Difference between revisions of "2015 AMC 10A Problems/Problem 16"

Latest revision as of 23:09, 23 September 2024

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

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

Alternative Ending

Note that $x^2+y^2 = (x+y)^2 - 2xy$ . Using Vieta's, we find that $(x+y)^2 = 3^2 = 9$ and $-2xy = -2\cdot 3 = -6$ . Subtracting them gives $\boxed{\textbf{(B) } 15}$ .

~Andrew2019

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

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 and reasoned)

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 how we have two perfect squares in our equation: $(x-2)^2$ and $(y-2)^2,$ which motivates us to take advantage of the 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 squares, so we can rewrite as $a^2-b^2 = -(x-y).$

Now we can finally use our difference of squares to get $(a-b)(a+b) = -(x-y).$

Let's now substitute $a$ and $b$ back to get information solely for $x$ and $y.$

Doing this we get, $(x-y)(x+y-4) = -(x-y).$

Now let us divide $x-y$ on both sides and clean up the equation. When we do this, we get that $x+y = 3.$

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 essence.

Doing that, we get $x+y +8 = (x-2)^2+ (y-2)^2.$ Also, when we set this equation up we not only do we 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 it 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 $11 = x^2+y^2-4(x+y)+8.$ $x+y=3,$ so we can plug that in and we get $x^2+y^2 = 15$ or $\boxed{\textbf{(B) } 15}$ after cleaning up some algebra.

~triggod

Solution 6 (Quickest)

Subtract the two equations to get $(x-y)= (x+y-4)(y-x) \iff x+y=3$ . Plugging back into any of the original equations yields $x^2-3x=3$ . However, we know $x^2+y^2= x^2+(3-x)^2= 2x^2-6x+9$ . Then, we can just multiply the first by two and add $9$ to get $\boxed{\textbf{(B) } 15}$

Video Solutions

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution 2

https://youtu.be/WsRGBqCJwGA

~savannahsolver

@@ Line 62: / Line 62: @@
 <math>\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}</math>.
+====Alternative Ending====
+Note that <math>x^2+y^2 = (x+y)^2 - 2xy</math>. Using Vieta's, we find that <math>(x+y)^2 = 3^2 = 9</math> and <math>-2xy = -2\cdot 3 = -6</math>. Subtracting them gives <math>\boxed{\textbf{(B) } 15}</math>.
+~Andrew2019
 ===Solution 3===
@@ Line 114: / Line 119: @@
 ~triggod
-==Solution 6 (I would never trust myself to do this on test day) ==
+===Solution 6 (Quickest) ===
+Subtract the two equations to get <math>(x-y)= (x+y-4)(y-x) \iff x+y=3</math>. Plugging back into any of the original equations yields <math>x^2-3x=3</math>. However, we know <math>x^2+y^2= x^2+(3-x)^2= 2x^2-6x+9</math>. Then, we can just multiply the first by two and add <math>9</math> to get <math>\boxed{\textbf{(B) } 15}</math>
-Expanding the given equations and subtracting a <math>4</math> from both equations gives us
-<math>y=x^2-4x</math> and <math>x=y^2-4y</math>.
-Adding these two and isolating <math>0</math> allows us to obtain the new equation
-<math>x^2-5x+y^2-5y=0.</math>
-We can complete the square twice here (creating an equation that represents a circle), both in <math>x</math> and <math>y</math>, by adding <math>\frac{25}{4}</math> to both sides and rearranging the equation. Doing so shows us that
-<math>\left(x-\frac{5}{2}\right)^2+\left(y-\frac{5}{2}\right)^2=\frac{25}{2}.</math>
-Recall that if a circle is centered at the origin, its equation is <math>x^2+y^2=r^2</math>. If we were to shift our circle's center down and to the left, each by <math>\frac{5}{2}</math>, its center would change, but its radius would remain constant.
-However, all of the points <math>(x, y)</math> within the circle are <math>\frac{5}{2}</math> less than that of the original circle, due to the reduction in both <math>x</math> and <math>y</math>.
-So now our equation to solve is <math>x^2+y^2-\frac{5}{2}=\frac{25}{2}</math>. Adding <math>\frac{5}{2}</math> gives us
-<math>x^2+y^2=\frac{30}{2}=\boxed{\textbf{(B) } 15}</math>
--Benedict T (countmath1)
 ==Video Solutions==
-=== Video Solution 1 ===
+=== Video Solution by OmegaLearn ===
 https://youtu.be/0W3VmFp55cM?t=15

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