1984 AIME Problems/Problem 15
Rewrite the system of equations as This equation is satisfied when , as then the equation is equivalent to the given equations. After clearing fractions, for each of the values , we have the equation . We can move the expression to the left hand side to obtain the difference of the polynomials: and
Since the polynomials are equal at , we can express the difference of the two polynomials with a quartic polynomial that has roots at , so
Note the leading coefficient of the RHS is because it must match the leading coefficient of the LHS, which is .
Now we can plug in into the polynomial equation. Most terms drop, and we end up with
Similarly, we can plug in and get
Now adding them up,
with a sum of
/*Lengthy proof that any two cubic polynomials in which are equal at 4 values of are themselves equivalent: Let the two polynomials be and and let them be equal at . Thus we have . Also the polynomial is cubic, but it equals 0 at 4 values of . Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into some nonzero polynomial which would have a degree greater than or equal to 4, contradicting the statement that is cubic. Because and are equivalent and must be equal for all .
Post script for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes and separately before adding them to obtain the final answer is appealing because it gives the individual values of and which can be plugged into the given equations to check.
As in Solution 1, we have
Now the coefficient of on both sides must be equal. Therefore we have .
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