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

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

@@ Line 1: / Line 1: @@
 == Problem ==
+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 ==
+== Video Solution ==
+https://youtu.be/3dfbWzOfJAI?t=2785
+~ pi_is_3.14
+== Solution 1 ==
+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>
+This is just the definition for <math>-P(-3) = \boxed{23}</math>.
+Alternatively, we can expand the expression to get
+<cmath>\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*}</cmath>
+== Solution  2==
+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:
+<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>
+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>
+(Add an extra <math>abc</math>)
+<math>a^2b + a^2c + abc + b^2a + b^2c + abc + c^2a + c^2b + abc - abc =</math>
+<math>a(ab + bc + ac) + b(ab + bc + ac) + c(ab + bc + ac) - abc = </math>
+<math>(a + b + c)(ab + bc + ac) - abc = (-3)(4) - 11 = -23</math>.
+The value of <math>t</math> is the negation of this, which is <math>-(-23) = \boxed{23}</math>
+~Extremelysupercooldude
+== Solution 3 ==
+We have that <math>x^3+3x^2+4x-11=0</math> for roots <math>a,b,c.</math>
+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>
+By Vieta's formulae, we see that <math>t= -(a+b)(b+c)(a+c).</math>
+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.
+Using this fact, we can rewrite <math>t</math> as <math>-(-3-a)(-3-b)(-3-c) = (a+3)(b+3)(c+3).</math>
+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 <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>.
+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{23}.</math>
+~MathWhiz35
 == See also ==
-* [[1996 AIME Problems]]
+{{AIME box|year=1996|num-b=4|num-a=6}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

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