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Difference between revisions of "1996 AIME Problems/Problem 5"
m (→‎Solution)
Line 8: Line 8:
  
 
Alternatively, we can expand the expression to get
 
Alternatively, we can expand the expression to get
−
<center><math>\begin{align*}
+
<cmath>\begin{align*}
 
t &= -(-3-a)(-3-b)(-3-c)\\
 
t &= -(-3-a)(-3-b)(-3-c)\\
 
  &= (a+3)(b+3)(c+3)\\
 
  &= (a+3)(b+3)(c+3)\\
 
  &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\
 
  &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\
−
t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}</math></center>
+
t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}</cmath>
  
  

Revision as of 11:45, 13 March 2015

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*}


A third solution arises if it is seen that each term in the expansion of $(a+b)(b+c)(c+a)$ has a total degree of 3. Another way to get terms with degree 3 is to multiply out $(a+b+c)(ab+bc+ca)$. Expanding both of these expressions and comparing them shows that:

$(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc$ $t = -(a+b)(b+c)(c+a) = abc-(ab+bc+ca)(a+b+c) = 11-(4)(-3) = 23$

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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