2002 AMC 12P Problems/Problem 4
- The following problem is from both the 2002 AMC 12P #4 and 2002 AMC 10P #10, so both problems redirect to this page.
Contents
[hide]Problem
Let and
be distinct real numbers for which
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
for the AMC 12P and
for the AMC 10P.
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
for the AMC 12P and
for the AMC 10P.
Note
For some unknown reason, the answer choices for the 2002 AMC 10P are different from the answer choices for the 2002 AMC 12P, even though the question is exactly the same. Indeed, is the correct answer choice for the 2002 AMC 10P.
See also
2002 AMC 10P (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.