2002 AMC 12P Problems/Problem 18
(Redirected from 2002 AMC 10P Problems/Problem 19)
- The following problem is from both the 2002 AMC 12P #18 and 2002 AMC 10P #19, so both problems redirect to this page.
Problem
If are real numbers such that , and , find
Solution 1
Adding all of the equations gives Adding 14 on both sides gives Notice that 14 can split into and which coincidentally makes and into perfect squares. Therefore, An easy solution to this equation is and Plugging in that solution, we get
See also
2002 AMC 10P (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 |
2002 AMC 12P (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.