Difference between revisions of "2008 AIME I Problems/Problem 13"
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Revision as of 14:46, 23 March 2008
Problem
Let
.
Suppose that
.
There is a point for which
for all such polynomials, where
,
, and
are positive integers,
and
are relatively prime, and
. Find
.
Solution
Adding the above two equations gives , and so we can deduce that
.
Similarly, plugging in and
gives
and
. Now,
Therefore and
. Finally
So
.
Now .
In order for the above to be zero, we must have
and
. Canceling terms on the second equation gives us
x = 5/19, y = 16/19
5+16+19 = \boxed{040}$.
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |