Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(created solution page)
 
(→‎Solution)
 
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
<math>\boxed{015}</math>.
+
 
 +
For some integer <math>n</math>, let <math>s_n</math> be <math>a^n+b^n+c^n</math>.
 +
 
 +
Note that <math>(X+1)(Y+1) = XY+X+Y+1</math>, so, to answer the problem, it suffices to know <math>XY</math> and <math>X+Y</math>.
 +
 
 +
Let <math>f(x)=x^3+2x-1=0</math>. First, note the arbitrary decision in the expression <math>\tfrac a b + \tfrac b c + \tfrac c a</math>. Why must <math>a</math> be over <math>b</math>, but not <math>b</math> over <math>a</math>? From this observation, we can deduce that the aforementioned two possible values of this sum are <math>\tfrac a b + \tfrac b c + \tfrac c a</math> ([[WLOG]] let this be <math>X</math>) and <math>\tfrac b a + \tfrac c b + \tfrac a c</math> (WLOG let this be <math>Y</math>). From these definitions and the knowledge from [[Vieta's Formulas]] that <math>abc=1</math>, 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 <math>(a+b+c)^3</math>, so we look for a way to get a value for <math>a+b+c</math>. Fortunately, we can use Vieta again to see that <math>a+b+c=0</math>. Thus, <math>(a+b)^3=0</math>, and so <math>a^3+b^3+c^3+3(a^2b+a^2c+ab^2+b^2c+ac^2+bc^2)+6abc=0</math>, or, by substitution and recalling that <math>abc=1</math>, <math>s_3+3(X+Y)+6=0</math>. To get <math>s_3</math>, 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 <math>s_3+3(X+Y)+6=0</math> becomes <math>3+3(X+Y)+6=0</math>, so <math>X+Y=-3</math>.
 +
 
 +
Now that we have <math>X+Y</math>, we desire to find <math>XY</math>. Note that Vieta gives us <math>ab+bc+ac=2</math>, so, because <math>abc=1</math>, by substitution <math>\tfrac1 c + \tfrac1 a +\tfrac1 b = s_{-1} = 2</math>. By substituting our values for <math>X</math> and <math>Y</math> into <math>XY</math>, 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 <math>s_{-3}</math>. To do this, we try to think of a function <math>g(x)</math> whose roots are <math>\tfrac1a, \tfrac1b,</math> and <math>\tfrac1c</math>. <math>g(x)=-x^3f(\tfrac1x)=x^3-2x^2-1</math> will work. Using Newton Sums again and recalling that  <math>s_{-1}=2</math>, 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 <math>s_{-3}=11</math> and <math>s_3=3</math>, we have that <math>XY=3+s_3+s_{-3}=3+3+11=17</math>.
 +
 
 +
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*}
  
 
== See also ==
 
== See also ==

Latest revision as of 16:31, 30 July 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

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 $XY=3+s_3+s_{-3}=3+3+11=17$.

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*}

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_Mock_AIME_I_Problems/Problem_11&oldid=223999"