AMC 10 · 2018 · #12

Grade 8 algebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area

absolute-valuesystems-of-equationscoordinate-geometry caseworkconvert-to-algebra ↑ Prerequisites: absolute-valuesystems-of-equations

📏 Long solution 💡 3 insights

Problem

Two conditions must hold at the same time for a point (x,y): it lies on the straight line x+3y=3, and the nested absolute value ||x|-|y|| equals 1. Count how many real points (x,y) meet both conditions, and match that count to one of the five choices.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram): the cleanest way to see how many solutions exist is to picture the line x+3y=3 crossing the absolute-value figure; each crossing is one solution. To turn that picture into exact counts, Tool #4 (Introduce a Variable) reduces the system to a single variable by substituting x=3-3y. Tool #7 (Identify Subproblems) then splits the work by the sign of each absolute value: the points where 3-3y and y change sign cut the number line into a few regions, and inside each region the bars become ordinary plus/minus signs. Tool #3 (Eliminate Possibilities) finishes by discarding any repeated boundary point and counting the distinct survivors.

1STEP 1

Picture the absolute-value figure

||x|-|y||=1 means |x|-|y|=±1; in quadrant one that's y=x-1 and y=x+1, mirrored across both axes into an ×-shaped band the line cuts.

||x|-|y||=1 ⇔ |x|-|y|=± 1

2STEP 2

Reduce to one variable

Solve the line for x: x=3-3y, so every point is one number y and the condition becomes ||3-3y|-|y||=1 — solutions become y-values.

x=3-3y → ||3-3y|-|y||=1

3STEP 3

Split the line into sign regions

The insides flip sign at y=1 and y=0, cutting the y-axis into three ranges; for y ＜ 0, (3-3y)-(-y)=3-2y ＞ 3, never ±1 — no solutions here.

y ＜ 0: (3-3y)-(-y)=3-2y ＞ 3

4STEP 4

Solve the middle region 0 ≤ y ≤ 1

In 0 ≤ y ≤ 1 the bars give 3-4y; 3-4y=1→( $\frac{3}{2}$ , $\frac{1}{2}$ ) and 3-4y=-1→(0,1), both in range — solutions ( $\frac{3}{2}$ , $\frac{1}{2}$ ) and (0,1).

3-4y=1→(3/2,1/2); 3-4y=-1→(0,1)

5STEP 5

Solve the region y ＞ 1

For y ＞ 1 the bars give 2y-3; 2y-3=1→y=2, x=-3, so (-3,2); 2y-3=-1→y=1, the boundary already counted — one new point.

2y-3=1→(-3,2); 2y-3=-1→ y=1 (already counted)

6STEP 6

Count the distinct solutions

Collect the survivors ( $\frac{3}{2}$ , $\frac{1}{2}$ ), (0,1), (-3,2); the repeated y=1 is one pair (0,1), counted once — three distinct pairs, choice (C).

{(3/2,1/2), (0,1), (-3,2)}→ 3 pairs→(C)

Answer

Check each pair in both original equations. (

\frac{3}{2}

\frac{1}{2}

\frac{3}{2}

+3·

\frac{1}{2}

=3 and ||

\frac{3}{2}

|-|

\frac{1}{2}

||=1. (0,1): 0+3=3 and ||0|-|1||=1. (-3,2): -3+6=3 and ||{-3}|-|2||=|3-2|=1. All three work, and the region y ＜ 0 was shown to give nothing, so no pair was missed. Three crossings also matches the picture: a line of negative slope passing through the ×-shaped band of diagonals meets it a small, finite number of times.

💡Key takeaway

When absolute-value bars block you, substitute to one variable, split the number line where each inside hits zero, solve the plain line in each piece, and count the different answers — here that gives 3, choice (C).

Picture the absolute-value figure
Reduce to one variable
Split the line into sign regions
Solve the middle region 0 ≤ y ≤ 1
Solve the region y ＞ 1
Count the distinct solutions