Difference between revisions of "2015 AMC 10A Problems/Problem 16"

Revision as of 18:16, 4 February 2015

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

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

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

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

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 ==Solution==
-Our equations simplify to (after subtracting 4 from both sides):
-<cmath>y = x^2 - 4x,</cmath>
-<cmath>x = y^2 - 4y.</cmath>
-Subtract the equations to obtain <math>y - x = x^2 - y^2 - 4x + 4y</math>, so <math>x^2 - y^2 = 3x - 3y</math>. This factors as <math>(x - y)(x + y) = 3(x - y)</math>, and so because <math>x \neq y</math>, we have <math>x + y = 3</math>.
-Add the equations to yield <math>x + y = x^2 + y^2 - 4(x + y)</math>. Hence, <math>x^2 + y^2 = 5(x + y) = 15</math>, so our answer is <math>\boxed{\textbf{(B)}}</math>.
+Note that we can add the two equations to yield the equation
+<math>x^2 + y^2 - 4x - 4y + 8 = x + y + 8.</math>
+Moving terms gives the equation
+<math>x^2+y^2=5 \left( x + y \right).</math>
+We can also subtract the two equations to yield the equation
+<math>x^2 - y^2 - 4x +4y = y - x.</math>
+Moving terms gives the equation
+<math>x^2 - y^2 = 3x - 3y.</math>
+Because <math>x \neq y,</math> we can divide both sides of the equation by <math>x - y</math> to yield the equation
+<math>x + y = 3.</math>
+Substituting this into the equation for <math>x^2 + y^2</math> that we derived earlier gives
+<math>x^2 + y^2 = 5 \left( x + y \right) = 5 \left( 3 \right) = \boxed{\left( \text{B} \right) 15}.</math>