Difference between revisions of "2002 AMC 12A Problems/Problem 8"

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m (Fixed inconsistency in diagram and problem)
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==Problem==
 
==Problem==
Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let <math>B</math> be the total area of the blue triangles, <math>W</math> the total area of the white squares, and <math>P</math> the area of the pink square. Which of the following is correct?
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Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let <math>B</math> be the total area of the blue triangles, <math>W</math> the total area of the white squares, and <math>P</math> the area of the red square. Which of the following is correct?
  
 
<asy>
 
<asy>
 
unitsize(3mm);
 
unitsize(3mm);
 
fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue);
 
fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue);
fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,pink);
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fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red);
 
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 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;
 
path divider=(-2,2)--(-3,3)--cycle;

Revision as of 18:16, 6 January 2019

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 red 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,red); 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]

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

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 $\boxed{B=W\Rightarrow \text{(A)}}$.

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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