1984 USAMO Problems/Problem 1
Contents
Problem
In the polynomial , the product of of its roots is . Find .
Solution 1 (ingenious)
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 (cool)
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
Let the roots of the equation be and . By Vieta's, \begin{align*} a+b+c+d &= 18\\ ab+ac+ad+bc+bd+cd &= k\\ abc+abd+acd+bcd &=-200\\ abcd &=-1984.\\ \end{align*} Since and , then, . Notice thatcan be factored intoFrom the first equation, . Substituting it back into the equation,Expanding,So, and . Notice thatPlugging all our values in,
~ kante314
Video Solution
https://youtu.be/5QdPQ3__a7I?t=589
~ pi_is_3.14
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.