AMC 10 · 2007 · #19
Grade 8 geometry-2dA paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The brush is swept along both diagonals of the square, making two crossing stripes; The brush has a constant width, measured perpendicular to the direction of the stroke; Exactly half the area of the square is painted; The picture is symmetric across both diagonals and both midlines of the square; Answer choices: (A) $2\sqrt{2}+1$, (B) $3\sqrt{2}$, (C) $2\sqrt{2}+2$, (D) $3\sqrt{2}+1$, (E) $3\sqrt{2}+2$
Unknowns: The ratio of the side length of the square to the brush width
Understand
Restated: 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.
Givens: The brush is swept along both diagonals of the square, making two crossing stripes; The brush has a constant width, measured perpendicular to the direction of the stroke; Exactly half the area of the square is painted; The picture is symmetric across both diagonals and both midlines of the square; Answer choices: (A) $2\sqrt{2}+1$, (B) $3\sqrt{2}$, (C) $2\sqrt{2}+2$, (D) $3\sqrt{2}+1$, (E) $3\sqrt{2}+2$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #1 Draw a Diagram, #4 Introduce a Variable, #7 Identify Subproblems
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.
Execute — Answer: C
6.G.A.3 Step 1 Set up and spot the white triangles
- Let the square be the unit square with corners $(0,0),(1,0),(1,1),(0,1)$, and let $w$ be the brush width.
- Each stroke is a stripe of width $w$ centered on a diagonal.
- The two stripes cross in the middle to make the painted X.
- What is left unpainted is four congruent triangles, one pressed against the middle of each side of the square.
💡 Naming the side $1$ and the width $w$ turns the picture into numbers you can chase.
6.G.A.1 Step 2 Count the white half instead
- Adding the two stripe areas would double count where they overlap, so measure the easy leftover part instead.
- The painted region is half the square, so the unpainted white region is the other half, with area $\tfrac12$.
- By the four-fold symmetry the four white triangles are identical, so each one has area $\tfrac12 \div 4 = \tfrac18$.
💡 If the shape you want is tangled, measure what's left over instead.
8.G.B.8 Step 3 Measure one white triangle's base
- Look at the white triangle sitting on the bottom edge.
- Its two slanted sides lie along the inner edges of the two stripes, which run at slopes $+1$ and $-1$, so they meet at a right angle at the top of the triangle: it is an isosceles right triangle, and its base on the bottom edge is its hypotenuse.
- Now find how much of the bottom edge each stripe paints.
- The stripe along $y=x$ has its edge cross the bottom side at $x=\tfrac{w\sqrt2}{2}$, because the point $\left(\tfrac{w\sqrt2}{2},\,0\right)$ sits at perpendicular distance $\tfrac{w\sqrt2/2}{\sqrt2}=\tfrac{w}{2}$ from that diagonal — exactly half the stripe's width.
- The other stripe paints an equal piece by the right corner, so the white base runs from $x=\tfrac{w\sqrt2}{2}$ to $x=1-\tfrac{w\sqrt2}{2}$.
💡 A stripe crossing an edge at $45^\circ$ paints a corner chunk set by half its width.
8.G.B.7 Step 4 Turn the area into an equation for w
- An isosceles right triangle with hypotenuse $b$ has legs $\tfrac{b}{\sqrt2}$, so its area is $\tfrac12\left(\tfrac{b}{\sqrt2}\right)^2=\tfrac{b^2}{4}$.
- Here $b=1-w\sqrt2$, and this area must equal $\tfrac18$.
- So $(1-w\sqrt2)^2=\tfrac12$, which gives $1-w\sqrt2=\tfrac{\sqrt2}{2}$ (the positive root; the other root would make the brush wider than the square, which is impossible).
- Then $w\sqrt2 = 1-\tfrac{\sqrt2}{2}=\tfrac{2-\sqrt2}{2}$, so $w=\tfrac{2-\sqrt2}{2\sqrt2}=\tfrac{\sqrt2-1}{2}$.
💡 One clean area equation squeezes the whole picture down to a single value of $w$.
8.EE.A.2 Step 5 Read off the ratio
- The side is $1$ and the brush width is $w=\tfrac{\sqrt2-1}{2}$, so the ratio of side to width is $\dfrac{1}{w}=\dfrac{2}{\sqrt2-1}$.
- Clear the radical from the bottom by multiplying top and bottom by $\sqrt2+1$: $\dfrac{2(\sqrt2+1)}{(\sqrt2-1)(\sqrt2+1)}=\dfrac{2(\sqrt2+1)}{2-1}=2\sqrt2+2$.
- That is choice (C).
💡 Flipping the width and clearing the radical from the bottom gives a clean form that matches a choice.
6.G.A.3 Let the square be the unit square with corners $(0,0),(1,0),(1,1),(0,1)$, and le 6.G.A.1 Adding the two stripe areas would double count where they overlap, so measure th 8.G.B.8 Look at the white triangle sitting on the bottom edge. Its two slanted sides lie 8.G.B.7 An isosceles right triangle with hypotenuse $b$ has legs $\tfrac{b}{\sqrt2}$, so 8.EE.A.2 The side is $1$ and the brush width is $w=\tfrac{\sqrt2-1}{2}$, so the ratio of Review
Reasonableness: Check the size. With $\sqrt2\approx1.414$, the width is $w=\tfrac{0.414}{2}\approx0.207$, so side-to-width is about $1/0.207\approx4.83$. Choice (C) is $2\sqrt2+2\approx2.83+2=4.83$ — 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) $2\sqrt2+1\approx3.83$ and (B) $3\sqrt2\approx4.24$ make the brush too wide (more than half would be painted), while (D) $\approx5.24$ and (E) $\approx6.24$ make it too thin.
Alternative: Use inclusion-exclusion on the stripes directly instead of the complement. Each stripe, clipped to the square, has area $1-\left(1-\tfrac{w\sqrt2}{2}\right)^2$, and the two stripes overlap in a central square of area $w^2$. The painted area is $2\left[1-\left(1-\tfrac{w\sqrt2}{2}\right)^2\right]-w^2$. Setting this equal to $\tfrac12$ and simplifying gives $8t^2-8t+1=0$ where $t=\tfrac{w\sqrt2}{2}$; the smaller root is $t=\tfrac{2-\sqrt2}{4}$, and back-substituting returns the same $w=\tfrac{\sqrt2-1}{2}$ and ratio $2\sqrt2+2$.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing the unit square on a grid and reading where each stripe's edge crosses a side.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Splitting the unpainted half of the square into four equal triangles, each of area $\tfrac18$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Locating where each stripe's edge meets the bottom side, fixing the white triangle's base at $1-w\sqrt2$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Relating the isosceles right triangle's hypotenuse $b$ to its area $\tfrac{b^2}{4}$ to solve for $w$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Solving $(1-w\sqrt2)^2=\tfrac12$ and rationalizing $\tfrac{2}{\sqrt2-1}$ into $2\sqrt2+2$.)
⭐ When the painted shape overlaps itself, measure the leftover white instead: four equal triangles turned 'half painted' into one solvable equation.
⭐ When the painted shape overlaps itself, measure the leftover white instead: four equal triangles turned 'half painted' into one solvable equation.
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