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2015 AMC 10A Problems/Problem 16

Revision as of 19:10, 4 February 2015 by Suli (talk | contribs)

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 $\boxed{\textbf{(B)}}$.

Solution 2

First simplify the equations

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

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

$y=x^2-4x$ and the the other equation will become $x=y^2-4y$

Substitute $y$ into $x=y^2-4y$ to get

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

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

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

$1=x^3-8x^2+12x+16$

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

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