2008 AIME I Problems/Problem 13
There is a point for which for all such polynomials, where , , and are positive integers, and are relatively prime, and . Find .
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 .
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
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