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

(Created page with '1984 USAMO Problem #1 Problem: In the polynomial <math>x^4 - 18x^3 + kx^2 + 200x - 1984 = 0</math>, the product of <math>2</math> of its roots is <math>- 32</math>. Find <math>…')
 
Line 9: Line 9:
 
Let the four roots be <math>a,b,c,d</math>. By Vieta's Formulas, we have the following:
 
Let the four roots be <math>a,b,c,d</math>. By Vieta's Formulas, we have the following:
  
<math>ab = - 32</math>
+
*<math>ab = - 32</math>
<math>abcd = - 1984</math>
+
*<math>abcd = - 1984</math>
<math>cd = 62</math>
+
*<math>cd = 62</math>
<math>a + b + c + d = 18</math>*
+
*<math>a + b + c + d = 18</math>*
<math>ab + ac + ad + bc + bd + cd = k</math>
+
*<math>ab + ac + ad + bc + bd + cd = k</math>
<math>abc + abd + acd + bcd = - 200</math>
+
*<math>abc + abd + acd + bcd = - 200</math>
  
 
Substituting given values obtains the following:
 
Substituting given values obtains the following:
  
<math>ac + ad + bc + bd = k - 62 + 32 = k - 30</math>
+
*<math>ac + ad + bc + bd = k - 62 + 32 = k - 30</math>
<math>(a + b)(c + d) = k - 30</math>
+
*<math>(a + b)(c + d) = k - 30</math>
<math>- 32c + 62b + 62a - 32d = - 200</math>
+
*<math>- 32c + 62b + 62a - 32d = - 200</math>
<math>31(a + b) - 16(c + d) = - 100</math>
+
*<math>31(a + b) - 16(c + d) = - 100</math>
*Multiplying the equation by 16 gives
+
 
 +
Multiplying the * equation by 16 gives
  
 
<math>16(a + b) + 16(c + d) = 288</math>. Adding it to the previous equation gives
 
<math>16(a + b) + 16(c + d) = 288</math>. Adding it to the previous equation gives
  
<math>47(a + b) = 188</math>
+
*<math>47(a + b) = 188</math>
<math>a + b = 4</math>
+
*<math>a + b = 4</math>
<math>c + d = 18 - 4 = 14</math>
+
*<math>c + d = 18 - 4 = 14</math>
<math>(a + b)(c + d) = 4(14) = 56 = k - 30</math>
+
*<math>(a + b)(c + d) = 4(14) = 56 = k - 30</math>
<math>k = \boxed{86}</math>
+
*<math>k = \boxed{86}</math>

Revision as of 11:35, 11 August 2009

1984 USAMO 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:

Let the four roots be $a,b,c,d$. By Vieta's Formulas, we have the following:

  • $ab = - 32$
  • $abcd = - 1984$
  • $cd = 62$
  • $a + b + c + d = 18$*
  • $ab + ac + ad + bc + bd + cd = k$
  • $abc + abd + acd + bcd = - 200$

Substituting given values obtains the following:

  • $ac + ad + bc + bd = k - 62 + 32 = k - 30$
  • $(a + b)(c + d) = k - 30$
  • $- 32c + 62b + 62a - 32d = - 200$
  • $31(a + b) - 16(c + d) = - 100$

Multiplying the * equation by 16 gives

$16(a + b) + 16(c + d) = 288$. Adding it to the previous equation gives

  • $47(a + b) = 188$
  • $a + b = 4$
  • $c + d = 18 - 4 = 14$
  • $(a + b)(c + d) = 4(14) = 56 = k - 30$
  • $k = \boxed{86}$