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

Note by Yan Yau Cheng
4 years, 3 months ago

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

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.

- 4 years, 3 months ago

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

- 4 years, 3 months ago

Me too...

- 4 years, 3 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.

- 4 years, 3 months ago

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

- 4 years, 3 months ago