AMC 10 · 2007 · #19

Grade 8 geometry-2d

Archetype Overlap / Complement Counting └ Sub-archetype Geometric Area via Inclusion-Exclusion or Area Balance

area-trianglesisosceles-trianglequadratic-equationssymmetry-argument convert-to-algebra ↑ Prerequisites: area-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A square is painted by sweeping a brush of fixed width along each of its two diagonals, leaving a symmetric crossing pattern. Exactly half of the square's area ends up painted. Find the ratio of the square's side length to the brush width.

Pick an answer.

(A)

$2\sqrt {2} + 1$

(B)

$3\sqrt {2}$

(C)

$2\sqrt {2} + 2$

(D)

$3\sqrt {2} + 1$

(E)

$3\sqrt {2} + 2$

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

The painted X-shape is awkward: its two stripes overlap, so adding their areas double counts the middle. Tool #16 (Change Focus / Count the Complement) flips the problem — since half the square is painted, the other half is unpainted, and that unpainted half is just four clean, congruent triangles. Tool #1 (Draw a Diagram) with coordinates fixes exactly where the stripe edges cut the sides. Tool #4 (Introduce a Variable) sets the side to 1 and names the brush width w, turning 'half the area' into one equation. Tool #7 (Identify Subproblems) isolates a single white triangle, whose area in terms of w is all that is needed.

1STEP 1

Set up and spot the white triangles

Take the side as $1$ and the brush width as $w$ : two stripes cross in the middle, leaving four congruent white triangles, one per side.

side=1, brush width=w

2STEP 2

Count the white half instead

Painted is half, so white is the other half, $\frac{1}{2}$ ; the four triangles are identical, so each has area $\frac{1}{8}$ .

each white triangle = 1/4·1/2 = 1/8

3STEP 3

Measure one white triangle's base

The stripes meet a side at 45°, painting $\frac{w\sqrt{2}}{2}$ at each end, so the white triangle's hypotenuse is $1-w\sqrt{2}$ .

base = 1 - 2·w√2/2 = 1 - w√2

4STEP 4

Turn the area into an equation for w

Hypotenuse $b$ means area $\frac{b^2}{4}$ , so $(1-w\sqrt{2})^2=\frac{1}{2}$ ; the root that fits inside the square is $w=\frac{\sqrt{2}-1}{2}$ .

((1-w√2)²)/4=1/8 → 1-w√2=√2/2 → w=(√2-1)/2

5STEP 5

Read off the ratio

Side over width is $\frac{1}{w}=\frac{2}{\sqrt{2}-1}$ ; clearing the radical below gives $2\sqrt{2}+2$ , choice (C).

1/w=2/(√2-1)=(2(√2+1))/1=2√2+2 → (C)

Answer

2√(2) + 2

Check the size. With √2≈1.414, the width is w=0.414/2≈0.207, so side-to-width is about 1/0.207≈4.83. Choice (C) is 2√2+2≈2.83+2=4.83 — a match. It also passes a common-sense check: a stripe about a fifth of the side, swept along a diagonal, covers a little under half the square by itself, and the two crossing stripes overlap in the middle, so landing at exactly half painted is believable. The wrong choices are each off in a telling way: (A) 2√2+1≈3.83 and (B) 3√2≈4.24 make the brush too wide (more than half would be painted), while (D) ≈5.24 and (E) ≈6.24 make it too thin.

💡Key takeaway

When the painted shape overlaps itself, measure the leftover white instead: four equal triangles turned 'half painted' into one solvable equation.

Set up and spot the white triangles
Count the white half instead
Measure one white triangle's base
Turn the area into an equation for w
Read off the ratio

Set up and spot the white triangles

Count the white half instead

Measure one white triangle's base

Turn the area into an equation for w

Read off the ratio

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