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