2013 AMC 12B Problems/Problem 6
- The following problem is from both the 2013 AMC 12B #6 and 2013 AMC 10B #11, so both problems redirect to this page.
Problem
Real numbers 
and 
satisfy the equation 
. What is 
?
Solution 1
If we move every term dependent on or to the LHS, we get 
. Adding 
to both sides, we have 
. Notice this is a circle with radius 
, which only contains one point. We can split the into 
and 
to get 
. So, the only point is 
, so the sum is 
. ~ asdf334
Solution 2
If we move every term including or to the LHS, we get ![\[x^2 - 10x + y^2 + 6y = -34.\]](http://latex.artofproblemsolving.com/d/d/1/dd18bde5b66627a6df1efd5dc46ed9e3700c99e4.png)
We can complete the square to find that this equation becomes ![\[(x - 5)^2 + (y + 3)^2 = 0.\]](http://latex.artofproblemsolving.com/b/5/a/b5a35b78809aaa8554858d204ba654a09ff65f46.png)
Since the square of any real number is nonnegative, we know that the sum is greater than or equal to . Equality holds when the value inside the parhentheses is equal to . We find that ![\[(x,y) = (5,-3)\]](http://latex.artofproblemsolving.com/a/c/5/ac52edccd81507a34589475cb65faa21d649d07f.png)
and the sum we are looking for is ![\[5+(-3)=2 \implies \boxed{\textbf{(B)}}.\]](http://latex.artofproblemsolving.com/7/9/7/7973ca732d2cb205e20ea299a2f6bfd784a9cd2a.png)
- Honestly
Video Solution
See Also
| 2013 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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 | |
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
