1984 AIME Problems/Problem 15
Contents
Problem
Determine if
Solution 1
Rewrite the system of equations as This equation is satisfied when . After clearing fractions, for each of the values , we have the equation where and , for .
Since the polynomials on each side are equal at , we can express the difference of the two polynomials by a quartic polynomial that has roots at , so The leading coefficient of the RHS is because the leading coefficient of the LHS is .
Plug in in succession, into . In each case, most terms drop, and we end up with Adding them up we get the sum as .
Postscript 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.
Solution 2
As in Solution 1, we have where and , for .
Now the coefficient of on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the and terms are, so we can eventually apply Vieta's. We can write the long equation as Rearranging gives us By Vieta's, we know that the sum of the roots of this equation is (recall that the roots of the original and this manipulated form of it had roots and ). Thus,
Solution 3 (Highly Unrecommended)
Before starting this solution, I highly recommend never following this unless you have no idea what to do with an hour of your time. Even so, learning the above solutions will be more beneficial.
can be rewritten as You might be able to see where this is going. First off, find and . Then, multiply by the respective lcm to clear all of the denominators. Once you do that, manipulate the equations to solve for .
Now, most of this is just a brainless bash, and reemphasizing, please try to learn the above solutions. This is only a last resort and only to be used if you have too much time left. The exact amount of time this bash takes depends on the person and how quickly they can manipulate the equations.
Solution 4 (Fast, Efficient)
Notice how on each line, we have equations of the form . We can let this be a polynomial, with respect to . We can say that , , , and are coefficients, and not variables. So, we can now expand the fractions to get .
Now, we have arrived at this huge expression, but what do we do with it?
Well, we can look at what we want to find - . So, we want the sum of , , , and . Looking back to our expression, we can note how on the right hand side, the terms add to . Also, on the left hand side, the coefficient is (which is achievable by Vieta's formulas rather than expanding if you want to save a few seconds). So, moving all the terms to the left hand side, then we have that by Vieta's formulas, the sum of the roots is . Then, we can solve to find that .
See also
1984 AIME (Problems • Answer Key • Resources) | ||
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