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 "2015 AMC 10A Problems/Problem 16"

(→‎Solution)
(→‎Solution 2)
Line 30: Line 30:
  
 
<math>x^2 + y^2 = 5 \left( x + y \right) = 5 \left( 3 \right) = \boxed{\textbf{(B) } 15}</math>
 
<math>x^2 + y^2 = 5 \left( x + y \right) = 5 \left( 3 \right) = \boxed{\textbf{(B) } 15}</math>
 +
 +
==Solution 2 (Algebraic) ==
 +
 +
Subtract <math>4</math> from the LHS of both equations, and use difference of squares to yield the equations
 +
 +
<math>x = y(y-4)</math>  and  <math>y = x(x-4)</math>.
 +
 +
It may save some time to find two solutions, <math>(0, 0)</math> and <math>(5, 5)</math>, at this point. However, <math> x = y</math> in these solutions.
 +
 +
 +
Substitute <math>y = x(x-4)</math> into <math>x = y(y-4)</math>.
 +
 +
 +
This gives the equation
 +
 +
<math>x = x(x-4)(x^2-4x-4)</math>
 +
 +
which can be simplified to
 +
 +
<math>x(x^3 - 8x^2 +12x + 15) = 0</math>.
 +
 +
Knowing <math>x = 0</math> and <math> x = 5</math> are solutions now is helpful, as you divide both sides by <math>x(x-5)</math>. This can also be done using polynomial division to find <math>x = 5</math> as a factor. This gives
 +
 +
<math>x^2 - 3x -3 = 0</math>.
 +
 +
Because the two equations <math>x = y(y-4)</math> and <math>y = x(x-4)</math> are symmetric, the <math>x</math> and <math>y</math> values are the roots of the equation, which are <math>x = \frac{3 + \sqrt{21}}{2}</math> and <math>x = \frac{3 - \sqrt{21}}{2}</math>.
 +
 +
Squaring these and adding the together gives
 +
 +
<math>\frac{3^2 + 21 + 6\sqrt{21}}{4} + \frac{3^2 + 21 - 6\sqrt{21}}{4} = \frac{2(3^2 +21)}{4} = (B) 15</math>.
  
 
==See Also==
 
==See Also==

Revision as of 17:42, 13 December 2015

Problem

If $y+4 = (x-2)^2, x+4 = (y-2)^2$, and $x \neq y$, what is the value of $x^2+y^2$?

$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }\text{30}$

Solution 1

Note that we can add the two equations to yield the equation

$x^2 + y^2 - 4x - 4y + 8 = x + y + 8.$

Moving terms gives the equation

$x^2+y^2=5 \left( x + y \right).$

We can also subtract the two equations to yield the equation

$x^2 - y^2 - 4x +4y = y - x.$

Moving terms gives the equation

$x^2 - y^2 = 3x - 3y.$

Because $x \neq y,$ we can divide both sides of the equation by $x - y$ to yield the equation

$x + y = 3.$

Substituting this into the equation for $x^2 + y^2$ that we derived earlier gives

$x^2 + y^2 = 5 \left( x + y \right) = 5 \left( 3 \right) = \boxed{\textbf{(B) } 15}$

Solution 2 (Algebraic)

Subtract $4$ from the LHS of both equations, and use difference of squares to yield the equations

$x = y(y-4)$ and $y = x(x-4)$.

It may save some time to find two solutions, $(0, 0)$ and $(5, 5)$, at this point. However, $x = y$ in these solutions.


Substitute $y = x(x-4)$ into $x = y(y-4)$.


This gives the equation

$x = x(x-4)(x^2-4x-4)$

which can be simplified to

$x(x^3 - 8x^2 +12x + 15) = 0$.

Knowing $x = 0$ and $x = 5$ are solutions now is helpful, as you divide both sides by $x(x-5)$. This can also be done using polynomial division to find $x = 5$ as a factor. This gives

$x^2 - 3x -3 = 0$.

Because the two equations $x = y(y-4)$ and $y = x(x-4)$ are symmetric, the $x$ and $y$ values are the roots of the equation, which are $x = \frac{3 + \sqrt{21}}{2}$ and $x = \frac{3 - \sqrt{21}}{2}$.

Squaring these and adding the together gives

$\frac{3^2 + 21 + 6\sqrt{21}}{4} + \frac{3^2 + 21 - 6\sqrt{21}}{4} = \frac{2(3^2 +21)}{4} = (B) 15$.

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 10 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=2015_AMC_10A_Problems/Problem_16&oldid=73694"