2017 AMC 12A Problems/Problem 9

Problem

Let $S$ be the set of points $(x,y)$ in the coordinate plane such that two of the three quantities $3$ , $x+2$ , and $y-4$ 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 $S$ ?

$\textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point}$

Solution

If the two equal values are $3$ and $x+2$ , then $x=1$ . Also, $y-4<3$ because 3 is the common value. Solving for $y$ , we get $y<7$ . Therefore the portion of the line $x=1$ where $y<7$ is part of $S$ . This is a ray with an endpoint of $(1, 7)$ .

Similar to the process above, we assume that the two equal values are $3$ and $y-4$ . Solving the equation $3=y-4$ then $y=7$ . Also, $x+2<3$ because 3 is the common value. Solving for $x$ , we get $x<1$ . Therefore the portion of the line $y=7$ where $x<1$ is also part of $S$ . This is another ray with the same endpoint as the above ray: $(1, 7)$ .

If $x+2$ and $y-4$ are the two equal values, then $x+2=y-4$ . Solving the equation for $y$ , we get $y=x+6$ . Also $3<y-4$ because $y-4$ is one way to express the common value. Solving for $y$ , we get $y>7$ . Therefore the portion of the line $y=x+6$ where $y>7$ is part of $S$ like the other two rays. The lowest possible value that can be achieved is also $(1, 7)$ .

Since $S$ is made up of three rays with common endpoint $(1, 7)$ , the answer is $\boxed{E}$

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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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

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2017 AMC 12A Problems/Problem 9

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