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

 
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== Problem ==
 
== Problem ==
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For real numbers <math>a</math> and <math>b</math>, define <math>a \diamond b = \sqrt{a^2 + b^2}</math>. What is the value of
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<math>(5 \diamond 12) \diamond ((-12) \diamond (-5))</math>?
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<math>\mathrm{(A)} 0 \qquad \mathrm{(B)} \frac{17}{2} \qquad \mathrm{(C)} 13 \qquad \mathrm{(D)} 13\sqrt{2} \qquad \mathrm{(E)} 26</math>
 
== Solution ==
 
== Solution ==
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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>
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Note that the negative signs did not matter and any number squared times two is that number times the square root of 2.
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== See Also ==
 
== See Also ==
*[[2005 AMC 10B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 20:34, 23 November 2018

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.

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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