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Difference between revisions of "1996 AIME Problems/Problem 5"
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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
 
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
 
<center><math>t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)</math></center>
 
<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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This is just the definition for <math>-P(-3) = \boxed{23}</math>.
  
 
Alternatively, we can expand the expression to get
 
Alternatively, we can expand the expression to get
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<math>(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc</math>
 
<math>(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc</math>
 
<math>t = -(a+b)(b+c)(c+a) = abc-(ab+bc+ca)(a+b+c) = 11-(4)(-3) = 23</math>
 
<math>t = -(a+b)(b+c)(c+a) = abc-(ab+bc+ca)(a+b+c) = 11-(4)(-3) = 23</math>
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 +
 +
A way to realize <math>(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc</math>:
 +
 +
<math>(a + b)(b+c)(c+a) = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc + abc</math>
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 +
(Add an extra <math>abc</math>)
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<math>a^2b + a^2c + abc + b^2a + b^2c + abc + c^2a + c^2b + abc - abc =</math>
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<math>a(ab + bc + ac) + b(ab + bc + ac) + c(ab + bc + ac) - abc = </math>
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<math>(a + b + c)(ab + bc + ac) - abc = (-3)(4) - 11 = -23</math>.
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The value of <math>t</math> is the negation of this, which is <math>-(-23) = \boxed{23}</math>
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~Extremelysupercooldude
  
 
== Solution 3 ==
 
== Solution 3 ==
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After simplifying, we get that the polynomial is <math>x'^3 - 6x'^2 + 13x' - 23 = 0.</math>
 
After simplifying, we get that the polynomial is <math>x'^3 - 6x'^2 + 13x' - 23 = 0.</math>
−
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>
+
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{23}.</math>
  
 
~MathWhiz35
 
~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}}

Latest revision as of 18:39, 8 September 2024

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 $-P(-3) = \boxed{23}$.

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:

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


A way to realize $(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc$:

$(a + b)(b+c)(c+a) = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc + abc$

(Add an extra $abc$)

$a^2b + a^2c + abc + b^2a + b^2c + abc + c^2a + c^2b + abc - abc =$

$a(ab + bc + ac) + b(ab + bc + ac) + c(ab + bc + ac) - abc =$

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

The value of $t$ is the negation of this, which is $-(-23) = \boxed{23}$

~Extremelysupercooldude

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 $t = -(\frac{-23}{1}) = \boxed{23}.$

~MathWhiz35

See also

1996 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions

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