2013 Mock AIME I Problems/Problem 11
Problem
Let and be the roots of the equation , and let and be the two possible values of Find .
Solution
For some integer , let be .
Note that , so, to answer the problem, it suffices to know and .
Let . First, note the arbitrary decision in the expression . Why must be over , but not over ? From this observation, we can deduce that the aforementioned two possible values of this sum are (WLOG let this be ) and (WLOG let this be ). From these definitions and the knowledge from Vieta's Formulas that , 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 , so we look for a way to get a value for . Fortunately, we can use Vieta again to see that . Thus, , and so , or, by substitution and recalling that , . To get , 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 becomes , so .
Now that we have , we desire to find . Note that Vieta gives us , so, because , by substitution . By substituting our values for and into , 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 . To do this, we try to think of a function whose roots are and . will work. Using Newton Sums again and recalling that , 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 and , we have that .
Thus, we can now find our desired answer: \begin{align*} (X+1)(Y+1) &= XY + (X+Y) + 1 \\ &= 17 - 3 + 1 \\ &= \boxed{015}. \end{align*}