Difference between revisions of "2013 AMC 12B Problems/Problem 6"
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Revision as of 12:59, 4 January 2020
- The following problem is from both the 2013 AMC 12B #6 and 2013 AMC 10B #11, so both problems redirect to this page.
Contents
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 We can complete the square to find that this equation becomes 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 and the sum we are looking for is - Honestly
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 |
{{AMC10 box|year=2013|ab=B|num-b=6|num-a=8} The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.