AMC 10 · 2008 · #14

Grade 9 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Axis-Aligned Figure Area from Coordinates
absolute-valuecoordinate-geometryarea-triangles caseworksymmetry-argument ↑ Prerequisites: absolute-value
📏 Medium solution 💡 2 insights
Problem
A region is every point where a sum of two absolute values stays under a fixed bound. Find its area.

Pick an answer.

(A)
3
(B)
$\frac{7}{2}$
(C)
4
(D)
$\frac{9}{2}$
(E)
5
How to solve
Strategy Introduce a Variable

Absolute value bars are sign cases in disguise, and the fastest way to expose the sign cases is to name the quantities the bars actually measure. Pull the positive constants out front, then let u and v stand for the shifted coordinates so the two bars become |u| and |v|. That turns the condition into 3|u|+2|v| ≤ 3, which is centered at the origin. From there, splitting on the signs of u and v breaks the region into four separate pieces, each governed by a plain linear inequality, and a linear inequality has a straight boundary. So instead of assuming the region is some nice shape, each piece is shown to be a triangle and the four triangles are added.

1STEP 1

Pull the coefficients out of the bars

Pulling the coefficients out leaves clean absolute values.

|3x-18|+|2y+7| = 3|x-6| + 2|y+7/2| ≤ 3
2STEP 2

Slide the region to the origin

Sliding to the origin costs no area.

u = x-6, v = y+7/2 ⟹ 3|u| + 2|v| ≤ 3
3STEP 3

Fix the signs and the bars vanish

Fixing the signs makes one quarter a plain triangle.

u ≥ 0, v ≥ 0: 3u+2v ≤ 3, corners (0,0), (1,0), (0,3/2)
4STEP 4

The other three quarters are mirror images

The other three are mirror images.

3|{-u}|+2|v| = 3|u|+2|v| = 3|u|+2|{-v}|
5STEP 5

Add the four triangles

Adding the four gives 3, choice (E).

4·1/2 · 1·3/2 = 4·3/4 = 3
Answer
3
Every point of the region has 3|u| ≤ 3 and 2|v| ≤ 3, so the whole thing fits inside the rectangle running from -1 to 1 in u and from -3/2 to 3/2 in v, an area of 2 times 3, or 6. The answer 3 is exactly half of that, which is what a four-cornered region whose corners are the midpoints of the rectangle's sides should give. A second and independent count: fix any u with |u| ≤ 1. Then 2|v| ≤ 3-3|u|, so the allowed v form a segment of length 3-3|u|. That length is 3 at u=0 and falls in a straight line to 0 at u=± 1, so the stacked segments make a triangle of base 2 and height 3, giving 1/2(2)(3)=3. Both checks agree with (A) 3.
💡Key takeaway

Absolute-value bars are sign cases in disguise: fix the signs, the bars disappear, every corner piece turns into a plain triangle, and you just add them up.

  • Pull the coefficients out of the bars
  • Slide the region to the origin
  • Fix the signs and the bars vanish
  • The other three quarters are mirror images
  • Add the four triangles

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