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

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

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

Solution

$(5 \diamond 12) \diamond ((-12) \diamond (-5))\\ (\sqrt{5^2+12^2}) \diamond (\sqrt{(-12)^2+(-5)^2})\\ (\sqrt{169})\diamond(\sqrt{169})\\13\diamond13\\ \sqrt{13^2+13^2}\\ \sqrt{338}\\ \boxed{\mathrm{(D)\,13\sqrt{2}}}$

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 4=5
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2005 AMC 10B Problems/Problem 4

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