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 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
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=1810
~ pi_is_3.14
Video Solution
–no one
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.