 11:46 PM
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Quote:

Originally Posted by GGGGGGGGGGGGGGGGGG
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.
• July 20th 2010, 01:26 AM
CaptainBlack
Quote:

Originally Posted by GGGGGGGGGGGGGGGGGG
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
• July 20th 2010, 02:45 AM
Quote:

Originally Posted by GGGGGGGGGGGGGGGGGG
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

$2|OA|^2=|AC|^2\ \Rightarrow\ |OA|=\frac{|AC|}{\sqrt{2}}\ \Rightarrow\ 2|OA|=\sqrt{2}|AC|$