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

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==Solution==
 
==Solution==
  
Let the four roots be <math>a,b,c,d</math>. By Vieta's Formulas, we have the following:
+
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.
  
*<math>ab = - 32</math>
+
Solution #1
*<math>abcd = - 1984</math>
 
*<math>cd = 62</math>
 
*<math>a + b + c + d = 18</math>*
 
*<math>ab + ac + ad + bc + bd + cd = k</math>
 
*<math>abc + abd + acd + bcd = - 200</math>
 
  
Substituting given values obtains the following:
+
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.
  
*<math>ac + ad + bc + bd = k - 62 + 32 = k - 30</math>
+
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.
*<math>(a + b)(c + d) = k - 30</math>
 
*<math>- 32c + 62b + 62a - 32d = - 200</math>
 
*<math>31(a + b) - 16(c + d) = - 100</math>
 
  
Multiplying the * equation by 16 gives
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Therefore, we have (\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30, yielding k=4\cdot 14+30 = \boxed{86}.
  
<math>16(a + b) + 16(c + d) = 288</math>. Adding it to the previous equation gives
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Solution #2 (sketch)  
  
*<math>47(a + b) = 188</math>
+
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.
*<math>a + b = 4</math>
+
 
*<math>c + d = 18 - 4 = 14</math>
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Now \begin{align*} x^4 - 18x^3 + kx^2 + 200x - 1984 &= (x^2 + rx - 32)(x^2 + sx + 62) <br /> & = 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}.
*<math>(a + b)(c + d) = 4(14) = 56 = k - 30</math>
 
*<math>k = \boxed{86}</math>
 
  
 
==See Also==
 
==See Also==

Revision as of 21:22, 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.
\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.

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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