Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.8 Junior Find the answer

For real numbers $a$ and $b$ , define $a \diamond b = \sqrt{a^2 + b^2}$ . What is the value of
$(5 \diamond 12) \diamond ((-12) \diamond (-5))$ ?

Pick one

Solution

$\begin{align*} (5 \diamond 12) \diamond ((-12) \diamond (-5))&=(\sqrt{5^2+12^2}) \diamond (\sqrt{(-12)^2+(-5)^2})\\ &=(\sqrt{169})\diamond(\sqrt{169})\\ &=13\diamond13\\ &=\sqrt{13^2+13^2}\\ &=\sqrt{338}\\ &=\boxed{\mathrm{(D)\,13\sqrt{2}}}\\ \end{align*}$
Note that the negative signs did not matter and any number squared times two is that number times the square root of 2.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.