2007 AMC 10A Problem 19

Let the side of the square be $s$ and the brush width $w$. Every edge of paint runs parallel to a diagonal, so it meets the sides of the square at $45^\circ$. That means every piece of the picture is a right triangle with two equal legs.

1. Look at the unpainted part. It is four identical triangles, one against the middle of each side. Each is a right triangle with two equal legs, and its long side lies along a side of the square.
2. Size of one unpainted triangle. Half the square is unpainted, so each triangle has area $\frac{s^2}{2} \div 4 = \frac{s^2}{8}$. A right triangle with equal legs $L$ has area $\frac{L^2}{2}$, so $L^2 = \frac{s^2}{4}$ and $L = \frac{s}{2}$. Its long side is $L\sqrt2 = \frac{s\sqrt2}{2}$.
3. Painted part of one side. The rest of that side is painted:
$$s - \frac{s\sqrt2}{2}.$$
It is made of two equal pieces, one at each corner.
4. One corner piece. At a corner the brush stripe runs along the diagonal. Its two edges are $\frac{w}{2}$ from the diagonal. An edge meets the side at $45^\circ$, so it hits the side $\frac{w}{2} \times \sqrt2 = \frac{w\sqrt2}{2}$ from the corner. That is the painted piece.
5. Set up the equation. Two corner pieces fill the painted part of the side:
$$2 \times \frac{w\sqrt2}{2} = s - \frac{s\sqrt2}{2}, \qquad w\sqrt2 = s \cdot \frac{2 - \sqrt2}{2}.$$
6. Solve for the ratio. Divide both sides by $\sqrt2$: $w = s \cdot \frac{2 - \sqrt2}{2\sqrt2} = s \cdot \frac{\sqrt2 - 1}{2}$. So
$$\frac{s}{w} = \frac{2}{\sqrt2 - 1} = \frac{2(\sqrt2 + 1)}{(\sqrt2 - 1)(\sqrt2 + 1)} = \frac{2(\sqrt2 + 1)}{1} = 2\sqrt2 + 2.$$

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

Everything in this figure is built from $45^\circ$ angles, so each piece is a right triangle with equal legs, and one length tells you the others. The unpainted part is four neat triangles, which is much simpler to measure than the painted cross. When a shape is hard to describe, try measuring what is left over instead.

Let $s = 1$ and measure the paint directly. One stripe along a diagonal is a strip of length $\sqrt2$ and width $w$, minus two small triangles that stick out past the corners, each of area $\frac{w^2}{4}$. So one stripe has area $\sqrt2\,w - \frac{w^2}{2}$. The two stripes overlap in a $w \times w$ square in the middle, so
$$2\left(\sqrt2\,w - \frac{w^2}{2}\right) - w^2 = \frac12, \qquad 2\sqrt2\,w - 2w^2 = \frac12.$$
Solving this (the smaller solution, since the brush is narrow) gives $w = \frac{\sqrt2 - 1}{2}$, so $\frac{1}{w} = 2\sqrt2 + 2$.