1984 USAMO Problems/Problem 1

Revision as of 21:22, 27 April 2014 by Gamjawon (talk | contribs) (Solution)

Problem

In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$, the product of $2$ of its roots is $- 32$. Find $k$.

Solution

Let the four roots be a, b, c, and d, so that ab=-32. 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.
\en... From the last of these equations, we see that cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62. Thus, the second equation becomes -32+ac+ad+bc+bd+62=k, and so ac+ad+bc+bd=k-30. The key insight is now to factor the left-hand side as a product of two binomials: (a+b)(c+d)=k-30, so that we now only need to determine a+b and c+d rather than all four of a,b,c,d.

Let p=a+b and q=c+d. Plugging our known values for ab and cd into the third Vieta equation, -200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b), we have -200 = -32(c+d) + 62(a+b) = 62p-32q. 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 (ProblemsResources)
Preceded by
First
Problem
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

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