Difference between revisions of "1984 USAMO Problems/Problem 1"
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− | Let the four roots be <math>a, b, c, and d</math>, so that <math>ab=-32</math>. From here we show two methods; the second is more slick, but harder to see. | + | Let the four roots be <math>a, b, c,</math> and <math>d</math>, so that <math>ab=-32</math>. From here we show two methods; the second is more slick, but harder to see. |
Solution #1 | Solution #1 |
Revision as of 10:52, 22 June 2014
Problem
In the polynomial , the product of of its roots is . Find .
Solution
Let the four roots be and , so that . From here we show two methods; the second is more slick, but harder to see.
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 (sketch)
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
$<cmath>
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
See Also
1984 USAMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.