For all positive reals \(a\) and \(b\), Let

\[\sqrt{\frac{(a+b)^2+(a-b)^2}2} = c\]

Prove that \(a\), \(b\), and \(c\) are side lengths of a right angled triangle

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## Comments

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TopNewest\((a+b)^2+(a-b)^2=2(a^2+b^2)\).

So the whole thing can be re-written as \(\sqrt{a^2+b^2}=c\) or \(a^2+b^2=c^2\).

Since \(a\), \(b\) are positive reals, by the converse of the Pythagorean theorem, they are the side legths of a right triangle.

@Yan Yau Cheng As an aside, are you looking for problem writers? Because I'd really love to join. It's completely okay if you say no. You are currently in a great place with wonderful staff members and you have successfully hosted three contests. So, it is completely natural for you say no. To me, it'd be a great experience to be able to write problems for a competition and I'd love to be a part of it.

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Me too. @Yan Yau Cheng I'm interested in writing problems as well.

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Me too...

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As for me I just free-lance write these random problems :P

I was also wondering, would you please grade my questions' overall difficulty from 1 to 10? 10 being the hardest. Thanks.

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OH GOD EASIEST QUESTION I HAVE EVER DONE IN MY LIFE.

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