2002 AMC 10A Problem 8

The blue triangles are awkward shapes, so measure the easy pieces (the whole flag, the red square, the white squares) and let subtraction give the blue.

1. Choose a size. Say the flag is $8$ by $8$, so its area is $64$. In the picture the red square reaches halfway across the flag, so its side is $4$ and
$$P = 4 \times 4 = 16.$$

1. One white square. Each white square is tilted, standing on a corner. From the picture, its corners touch the flag's edge and the red square, so its height from top corner to bottom corner is $2$, and its width is also $2$. It fits exactly inside a $2$ by $2$ box and fills half of it (the four corner triangles of the box are outside it). So one white square has area $\frac{2 \times 2}{2} = 2$.

3. All white squares. There are $3$ along each of the $4$ sides, so $12$ in all:
$$W = 12 \times 2 = 24.$$

4. Blue is everything else:
$$B = 64 - 16 - 24 = 24.$$

1. So $B = W$. (The other choices all fail: $P = 16$ is not $24$, and $3B = 72$ is not $2P = 32$, and $2P = 32$ is not $W = 24$.)

The answer is $\boxed{\textbf{(A)}\ B = W}$.

When some pieces are irregular, measure the regular ones and find the rest by subtracting from the whole. Choosing a convenient size for the flag turns a picture question into arithmetic. A tilted square has half the area of the box it fits in, which is worth remembering.

Cut the $8$ by $8$ flag into a $4 \times 4$ grid of $2$ by $2$ squares. The middle $4$ squares make up the red square. Each of the other $12$ squares, around the border, contains two halves of white squares and blue everywhere else, so it is exactly half white and half blue. Twelve squares that are each half white and half blue give the same amount of white as blue, so $B = W$.