2015 AMC 10A Problems/Problem 16

Problem

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

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

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

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

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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