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

Revision as of 00:31, 3 November 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 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 to

the test.

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. :/

@@ Line 68: / Line 68: @@
 ====Note for solution 3:====
 This is risky, as <math>20</math> 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 to
+ the test.
+Looking at the first given equation, begin searching for solutions. Notice how <math>(4,0)</math> works but when plugged into the
+second equation, you get <math>8=4</math>. Now, if you decrease the <math>y</math> value by <math>1</math> to obtain <math>(2+\sqrt3,-1)</math>, plugging it into
+the second equation will yield <math>6+\sqrt3=9</math>. Now, notice how the LHS is now less than the RHS as opposed to what it had
+been when we plugged <math>(4,0)</math> into the second given equation. We then conclude that there must be a solution between
+<math>y=0</math> and <math>y=-1</math>, so calculating <math>x^2+y^2</math> of <math>(4,0)</math> and <math>(2+\sqrt3,-1)</math> we obtain
+<cmath>4^2+0^2=16</cmath> and
+<cmath>(2+\sqrt3)^2+(-1)^2=8+4\sqrt3\approx14.9</cmath>
+Thus we know the answer to be <math>\boxed{\textbf{(B) } 15}</math>.
+This is a really crappy solution, but it works. :/
 ==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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