1984 USAMO Problems/Problem 1
Problem
In the polynomial , the product of of its roots is . Find .
Solution
Let the four roots be , 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: \begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*}
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, a+b+c+d=18, gives p+q=18. Thus we have two linear equations in p and q, which we solve to obtain p=4 and q=14.
Therefore, we have (\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30, yielding k=4\cdot 14+30 = \boxed{86}.
Solution #2 (sketch)
We start as before: ab=-32 and cd=62. We now observe that a and b must be the roots of a quadratic, x^2+rx-32, 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 x^2+sx+62.
Now \begin{align*} x^4 - 18x^3 + kx^2 + 200x - 1984 &= (x^2 + rx - 32)(x^2 + sx + 62)
& = x^4 + (r + s)x^3 + (62 - 32 ... Equating the coefficients of x^3 and x 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 x^2 and get k=\boxed{86}.
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.