1984 USAMO Problems/Problem 1
Problem
In the polynomial , the product of of its roots is . Find .
Solution 1
Using Vieta's formulas, we have:
From the last of these equations, we see that . Thus, the second equation becomes , and so . The key insight is now to factor the left-hand side as a product of two binomials: , so that we now only need to determine and rather than all four of .
Let and . Plugging our known values for and into the third Vieta equation, , we have . Moreover, the first Vieta equation, , gives . Thus we have two linear equations in and , which we solve to obtain and .
Therefore, we have , yielding .
Solution 2
We start as before: and . We now observe that a and b must be the roots of a quadratic, , where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic .
Now
Equating the coefficients of and with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of and get
Solution 3
Call the roots of the polynomial and By Vieta's formula, we have and From the problem statement, we know that Dividing by gives Factoring the 3rd symmetric sum and substituting in the following values, we have Let From the first equation, we have Thus, we have the equation Substituting values and factoring the remaining values in the second symmetric sum gives ~coolmath2017
See Also
1984 USAMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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