Difference between revisions of "2013 AMC 10B Problems/Problem 14"
MathDragon2 (talk | contribs) m (→Solution 2) |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 4: | Line 4: | ||
<math> \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} </math> | <math> \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} </math> | ||
| − | == Solution== | + | == Solution 1== |
<math>x\clubsuit y = x^2y-xy^2 </math> and <math>y\clubsuit x = y^2x-yx^2</math>. Therefore, we have the equation <math> x^2y-xy^2 = y^2x-yx^2</math> Factoring out a <math>-1</math> gives <math> x^2y-xy^2 = -(x^2y-xy^2)</math> Factoring both sides further, <math>xy(x-y) = -xy(x-y)</math>. It follows that if <math>x=0</math>, <math>y=0</math>, or <math>(x-y)=0</math>, both sides of the equation equal 0. By this, there are 3 lines (<math>x=0</math>, <math>y=0</math>, or <math>x=y</math>) so the answer is <math>\boxed{\textbf{(E)}\text{ three lines}}</math>. | <math>x\clubsuit y = x^2y-xy^2 </math> and <math>y\clubsuit x = y^2x-yx^2</math>. Therefore, we have the equation <math> x^2y-xy^2 = y^2x-yx^2</math> Factoring out a <math>-1</math> gives <math> x^2y-xy^2 = -(x^2y-xy^2)</math> Factoring both sides further, <math>xy(x-y) = -xy(x-y)</math>. It follows that if <math>x=0</math>, <math>y=0</math>, or <math>(x-y)=0</math>, both sides of the equation equal 0. By this, there are 3 lines (<math>x=0</math>, <math>y=0</math>, or <math>x=y</math>) so the answer is <math>\boxed{\textbf{(E)}\text{ three lines}}</math>. | ||
== Solution 2== | == Solution 2== | ||
| − | Following from the previous solution, <math> x^2y-xy^2 = y^2x-yx^2</math>. Then, <math>2x^2y-2xy^2=0</math>. Factoring, <math>2xy(x-y)=0</math>. Now, the solutions are obviously <math>x=0</math>, <math>y=0</math>, or <math>x=y</math>, which each correspond to a line. Thus, the answer is <math>\boxed{\textbf{(E)}\text{ three lines}}</math>. | + | Following from the previous solution, <math>x^2y-xy^2 = y^2x-yx^2</math>. Then, <math>2x^2y-2xy^2=0</math>. Factoring, <math>2xy(x-y)=0</math>. Now, the solutions are obviously <math>x=0</math>, <math>y=0</math>, or <math>x=y</math>, which each correspond to a line. Thus, the answer is <math>\boxed{\textbf{(E)}\text{ three lines}}</math>. |
== See also == | == See also == | ||
{{AMC10 box|year=2013|ab=B|num-b=13|num-a=15}} | {{AMC10 box|year=2013|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 14:09, 1 May 2021
Contents
Problem
Define 
. Which of the following describes the set of points 
for which 
?
Solution 1
and 
. Therefore, we have the equation 
Factoring out a 
gives 
Factoring both sides further, 
. It follows that if 
, 
, or 
, both sides of the equation equal 0. By this, there are 3 lines (, , or 
) so the answer is 
.
Solution 2
Following from the previous solution, . Then, 
. Factoring, 
. Now, the solutions are obviously , , or , which each correspond to a line. Thus, the answer is .
See also
| 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. 
