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1996 AIME Problems/Problem 5: Difference between revisions

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== Solution ==
== Solution ==
Use [[Vieta's formulas]]. These tell us that <math>a + b + c = s = -3</math>, <math>ab + bc + ca = 4</math>, and <math>abc = 11</math> for the first polynomial. Then:
By [[Vieta's formulas]] on the polynomial <math>P(x) = x^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0</math>, we have <math>a + b + c = s = -3</math>, <math>ab + bc + ca = 4</math>, and <math>abc = 11</math>. Then
<div style="text-align:center;">
<center><math>t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)</math></center>
<math>\begin{eqnarray*}
This is just the definition for <math>-P(-3) = \boxed{023}</math>.
t &=& -(a+b)(b+c)(c+a)\\
 
&=& -[(s-a)(s-b)(s-c)]\\
Alternatively, we can expand the expression to get
&=& -[(-3-a)(-3-b)(-3-c)]\\
<center><math>\begin{align*}
  &=& [(a+3)(b+3)(c+3)]\\
t &= -(-3-a)(-3-b)(-3-c)\\
  &=& abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\
  &= (a+3)(b+3)(c+3)\\
t &=& 11 + 3(4) + 9(-3) + 27 = 23</math></div>
  &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\
t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}</math></center>


== See also ==
== See also ==

Revision as of 15:18, 21 April 2008

Problem

Suppose that the roots of $x^3+3x^2+4x-11=0$ are $a$, $b$, and $c$, and that the roots of $x^3+rx^2+sx+t=0$ are $a+b$, $b+c$, and $c+a$. Find $t$.

Solution

By Vieta's formulas on the polynomial $P(x) = x^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0$, we have $a + b + c = s = -3$, $ab + bc + ca = 4$, and $abc = 11$. Then

$t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)$

This is just the definition for $-P(-3) = \boxed{023}$.

Alternatively, we can expand the expression to get

$\begin{align*}

t &= -(-3-a)(-3-b)(-3-c)\\ 

&= (a+3)(b+3)(c+3)\\
&= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\
t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

See also

1996 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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