Difference between revisions of "2013 AMC 10B Problems/Problem 14"

Revision as of 15:46, 3 July 2013

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 $\boxed{\textbf{(E)}\text{ three lines}}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See also ==
 {{AMC10 box|year=2013|ab=B|num-b=13|num-a=15}}
+{{MAA Notice}}

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Difference between revisions of "2013 AMC 10B Problems/Problem 14"

Revision as of 15:46, 3 July 2013

Problem

Solution

See also