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 $x^2+y^2$ ?

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

Solutions

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

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

Solution 3

By graphing the two equations on a piece of graph paper, we can see that the point where they intersect that is not on the line $y=x$ is close to the point $(4,-1)$ (or $(-1, 4)$ ). $(-1)^2+4^2=17$ , and the closest answer choice to $17$ is $\boxed{\textbf{(B) } 15}$ .

Note for solution 3:

This is risky, as $20$ could be a viable answer too. Do not use this method unless you're sure about the answer. Also, it's very time consuming to graph the equations, so don't attempt it with a time limit. In other words, this solution is less reliable than the others, so only use it if you can't do the other methods.

Solution 4 (When you can't algebraically manipulate)

This, by the way, is a really cheap way of solving the problem but as long as you get an answer, it doesn't matter.

Looking at the first given equation, begin searching for solutions. Notice how $(4,0)$ works but when plugged into the

second equation, you get $8=4$ . Now, if you decrease the $y$ value by $1$ to obtain $(2+\sqrt3,-1)$ , plugging it into

the second equation will yield $6+\sqrt3=9$ . Now, notice how the LHS is now less than the RHS as opposed to what it had

been when we plugged $(4,0)$ into the second given equation. We then conclude that there must be a solution between

$y=0$ and $y=-1$ , so calculating $x^2+y^2$ of $(4,0)$ and $(2+\sqrt3,-1)$ we obtain

$4^2+0^2=16$ and $(2+\sqrt3)^2+(-1)^2=8+4\sqrt3\approx14.9$

Thus we know the answer to be $\boxed{\textbf{(B) } 15}$ .

This is a really crappy solution, but it works. :/

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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