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

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==Solutions==
 
==Solutions==
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==Solution 1==
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Note that we can add the two equations to yield the equation
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<math>x^2 + y^2 - 4x - 4y + 8 = x + y + 8.</math>
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Moving terms gives the equation
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<math>x^2+y^2=5 \left( x + y \right).</math>
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We can also subtract the two equations to yield the equation
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<math>x^2 - y^2 - 4x +4y = y - x.</math>
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Moving terms gives the equation
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<math>x^2 - y^2 = 3x - 3y.</math>
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Because <math>x \neq y,</math> we can divide both sides of the equation by <math>x - y</math> to yield the equation
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<math>x + y = 3.</math>
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Substituting this into the equation for <math>x^2 + y^2</math> that we derived earlier gives
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<math>x^2 + y^2 = 5 \left( x + y \right) = 5 \left( 3 \right) = \boxed{\textbf{(B) } 15}</math>
  
 
==Solution 1==
 
==Solution 1==

Revision as of 17:13, 5 January 2019

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 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: This is risky, as $20$ could be a viable answer too. Do not use this method unless sure about the answer. Also, very time consuming to graph the equations, do not attempt with time limit.

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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