Difference between revisions of "2005 AMC 12B Problems/Problem 23"
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== Solution == | == Solution == | ||
| + | Call x + y = s and x^2 + y^2 = t. Then, we note that log(s)=z which implies that log(10s) = z+1= log(t). Therefore, t=10s. Let us note that x^3 + y^3 = (s)((3t/2)-((s^2)/2)) = (s)(15s-((s^2)/2). Since s = 10^z, we find that x^3 + y^3 = 15 x 10^2 - (1/2) x 10^3. Thus, a+b = 29/2. B is the answer. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2005|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2005|ab=B|num-b=22|num-a=24}} | ||
Revision as of 19:59, 5 February 2011
Problem
Let 
be the set of ordered triples 
of real numbers for which
![\[\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.\]](http://latex.artofproblemsolving.com/c/1/1/c117b87dca578c53d5694e7f783975909325a5bc.png)
There are real numbers 
and 
such that for all ordered triples 
in we have 
What is the value of 
Solution
Call x + y = s and x^2 + y^2 = t. Then, we note that log(s)=z which implies that log(10s) = z+1= log(t). Therefore, t=10s. Let us note that x^3 + y^3 = (s)((3t/2)-((s^2)/2)) = (s)(15s-((s^2)/2). Since s = 10^z, we find that x^3 + y^3 = 15 x 10^2 - (1/2) x 10^3. Thus, a+b = 29/2. B is the answer.
See Also
| 2005 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 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 12 Problems and Solutions | |
