2015 AMC 10A Problems/Problem 16

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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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