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

(Solution 2)
(Solution 3)
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=== Solution 3 ===
 
=== Solution 3 ===
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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==
 
==See Also==

Revision as of 22:45, 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 $r_1r_2r_3r_4 = -1984.$

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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