2005 AMC 10B Problems/Problem 4

Problem

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

$(5 \diamond 12) \diamond ((-12) \diamond (-5))$ ?

$\textbf{(A) } 0 \qquad \textbf{(B) } \frac{17}{2} \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13\sqrt{2} \qquad \textbf{(E) } 26$

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.

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 10 Problems and Solutions

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2005 AMC 10B Problems/Problem 4

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