2013 Mock AIME I Problems/Problem 11

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

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: \begin{align*} X = \frac a b + \frac b c + \frac c a &= \frac{a^2c+ab^2+bc^2}{abc} = a^2c+ab^2+bc^2 \\ Y = \frac b a + \frac c b + \frac a c &= \frac{b^2c+ac^2+a^2b}{abc} = b^2c+ac^2+a^2b \end{align*} 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: \begin{align*} s_3 + 0s_2 + 2s_1 + 3(-1) &= 0 \\ s_3 &= -2(a+b+c)+3 \\ s_3 &= 3 \end{align*} 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: \begin{align*} XY &= (\frac a b + \frac b c + \frac c a)(\frac b a + \frac c b + \frac a c) \\ &= 1 + \frac{ac}{b^2} + \frac{a^2}{bc} + \frac{b^2}{ac} + 1 + \frac{ab}{c^2} + \frac{bc}{a^2} + \frac{c^2}{ab} + 1 \\ \text{Recalling that } abc=1 &\text{, we have the following by substitution:} \\ XY &= 3 + a^3+b^3+c^3 + \frac1{a^3} + \frac1{b^3} + \frac1{c^3} \\ &= 3+s_3+s_{-3} \end{align*} 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: \begin{align*} s_{-2}+(-2)s_{-1}+2(0) &= 0 \\ s_{-2} &= 4 \end{align*} We use Newton Sums a third time: \begin{align*} s_{-3}+(-2)s_{-2}+0s_{-1}+3(-1) &= 0 \\ s_{-3} &= 8 + 3 = 11 \end{align*} Thus, because $s_{-3}=11$ and $s_3=3$ , we have that $XY=3+s_3+s_{-3}=3+3+11=17$ .