2017 AMC 10A Problem 12

There are three quantities, $3$, $x + 2$ and $y - 4$, so there are three ways to choose which two are equal. Handle each one separately.

1. $3$ and $x + 2$ are equal. Then $x = 1$. The third quantity must not be bigger than $3$: $y - 4 \le 3$, so $y \le 7$. These are the points on the vertical line $x = 1$ at or below $(1, 7)$. That is a ray starting at $(1, 7)$ and pointing down.
2. $3$ and $y - 4$ are equal. Then $y = 7$. The third quantity must not be bigger than $3$: $x + 2 \le 3$, so $x \le 1$. These are the points on the horizontal line $y = 7$ at or to the left of $(1, 7)$. That is a ray starting at $(1, 7)$ and pointing left.
3. $x + 2$ and $y - 4$ are equal. Then $y = x + 6$. Now $3$ must not be bigger than the common value: $3 \le x + 2$, so $x \ge 1$. These are the points on the line $y = x + 6$ with $x \ge 1$. At $x = 1$ the point is $(1, 7)$, and the rest goes up and to the right. That is a third ray starting at $(1, 7)$.
4. Check the shared point. At $(1, 7)$ all three quantities equal $3$, which is why all three rays start there.

So $S$ is three rays with the same starting point.

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

When a problem says "two of these are equal," split into one case per pair. Each equation draws a line, and an inequality attached to it cuts the line in half, leaving a ray. Finding the point where all three quantities agree tells you where the rays meet.

There are three ways to pick which two quantities are equal.

Case 1: $3 = x + 2$, so $x = 1$, and the third quantity satisfies $y - 4 \le 3$, so $y \le 7$. This is the part of the vertical line $x = 1$ at or below $(1, 7)$: a ray pointing down from $(1, 7)$.

Case 2: $3 = y - 4$, so $y = 7$, with $x + 2 \le 3$, so $x \le 1$. This is the part of the horizontal line $y = 7$ to the left of $(1, 7)$: a ray pointing left from $(1, 7)$.

Case 3: $x + 2 = y - 4$, so $y = x + 6$, with $3 \le x + 2$, so $x \ge 1$ (and then $y \ge 7$). This is the part of the line $y = x + 6$ from $(1, 7)$ going up and to the right: a ray.

All three rays begin at the same point $(1, 7)$, so $S$ is three rays with a common endpoint.

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