Difference between revisions of "2002 AMC 12A Problems/Problem 8"
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</asy> | </asy> | ||
| − | <math>\ | + | <math>\textbf{(A)}\ B = W \qquad \textbf{(B)}\ W = P \qquad \textbf{(C)}\ B = P \qquad \textbf{(D)}\ 3B = 2P \qquad \textbf{(E)}\ 2P = W</math> |
==Solution== | ==Solution== | ||
| − | The blue that's touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is | + | The blue that's touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is 1 square and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have <math>\boxed{B=W\Rightarrow \text{(A)}}</math>. |
| + | |||
| + | ==Solution 2== | ||
| + | Cut the pattern into 16 square regions and use symmetry. | ||
| + | <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); | ||
| + | draw((-4,-2)--(4,-2),black); | ||
| + | draw((-4,0)--(4,0),black); | ||
| + | draw((-4,2)--(4,2),black); | ||
| + | draw((-2,-4)--(-2,4),black); | ||
| + | draw((0,-4)--(0,4),black); | ||
| + | draw((2,-4)--(2,4),black); | ||
| + | </asy> | ||
| + | |||
| + | This clearly shows <math>\boxed{\textbf{(A) }B=W}</math>. | ||
| + | |||
| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/25qXgtEKULA | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
==See Also== | ==See Also== | ||
Latest revision as of 23:52, 25 July 2024
- 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 
be the total area of the blue triangles, 
the total area of the white squares, and 
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]](http://latex.artofproblemsolving.com/a/0/b/a0b90bdab50e7b1984850bd644f9291ddc2de34e.png)
Solution
The blue that's touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is 1 square and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have 
.
Solution 2
Cut the pattern into 16 square regions and use symmetry.
![[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); draw((-4,-2)--(4,-2),black); draw((-4,0)--(4,0),black); draw((-4,2)--(4,2),black); draw((-2,-4)--(-2,4),black); draw((0,-4)--(0,4),black); draw((2,-4)--(2,4),black); [/asy]](http://latex.artofproblemsolving.com/e/2/c/e2c65e1a95d3f58215e53f1192048a8ec706c1e8.png)
This clearly shows 
.
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
See Also
| 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. 
