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2007 AMC 10A Problem 19

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2007 AMC 10A Problem 19AMC 10 Step by Step
Name:                             Date: September 13, 2026
Quadrilaterals & Polygon Areas · difficulty 4 of 5 · about 6 min · position P16-20

A 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?

figure

  1. A)

    22+12\sqrt {2} + 1

  2. B)

    323\sqrt {2}

  3. C)

    22+22\sqrt {2} + 2

  4. D)

    32+13\sqrt {2} + 1

  5. E)

    32+23\sqrt {2} + 2

Answer

C

Key insight

The unpainted half is four isosceles right triangles, each with hypotenuse s/sqrt(2) along a side; the leftover painted edge is two corner legs of length w/sqrt(2).

Solution

Let the square have side ss and the brush width ww. Each painted strip has edges parallel to a diagonal, so every edge meets the sides of the square at 45∘45^\circ.

Work with the unpainted part first. It consists of four congruent isosceles right triangles, one against the middle of each side, with total area s22\frac{s^2}{2}. Each has area s28\frac{s^2}{8}, so its legs are s2\frac{s}{2} and its hypotenuse, which lies along a side of the square, is s22=s2\frac{s}{2}\sqrt2 = \frac{s}{\sqrt2}.

The rest of that side, of length s−s2s - \frac{s}{\sqrt2}, is painted and is split evenly between the two corners. At a corner the painted region is a small isosceles right triangle whose hypotenuse, perpendicular to the diagonal, is the brush width ww; so its leg along the side is w2\frac{w}{\sqrt2}. Two such legs fill the painted part of the side:
2w2=s−s2⟹2 w=s(1−12)⟹w=s⋅2−12.\frac{2w}{\sqrt2} = s - \frac{s}{\sqrt2} \quad\Longrightarrow\quad \sqrt2\, w = s\left(1 - \frac{1}{\sqrt2}\right) \quad\Longrightarrow\quad w = s \cdot \frac{\sqrt2 - 1}{2}.
Therefore
sw=22−1=2(2+1)=22+2.\frac{s}{w} = \frac{2}{\sqrt2 - 1} = 2(\sqrt2 + 1) = 2\sqrt2 + 2.

The answer is (C) 22+2\boxed{\textbf{(C)}\ 2\sqrt {2} + 2}.

Why this works

Everything in the figure is built from 45∘45^\circ angles, so each region is an isosceles right triangle or a square, and one length determines all the others. Using the unpainted half (four clean triangles) is far simpler than measuring the painted cross directly, and "half the area" then fixes the triangle size. Look for the simpler complement whenever a figure is described by what it is not.

Alternative approach

Set s=1s = 1 and compute the painted area directly: each strip is a rectangle of length 2\sqrt2 and width ww minus two corner triangles, area 2 w−w22\sqrt2\,w - \frac{w^2}{2}; the two strips overlap in a w×ww \times w square. So 22 w−2w2=122\sqrt2\,w - 2w^2 = \frac12, whose smaller root is w=2−12w = \frac{\sqrt2 - 1}{2}, giving 1w=22+2\frac1w = 2\sqrt2 + 2.

The trap

Treating the brush width as the length of painted edge at each corner; the painted edge piece is w/sqrt(2), the leg of a 45-degree triangle with hypotenuse w.

Common mistakes

  • Treating the brush width as the length of painted edge at each corner; the painted edge piece is w/2w/\sqrt2, the leg of a 4545-degree triangle with hypotenuse ww.
  • Adding the two strips' areas without subtracting their overlapping central square, which shifts the equation and the root.

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors) · Exploit symmetry to reduce work or pair up objects