2017 AMC 10A Problem 12

P11-15~4 minPrint

Coordinate Geometry Absolute Value & Inequalities Split into exhaustive cases and handle each

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 $S?$

Show solution without answering

Key insight

Make each pair of quantities equal. Each gives a line, and 'the third is not bigger' keeps only half of it. All three halves start at (1, 7).

Solution

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.

$3$ 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.
$3$ 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.
$x + 2$ 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 $(1, 7)$ .
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 $\boxed{\textbf{(E)}\ \text{three rays with a common endpoint}}$ .

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Algebra, Chapter 8: Graphing Lines