Difference between revisions of "2005 AMC 10B Problems/Problem 4"
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== Problem == | == Problem == | ||
| + | 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 | ||
| + | |||
| + | <math>(5 \diamond 12) \diamond ((-12) \diamond (-5))</math>? | ||
| + | |||
| + | <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 == | ||
| + | <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> | ||
| + | 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 == | == See Also == | ||
| − | + | {{AMC10 box|year=2005|ab=B|num-b=3|num-a=5}} | |
| + | {{MAA Notice}} | ||
Revision as of 21:34, 23 November 2018
Problem
For real numbers 
and 
, define 
. What is the value of
?
Solution
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 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
