2008 AIME I Problems/Problem 13
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
Solution 1
Adding the above two equations gives , and so we can deduce that .
Similarly, plugging in and gives and . Now, Therefore and . Finally, So, , or equivalently .
Substituting these equations into the original polynomial , we find that at , . The remaining coefficients and are now completely arbitrary because the original equations impose no more restrictions on them. Hence, for the final equation to hold for all possible , we must have .
As the answer format implies that the -coordinate of the root is non-integral, . The format also implies that is positive, so . Substituting into and reducing to a quadratic yields , in which the only non-integral root is , so .
The answer is .
Solution 2
Consider the cross section of on the plane . We realize that we could construct the lines/curves in the cross section such that their equations multiply to match the form of and they go over the eight given points. A simple way to do this would be to use the equations , , and , giving us
.
Another way to do this would to use the line and the ellipse, . This would give
.
At this point, we see that and both must have as a zero. A quick graph of the 4 lines and the ellipse used to create and gives nine intersection points. Eight of them are the given ones, and the ninth is . The last intersection point can be found by finding the intersection points of and . Finally, just add the values of , , and to get
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 |
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