Difference between revisions of "2005 AMC 10B Problems/Problem 4"
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<math>(5 \diamond 12) \diamond ((-12) \diamond (-5))</math>? | <math>(5 \diamond 12) \diamond ((-12) \diamond (-5))</math>? | ||
| − | <math>\ | + | <math>\textbf{(A) } 0 \qquad \textbf{(B) } \frac{17}{2} \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13\sqrt{2} \qquad \textbf{(E) } 26</math> |
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== Solution == | == Solution == | ||
| − | < | + | <cmath> |
| + | \begin{align*} | ||
| + | (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}}}\\ | ||
| + | \end{align*} | ||
| + | </cmath> | ||
| + | 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 | + | {{AMC10 box|year=2005|ab=B|num-b=3|num-a=5}} |
| + | {{MAA Notice}} | ||
Latest revision as of 12:58, 24 July 2022
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. 
