# Difference between revisions of "1984 USAMO Problems/Problem 1"

## 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 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 $cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62$. 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

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+rs)x^2\\ &+(62s-32r)x-1984.\end{align*}

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 $k=\boxed{86}.$

### Solution 3

Call the roots of the polynomial $r_1, r_2, r_3,$ and $r_4.$ By Vieta's formula, we have $$r_1 + r_2 + r_3 + r_4 = 18,$$ $$r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 +r_2r_4 + r_3r_4 = k,$$ $$r_1r_2r_3 + r_2r_3r_4 + r_1r_2r_4 + r_1r_3r_4 = -200,$$ and $$r_1r_2r_3r_4 = -1984.$$ From the problem statement, we know that $r_1r_2 = -32.$ Dividing $r_1r_2r_3r_4 = -1984$ by $r_1r_2 = -32$ gives $r_3r_4 = 62.$ Factoring the 3rd symmetric sum and substituting in the following values, we have $r_1r_2(r_3 + r_4) + r_3r_4(r_1 + r_2) = -200 \Rightarrow 62(r_1 + r_2) -32(r_3 + r_4) = -200.$ Let $r_1 + r_2 = x.$ From the first equation, we have $r_3+r_4 = 18-x.$ Thus, we have the equation $62(x) -32(18-x) = -200 \Rightarrow x=4.$ Substituting values and factoring the remaining values in the second symmetric sum gives $-32 + r_1(r_3 + r_4) + r_2(r_3 + r_4) +62 = k \Rightarrow -32 + (r_1+r_2)(r_3+r_4) + 62 = k \Rightarrow k = 86.$

~coolmath2017