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

Revision as of 21:45, 15 December 2020

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 $r_1r_2r_3r_4 = -1984.$

@@ Line 29: / Line 29: @@
 === Solution 3 ===
+Call the roots of the polynomial <math>r_1, r_2, r_3,</math> and <math>r_4.</math> By Vieta's formula, we have <cmath>r_1 + r_2 + r_3 + r_4 = 18,</cmath> <cmath>r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 +r_2r_4 + r_3r_4 = k,</cmath> <cmath>r_1r_2r_3 + r_2r_3r_4 + r_1r_2r_4 + r_1r_3r_4 = -200,</cmath> and <math>r_1r_2r_3r_4 = -1984.</math>
 ==See Also==

1984 USAMO (Problems • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

Page

Toolbox

Search