2013 AMC 10B Problems/Problem 14

Problem

Define $a\clubsuit b=a^2b-ab^2$ . Which of the following describes the set of points $(x, y)$ for which $x\clubsuit y=y\clubsuit x$ ?

$\textbf{(A)}\ \text{a finite set of points}\\ \qquad\textbf{(B)}\ \text{one line}\\ \qquad\textbf{(C)}\ \text{two parallel lines}\\ \qquad\textbf{(D)}\ \text{two intersecting lines}\\ \qquad\textbf{(E)}\ \text{three lines}$

Solution

$x\clubsuit y = x^2y-xy^2$ and $y\clubsuit x = y^2x-yx^2$ . Therefore, we have the equation $x^2y-xy^2 = y^2x-yx^2$ Factoring out a $-1$ gives $x^2y-xy^2 = -(x^2y-xy^2)$ Factoring both sides further, $xy(x-y) = -xy(x-y)$ . It follows that if $x=0$ , $y=0$ , or $(x-y)=0$ , both sides of the equation equal 0. By this, there are 3 lines ( $x=0$ , $y=0$ , or $x=y$ ) so the answer is $\boxed{\textbf{(E)}\text{ three lines}}$ .

Alternate Solution

Following from the previous solution, $x^2y-xy^2 = y^2x-yx^2$ . Then, $2x^2y-2xy^2=0$ . Factoring, $2xy(x-y)=0$ . Now, the solutions are obviously $x=0$ , $y=0$ , or $x=y$ , which each correspond to a line. Thus, the answer is $\boxed{\textbf{(E)}\text{ three lines}}.

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2013 AMC 10B Problems/Problem 14

Contents

Problem

Solution

Alternate Solution

See also