2002 AMC 12A Problems/Problem 8

The following problem is from both the 2002 AMC 12A #8 and 2002 AMC 10A #8, so both problems redirect to this page.

Problem

Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $P$ the area of the pink square. Which of the following is correct?

$[asy] unitsize(3mm); fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue); fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,pink); path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle; path divider=(-2,2)--(-3,3)--cycle; fill(onewhite,white); fill(rotate(90)*onewhite,white); fill(rotate(180)*onewhite,white); fill(rotate(270)*onewhite,white); [/asy]$

$\text{(A)}\ B = W \qquad \text{(B)}\ W = R \qquad \text{(C)}\ B = R \qquad \text{(D)}\ 3B = 2R \qquad \text{(E)}\ 2R = W$

Solution

The blue that's touching the center pink square makes up 8 triangles, or 4 squares. Each of the corners is 2 squares and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have $\boxed{B=W\Rightarrow \text{(A)}}$ .

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2002 AMC 12A Problems/Problem 8

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