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

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Solution #1  
 
Solution #1  
  
Using Vieta's formulas, we have: \begin{align*} a+b+c+d &= 18, <br /> ab+ac+ad+bc+bd+cd &= k, <br /> abc+abd+acd+bcd &= -200, <br /> abcd &= -1984. <br /> \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.
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Using Vieta's formulas, we have: <cmath> \begin{align*}a+b+c+d &= 18,\ ab+ac+ad+bc+bd+cd &= k,\ abc+abd+acd+bcd &=-200,\ abcd &=-1984.\ \end{align*} </cmath>
  
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.  
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From the last of these equations, we see that <math>cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62</math>. Thus, the second equation becomes <math>-32+ac+ad+bc+bd+62=k</math>, and so <math>ac+ad+bc+bd=k-30</math>. The key insight is now to factor the left-hand side as a product of two binomials: <math>(a+b)(c+d)=k-30</math>, so that we now only need to determine <math>a+b</math> and <math>c+d</math> rather than all four of <math>a,b,c,d</math>.
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Let <math>p=a+b</math> and <math>q=c+d</math>. Plugging our known values for <math>ab</math> and <math>cd</math> into the third Vieta equation, <math>-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)</math>, we have <math>-200 = -32(c+d) + 62(a+b) = 62p-32q</math>. 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}.  
 
Therefore, we have (\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30, yielding k=4\cdot 14+30 = \boxed{86}.  

Revision as of 20:24, 27 April 2014

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.\\ \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 (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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