Difference between revisions of "2005 AMC 10B Problems/Problem 4"

(See Also)
(Undo revision 40052 by Joshxiong (talk))
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<math>(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}}}</math>
 
<math>(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}}}</math>
 
== See Also ==
 
== See Also ==
{{AMC10 box|year=2005|ab=B|num-b=3|num-a=4=5}}
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{{AMC10 box|year=2005|ab=B|num-b=3|num-a=5}}

Revision as of 10:48, 29 June 2011

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

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

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 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
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