Right angled Triangle?

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

#JOMO #JOMO1

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Comments

$(a+b)^2+(a-b)^2=2(a^2+b^2)$ .

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.

Mursalin Habib - 7 years, 2 months ago

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

Trevor B. - 7 years, 2 months ago

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.

Daniel Liu - 7 years, 2 months ago

Me too...

敬全钟 - 7 years, 1 month ago

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

Kushagra Sahni - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$