# 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, $y-4\leqslant 3$ because $3$ is the common value. Solving for $y$, we get $y\leqslant 7$. Therefore the portion of the line $x=1$ where $y\leqslant 7$ 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, $x+2\leqslant 3$ because $3$ is the common value. Solving for $x$, we get $x\leqslant 1$. Therefore the portion of the line $y=7$ where $x\leqslant 1$ is also part of $S$. This is another ray with the same endpoint as the above ray: $(1, 7)$. (Note that the only answer choice which has rays in it is answer choice $E$.)

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 $3\leqslant y-4$ because $y-4$ is one way to express the common value (using $x-2$ as the common value works as well). Solving for $y$, we get $y\geqslant 7$. Therefore the portion of the line $y=x+6$ where $y\geqslant 7$ 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 $\boxed{E}$.

Solution by TheMathematicsTiger7

## See Also

 2017 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 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 byProblem 8 Followed byProblem 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. Invalid username
Login to AoPS