# Difference between revisions of "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

$(5 \diamond 12) \diamond ((-12) \diamond (-5))\\ (\sqrt{5^2+12^2}) \diamond (\sqrt{(-12)^2+(-5)^2})\\ (\sqrt{169})\diamond(\sqrt{169})\\13\diamond13\\ \sqrt{13^2+13^2}\\ \sqrt{338}\\ \boxed{\mathrm{(D)\,13\sqrt{2}}}$ 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 byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions