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2015 AMC 10A Problems/Problem 16

Revision as of 20:02, 16 January 2018 by Lukefly2 (talk | contribs) (→‎Solution 2 (Algebraic))

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 1

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

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

Solution 2 (Algebraic)

Subtract $4$ from the left hand side of both equations, and use difference of squares to yield the equations

$x = y(y-4)$ and $y = x(x-4)$.

It may save some time to find two solutions, $(0, 0)$ and $(5, 5)$, at this point. However, $x = y$ in these solutions.


Substitute $y = x(x-4)$ into $x = y(y-4)$.


This gives the equation

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

which can be simplified to

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

Knowing $x = 0$ and $x = 5$ are solutions is now helpful, as you divide both sides by $x(x-5)$. This can also be done using polynomial division to find $x = 5$ as a factor. This gives

$x^2 - 3x -3 = 0$.

Because the two equations $x = y(y-4)$ and $y = x(x-4)$ are symmetric, the $x$ and $y$ values are the roots of the equation, which are $x = \frac{3 + \sqrt{21}}{2}$ and $x = \frac{3 - \sqrt{21}}{2}$.

Squaring these and adding them together gives

$\frac{3^2 + 21 + 6\sqrt{21}}{4} + \frac{3^2 + 21 - 6\sqrt{21}}{4} = \frac{2(3^2 +21)}{4} = \boxed{\textbf{(B) } 15}$.

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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