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 .
In order for the above to be zero, we must have
Canceling terms on the second equation gives us . Plugging that into the first equation and solving yields , and .
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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