2017 AMC 10A Problem 12

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

"Two are equal and the third is no greater" is the definition of the maximum of three quantities being attained twice; geometrically it is the "fold lines" of the graph of $z = \max(3, x+2, y-4)$. Each equality is a line, the inequality keeps only half of it, and the three halves must share the point where all three quantities equal $3$.

• Ignoring the inequality and answering 'three lines' (C), or checking the inequality with the wrong direction and getting rays pointing the wrong way.
• Solving only the case where all three quantities are equal and answering 'a single point' (A).