Difference between revisions of "1996 AIME Problems/Problem 5"

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Suppose that the [[root]]s of <math>x^3+3x^2+4x-11=0</math> are <math>a</math>, <math>b</math>, and <math>c</math>, and that the roots of <math>x^3+rx^2+sx+t=0</math> are <math>a+b</math>, <math>b+c</math>, and <math>c+a</math>. Find <math>t</math>.
 
Suppose that the [[root]]s of <math>x^3+3x^2+4x-11=0</math> are <math>a</math>, <math>b</math>, and <math>c</math>, and that the roots of <math>x^3+rx^2+sx+t=0</math> are <math>a+b</math>, <math>b+c</math>, and <math>c+a</math>. Find <math>t</math>.
  
== Solution ==
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== Video 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:
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https://youtu.be/3dfbWzOfJAI?t=2785
<div style="text-align:center;">
 
<math>\begin{eqnarray*}
 
t &=& -(a+b)(b+c)(c+a)\\
 
&=& -[(s-a)(s-b)(s-c)]\\
 
&=& -[(-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</math></div>
 
  
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~ pi_is_3.14
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== Solution 1 ==
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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
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<center><math>t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)</math></center>
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This is just the definition for <math>-P(-3) = \boxed{023}</math>.
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Alternatively, we can expand the expression to get
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<cmath>\begin{align*}
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t &= -(-3-a)(-3-b)(-3-c)\\
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&= (a+3)(b+3)(c+3)\\
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&= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\
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t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}</cmath>
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== Solution  2==
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Each term in the expansion of <math>(a+b)(b+c)(c+a)</math> has a total degree of 3.  Another way to get terms with degree 3 is to multiply out <math>(a+b+c)(ab+bc+ca)</math>.  Expanding both of these expressions and comparing them shows that:
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<math>(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc</math>
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<math>t = -(a+b)(b+c)(c+a) = abc-(ab+bc+ca)(a+b+c) = 11-(4)(-3) = 23</math>
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== Solution 3 ==
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We have that <math>x^3+3x^2+4x-11=0</math> for roots <math>a,b,c.</math>
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In the second cubic function <math>x^3+rx^2+sx+t=0,</math> the roots are <math>a+b,b+c,c+a.</math>
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By Vieta's formulae, we see that <math>t= -(a+b)(b+c)(a+c).</math>
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As we know that the sum of the roots of the first polynomial, <math>a+b+c</math> is <math>-3</math> by applying Vieta's again.
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Using this fact, we can rewrite <math>t</math> as <math>-(-3-a)(-3-b)(-3-c) = (a+3)(b+3)(c+3).</math>
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Seeing this, we can find the value of the product of the roots by applying this to the first equation.
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This can be done by setting <math>x'=a+3,</math> so, from this, we see that we should plug in <math>x'-3</math> for <math>x</math> in <math>x^3+3x^2+4x-11=0</math>.
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After simplifying, we get that the polynomial is <math>x'^3 - 6x'^2 + 13x' - 23 = 0.</math>
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Given that the product of the roots of this equation is equivalent to our desired value for <math>t</math>, we can apply Vieta's formulae for a third time to find that <math>t = -(\frac{-23}{1}) = \boxed{023}.</math>
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~MathWhiz35
 
== See also ==
 
== See also ==
 
{{AIME box|year=1996|num-b=4|num-a=6}}
 
{{AIME box|year=1996|num-b=4|num-a=6}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 20:53, 14 February 2021

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 $t$.

Video Solution

https://youtu.be/3dfbWzOfJAI?t=2785

~ pi_is_3.14

Solution 1

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*}

Solution 2

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:

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

Solution 3

We have that $x^3+3x^2+4x-11=0$ for roots $a,b,c.$ In the second cubic function $x^3+rx^2+sx+t=0,$ the roots are $a+b,b+c,c+a.$

By Vieta's formulae, we see that $t= -(a+b)(b+c)(a+c).$ As we know that the sum of the roots of the first polynomial, $a+b+c$ is $-3$ by applying Vieta's again.

Using this fact, we can rewrite $t$ as $-(-3-a)(-3-b)(-3-c) = (a+3)(b+3)(c+3).$

Seeing this, we can find the value of the product of the roots by applying this to the first equation. This can be done by setting $x'=a+3,$ so, from this, we see that we should plug in $x'-3$ for $x$ in $x^3+3x^2+4x-11=0$.

After simplifying, we get that the polynomial is $x'^3 - 6x'^2 + 13x' - 23 = 0.$ Given that the product of the roots of this equation is equivalent to our desired value for $t$, we can apply Vieta's formulae for a third time to find that $t = -(\frac{-23}{1}) = \boxed{023}.$

~MathWhiz35

See also

1996 AIME (ProblemsAnswer KeyResources)
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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