AMC10 2005 10B, Problem 4

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

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}\\  &=\fbox{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.

Answer: 13\sqrt{2}