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 (ingenious)

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 (cool)

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

Let the roots of the equation be $a,b,c,$ and $d$ . 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 $abcd=-1984$ and $ab=-32$ , then, $cd=62$ . Notice that $abc + abd + acd + bcd = -200$ can be factored into $ab(c+d)+cd(a+b)=-32(c+d)+62(a+b).$ From the first equation, $c+d=18-a-b$ . Substituting it back into the equation, $-32(18-a-b)+62(a+b)=-200$ Expanding, $-576+32a+32b+62a+62b=-200 \implies 94a+94b=376$ So, $a+b=4$ and $c+d=14$ . Notice that $ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)$ Plugging all our values in, $-32+62+4(14)=\boxed{86}.$

~ kante314

Solution 4 (Alcumus)

Since two of the roots have product $-32,$ the equation can be factored in the form $x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + c).$ Expanding, we get $x^4 - 18x^3 + kx^2 + 200x - 1984 = x^4 + (a + b) x^3 + (ab + c - 32) x^2 + (ac - 32b) x - 32c = 0.$ Matching coefficients, we get \begin{align*} a + b &= -18, \\ ab + c - 32 &= k, \\ ac - 32b &= 200, \\ -32c &= -1984. \end{align*}Then $c = \frac{-1984}{-32} = 62,$ so $62a - 32b = 200.$ With $a + b = -18,$ we can solve to find $a = -4$ and $b = -14.$ Then $k = ab + c - 32 = \boxed{86}.$

Video Solution by Omega Learn

https://youtu.be/Dp-pw6NNKRo?t=316

~ pi_is_3.14

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