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

$x^2 - y^2 = (x + y)(x - y) = 3x - 3y = 3(x - y).$

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. 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 bad solution, but it works. :/

Solution 5 (Very in Depth) Note: In this solution, I explain my entire thought process and the reason why I did each step, so you can read a more intuitive solution rather than a contrived one.

We first notice that each equation has a 4, which inspires us to isolate the 4 on both equations and equate them to see if we can gather any useful information on $x and$ y. Doing this we get (x-2)^2 $-$ y = (y-2)^2 $-$ x. Now you might be tempted to expand, but notice we have to perfect squares in our equation: (x-2)^2 $and (y-2)^2$ , which motivates us to take advantage of difference of squares. For simplicity we suppose that (x-2) $= a and (y-2)$ =b; thus our equation is now a^2 $-$ y = b^2 $-$ x, we manipulate this equation by getting the LHS in a form that lets us do difference of equations, so we rewrite as a^2 $-b^2$ = -( $x-$ y). Now we can finally use our difference of squares to get ( $a-$ b)(a $+$ b) = -( $x-$ y). Lets now substitute $a and$ b back to get information solely for x and y (our substitution was to make manipulation easier for us, it isn't necessary though). Doing this we get, ( $x-$ y)( $x+$ y-4) = -( $x-$ y). Now let us divide by ( $x-$ y) on both sides and clean up the equation; when we do this, we get that $x+$ y = 3. Okay so what may we do from here? Well we got $x+$ y = $3, which means that this is fact is only of use if we can obtain it in another way, so we add the LHS and the RHS of both equations and equate them to make our fact of use. Doing that, we get ($ x+ $y +8) = (x-2)^2$ + (y-2)^2 $. Also, when we set this equation up we note only make our fact useful but we also can see that after expanding the RHS we get$ x^2+ $y^2 on the RHS too, which makes even a better move that we added both equations because remember our goal is to find$ x^2+ $y^2. Expanding and plugging in$ x+ $y = 3 we get 11 =$ x^2+ $y^2-4($ x+ $y)+$ 8. Oh wait but $x+$ y=3, so we can plug that in and we get $x^2+$ y^2 = $15 or$ \boxed{\textbf{(B) } after cleaning up some of the algebra. ~triggod

Video Solution

https://youtu.be/0W3VmFp55cM?t=15

~ pi_is_3.14

Video Solution

https://youtu.be/WsRGBqCJwGA

~savannahsolver

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2015 AMC 10A Problems/Problem 16

Contents

Problem

Solutions

Solution 1

Solution 2 (Algebraic)

Solution 3

Note for solution 3:

Solution 4 (When you can't algebraically manipulate)

Video Solution

Video Solution

See Also