Difference between revisions of "2017 AMC 12A Problems/Problem 9"
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| − | ==Problem== | + | == Problem == |
Let <math>S</math> be the set of points <math>(x,y)</math> in the coordinate plane such that two of the three quantities <math>3</math>, <math>x+2</math>, and <math>y-4</math> are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of <math>S</math>? | Let <math>S</math> be the set of points <math>(x,y)</math> in the coordinate plane such that two of the three quantities <math>3</math>, <math>x+2</math>, and <math>y-4</math> are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of <math>S</math>? | ||
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<math> \textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point} </math> | <math> \textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point} </math> | ||
| − | ==Solution== | + | == Solution == |
If the two equal values are <math>3</math> and <math>x+2</math>, then <math>x=1</math>. Also, <math>y-4\leqslant 3</math> because 3 is the common value. Solving for <math>y</math>, we get <math>y\leqslant 7</math>. Therefore the portion of the line <math>x=1</math> where <math>y\leqslant 7</math> is part of <math>S</math>. This is a ray with an endpoint of <math>(1, 7)</math>. | If the two equal values are <math>3</math> and <math>x+2</math>, then <math>x=1</math>. Also, <math>y-4\leqslant 3</math> because 3 is the common value. Solving for <math>y</math>, we get <math>y\leqslant 7</math>. Therefore the portion of the line <math>x=1</math> where <math>y\leqslant 7</math> is part of <math>S</math>. This is a ray with an endpoint of <math>(1, 7)</math>. | ||
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Solution by TheMathematicsTiger7 | Solution by TheMathematicsTiger7 | ||
| − | ==See Also== | + | == See Also == |
{{AMC10 box|year=2017|ab=A|num-b=11|num-a=13}} | {{AMC10 box|year=2017|ab=A|num-b=11|num-a=13}} | ||
{{AMC12 box|year=2017|ab=A|num-b=8|num-a=10}} | {{AMC12 box|year=2017|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:36, 7 June 2018
Problem
Let 
be the set of points 
in the coordinate plane such that two of the three quantities 
, 
, and 
are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of ?
Solution
If the two equal values are and , then 
. Also, 
because 3 is the common value. Solving for 
, we get 
. Therefore the portion of the line where is part of . This is a ray with an endpoint of 
.
Similar to the process above, we assume that the two equal values are and . Solving the equation 
then 
. Also, 
because 3 is the common value. Solving for 
, we get 
. Therefore the portion of the line where is also part of . This is another ray with the same endpoint as the above ray: .
If and are the two equal values, then 
. Solving the equation for , we get 
. Also 
because is one way to express the common value (using 
as the common value works as well). Solving for , we get 
. Therefore the portion of the line where is part of like the other two rays. The lowest possible value that can be achieved is also .
Since is made up of three rays with common endpoint , the answer is 
.
Solution by TheMathematicsTiger7
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
