2002 AMC 12P Problems/Problem 4
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, the
cannot be
so the latter option is correct, giving us our answer of
See also
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 |
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