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 AMC 12A Problems/Problem 8"

(→‎See also)
(→‎Solution 4)
 
(18 intermediate revisions by 6 users not shown)
Line 5: Line 5:
 
<math> \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad </math>
 
<math> \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad </math>
  
−
==Solution==
+
==Solution 1==
  
 
<math> x+\tfrac{2}{x}= y+\tfrac{2}{y} </math>
 
<math> x+\tfrac{2}{x}= y+\tfrac{2}{y} </math>
Line 12: Line 12:
  
 
Cross multiply in either equation, giving us <math>xy=2</math>.
 
Cross multiply in either equation, giving us <math>xy=2</math>.
 +
 +
<math>\boxed{\textbf{(D) }{2}}</math>
 +
 +
==Solution 2==
 +
 +
<math>x+\tfrac{2}{x}= y+\tfrac{2}{y}</math>
 +
 +
<math>x-y+\frac{2}{x}-\frac{2}{y} = 0</math>
 +
 +
<math>(x-y)+2(\frac{y-x}{xy}) = 0</math>
 +
 +
<math>(x-y)(1-\frac{2}{xy})=0</math>
 +
 +
Since <math>x\not=y</math>
 +
 +
<math>1 = \frac{2}{xy}</math>
 +
 +
<math>xy = 2</math>
 +
 +
== Solution 3 ==
 +
 +
Let <math>A = x + \frac{2}{x} = y + \frac{2}{y}.</math> Consider the equation <cmath>u + \frac{2}{u} = A.</cmath> Reorganizing, we see that <math>u</math> satisfies <cmath>u^2 - Au + 2 = 0.</cmath> Notice that there can be at most two distinct values of <math>u</math> which satisfy this equation, and <math>x</math> and <math>y</math> are two distinct possible values for <math>u.</math> Therefore, <math>x</math> and <math>y</math> are roots of this quadratic, and by Vieta’s formulas we see that <math>xy</math> thereby must equal <math>\boxed{2}.</math>
 +
 +
~ Professor-Mom
 +
 +
== Solution 4 ==
 +
 +
<cmath>x + \frac{2}{x} = y + \frac{2}{y}.</cmath>
 +
 +
Multiply both sides by xy to get
 +
 +
<cmath>x^2y + 2y = y^2x +2x </cmath>
 +
 +
Rearrange to get
 +
 +
<cmath>x^2y - y^2x = 2x - 2y</cmath>
 +
 +
Factor out <math>xy</math> on the left side and <math>2</math> on the right side to get
 +
 +
<cmath> xy(x-y) = 2(x-y) </cmath>
 +
 +
Divide by <math>(x-y)</math> {You can do this since x and y are distinct} to get
 +
 +
<math>\boxed{\textbf{(D) }{2}}</math>
 +
~ e__ (the goat)
 +
 +
== Video Solution ==
 +
https://youtu.be/ba6w1OhXqOQ?t=1129
 +
 +
~ pi_is_3.14
 +
 +
==Video Solution==
 +
 +
https://youtu.be/CCjcMVtkVaQ
 +
~sugar_rush
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2013|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:20, 7 March 2024

Problem

Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$, what is $xy$?

$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$

Solution 1

$x+\tfrac{2}{x}= y+\tfrac{2}{y}$

Since $x\not=y$, we may assume that $x=\frac{2}{y}$ and/or, equivalently, $y=\frac{2}{x}$.

Cross multiply in either equation, giving us $xy=2$.

$\boxed{\textbf{(D) }{2}}$

Solution 2

$x+\tfrac{2}{x}= y+\tfrac{2}{y}$

$x-y+\frac{2}{x}-\frac{2}{y} = 0$

$(x-y)+2(\frac{y-x}{xy}) = 0$

$(x-y)(1-\frac{2}{xy})=0$

Since $x\not=y$

$1 = \frac{2}{xy}$

$xy = 2$

Solution 3

Let $A = x + \frac{2}{x} = y + \frac{2}{y}.$ Consider the equation \[u + \frac{2}{u} = A.\] Reorganizing, we see that $u$ satisfies \[u^2 - Au + 2 = 0.\] Notice that there can be at most two distinct values of $u$ which satisfy this equation, and $x$ and $y$ are two distinct possible values for $u.$ Therefore, $x$ and $y$ are roots of this quadratic, and by Vieta’s formulas we see that $xy$ thereby must equal $\boxed{2}.$

~ Professor-Mom

Solution 4

\[x + \frac{2}{x} = y + \frac{2}{y}.\]

Multiply both sides by xy to get

\[x^2y + 2y = y^2x +2x\]

Rearrange to get

\[x^2y - y^2x = 2x - 2y\]

Factor out $xy$ on the left side and $2$ on the right side to get

\[xy(x-y) = 2(x-y)\]

Divide by $(x-y)$ {You can do this since x and y are distinct} to get

$\boxed{\textbf{(D) }{2}}$ ~ e__ (the goat)

Video Solution

https://youtu.be/ba6w1OhXqOQ?t=1129

~ pi_is_3.14

Video Solution

https://youtu.be/CCjcMVtkVaQ ~sugar_rush

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_8&oldid=216591"