Difference between revisions of "2002 AMC 12P Problems/Problem 4"
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{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:13, 14 July 2024
- The following problem is from both the 2002 AMC 12P #4 and 2002 AMC 10P #10, so both problems redirect to this page.
Contents
Problem
Let 
and 
be distinct real numbers for which
![\[\frac{a}{b} + \frac{a+10b}{b+10a} = 2.\]](http://latex.artofproblemsolving.com/3/0/3/303291e61ebd979d00d2fb126fd207adb3df898b.png)
Find 
Solution 1
For sake of speed, WLOG, let 
. This means that the ratio will simply be because 
Solving for with some very simple algebra gives us a quadratic which is 
Factoring the quadratic gives us 
. Therefore, 
or 
However, since and must be distinct, cannot be 
so the latter option is correct, giving us our answer of 
Solution 2
The only tricky part about this equation is the fact that the left-hand side has fractions. Multiplying both sides by 
gives us 
Moving everything to the left-hand side and dividing by 
gives 
which factors into 
Because 
giving us our answer of
See also
| 2002 AMC 10P (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 | ||
| 2002 AMC 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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. 
