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 (same degree of and in terms) and they include the eight given points. One 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
. (But
(Another way would be to use the hyperbola from Solution 1. Interesting that different curves both work.)
At this point, we consider 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
Solution 3 (Ansatz)
We can plug in the values to obtain
Now, this means that
After some simplifying, we obtain
Since , and we suspect that:
and
.
Plugging this into the first equation, and factoring, and cancelling , and simplifying, we get , so we find that
~~pinkpig
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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