AMC 10 · 2018 · #10

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

A point must meet two conditions at once: it lies on the straight line x plus three y equals three, and the nested absolute value of the difference of the absolute values of x and y equals one. Count how many real points meet both conditions, and match that count to one of the five choices.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

The nested absolute value splits into two cases.

||x|-|y||=1 ⇔ |x|-|y|=± 1

2STEP 2

Reduce to one variable

The line reduces it to one variable.

x=3-3y → ||3-3y|-|y||=1

3STEP 3

Split the line into sign regions

Split the line where the signs change.

y ＜ 0: (3-3y)-(-y)=3-2y ＞ 3

4STEP 4

Solve the middle region 0 ≤ y ≤ 1

The middle region gives two solutions.

3-4y=1→(3/2,1/2); 3-4y=-1→(0,1)

5STEP 5

Solve the region y ＞ 1

One upper-region root is already counted.

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

6STEP 6

Count the distinct solutions

There are 3 distinct solutions.

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

Answer

Check each pair in both original equations. (3/2,1/2): 3/2+3·1/2=3 and ||3/2|-|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