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 $x^2+y^2$ ?

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

Solution

Our equations simplify to (after subtracting 4 from both sides): $y = x^2 - 4x,$ $x = y^2 - 4y.$ Subtract the equations to obtain $y - x = x^2 - y^2 - 4x + 4y$ , so $x^2 - y^2 = 3x - 3y$ . This factors as $(x - y)(x + y) = 3(x - y)$ , and so because $x \neq y$ , we have $x + y = 3$ .

Add the equations to yield $x + y = x^2 + y^2 - 4(x + y)$ . Hence, $x^2 + y^2 = 5(x + y) = 15$ , so our answer is $\boxed{\textbf{(B)}}$ .

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2015 AMC 10A Problems/Problem 16

Problem

Solution