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Revision as of 12:53, 9 February 2017

Problem

Let $S$ be a 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 this common value. Which of the following is a correct description for $S?$

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

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 $\boxed{\textbf{(E) }\text{three rays with a common endpoint}}$

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

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