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Difference between revisions of "2013 Mock AIME I Problems/Problem 11"

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Let <math>a,b,</math> and <math>c</math> be the roots of the equation <math>x^3+2x-1=0</math>, and let <math>X</math> and <math>Y</math> be the two possible values of <math>\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}.</math>  Find <math>(X+1)(Y+1)</math>.
 
Let <math>a,b,</math> and <math>c</math> be the roots of the equation <math>x^3+2x-1=0</math>, and let <math>X</math> and <math>Y</math> be the two possible values of <math>\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}.</math>  Find <math>(X+1)(Y+1)</math>.
  
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== Solution ==
+
== Solution 1 ==
  
 
For some integer <math>n</math>, let <math>s_n</math> be <math>a^n+b^n+c^n</math>.
 
For some integer <math>n</math>, let <math>s_n</math> be <math>a^n+b^n+c^n</math>.
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&= \boxed{015}.
 
&= \boxed{015}.
 
\end{align*}
 
\end{align*}
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== Solution 2(Sketch) ==
 +
 +
This solution can also be solved by using Vieta's and getting <math>a+b+c=0</math>, <math>ab+bc+ac=2</math>, <math>abc=1</math>.
 +
 +
Then you can directly expand <math>(a^2c+b^2a+c^2b+1)(a^2b+ac^2+b^2c+1)</math>. And solve for the seperate parts.
 +
 +
A key step in the solution from there is realizing that <math>x^3=-2x+1</math>, which can be used to simplify by a great deal.
 +
 +
~Bigbrain_2009
  
 
== See also ==
 
== See also ==

Latest revision as of 12:59, 31 December 2024

Problem

Let $a,b,$ and $c$ be the roots of the equation $x^3+2x-1=0$, and let $X$ and $Y$ be the two possible values of $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}.$ Find $(X+1)(Y+1)$.

Solution 1

For some integer $n$, let $s_n$ be $a^n+b^n+c^n$.

Note that $(X+1)(Y+1) = XY+X+Y+1$, so, to answer the problem, it suffices to know $XY$ and $X+Y$.

Let $f(x)=x^3+2x-1=0$. First, note the arbitrary decision in the expression $\tfrac a b + \tfrac b c + \tfrac c a$. Why must $a$ be over $b$, but not $b$ over $a$? From this observation, we can deduce that the aforementioned two possible values of this sum are $\tfrac a b + \tfrac b c + \tfrac c a$ (WLOG let this be $X$) and $\tfrac b a + \tfrac c b + \tfrac a c$ (WLOG let this be $Y$). From these definitions and the knowledge from Vieta's Formulas that $abc=1$, we can now combine fractions to get the following: X=ab+bc+ca=a2c+ab2+bc2abc=a2c+ab2+bc2Y=ba+cb+ac=b2c+ac2+a2babc=b2c+ac2+a2b Notice that these terms appear in the expansion of $(a+b+c)^3$, so we look for a way to get a value for $a+b+c$. Fortunately, we can use Vieta again to see that $a+b+c=0$. Thus, $(a+b)^3=0$, and so $a^3+b^3+c^3+3(a^2b+a^2c+ab^2+b^2c+ac^2+bc^2)+6abc=0$, or, by substitution and recalling that $abc=1$, $s_3+3(X+Y)+6=0$. To get $s_3$, we think Newton Sums, which give us the following: s3+0s2+2s1+3(−1)=0s3=−2(a+b+c)+3s3=3 Thus, the equation $s_3+3(X+Y)+6=0$ becomes $3+3(X+Y)+6=0$, so $X+Y=-3$.

Now that we have $X+Y$, we desire to find $XY$. Note that Vieta gives us $ab+bc+ac=2$, so, because $abc=1$, by substitution $\tfrac1 c + \tfrac1 a +\tfrac1 b = s_{-1} = 2$. By substituting our values for $X$ and $Y$ into $XY$, we see the following: XY=(ab+bc+ca)(ba+cb+ac)=1+acb2+a2bc+b2ac+1+abc2+bca2+c2ab+1Recalling that abc=1, we have the following by substitution:XY=3+a3+b3+c3+1a3+1b3+1c3=3+s3+s−3 Now, we desire to find $s_{-3}$. To do this, we try to think of a function $g(x)$ whose roots are $\tfrac1a, \tfrac1b,$ and $\tfrac1c$. $g(x)=-x^3f(\tfrac1x)=x^3-2x^2-1$ will work. Using Newton Sums again and recalling that $s_{-1}=2$, we see that: s−2+(−2)s−1+2(0)=0s−2=4 We use Newton Sums a third time: s−3+(−2)s−2+0s−1+3(−1)=0s−3=8+3=11 Thus, because $s_{-3}=11$ and $s_3=3$, we have that $XY=3+s_3+s_{-3}=3+3+11=17$.

Thus, we can now find our desired answer: (X+1)(Y+1)=XY+(X+Y)+1=17−3+1=015.

Solution 2(Sketch)

This solution can also be solved by using Vieta's and getting $a+b+c=0$, $ab+bc+ac=2$, $abc=1$.

Then you can directly expand $(a^2c+b^2a+c^2b+1)(a^2b+ac^2+b^2c+1)$. And solve for the seperate parts.

A key step in the solution from there is realizing that $x^3=-2x+1$, which can be used to simplify by a great deal.

~Bigbrain_2009

See also

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