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

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 $B$ 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?

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

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

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

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

Video Solution by Daily Dose of Math

https://youtu.be/25qXgtEKULA

~Thesmartgreekmathdude

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.

@@ Line 17: / Line 17: @@
 </asy>
-<math>\text{(A)}\ B = W \qquad \text{(B)}\ W = P \qquad \text{(C)}\ B = P \qquad \text{(D)}\ 3B = 2P \qquad \text{(E)}\ 2P = W</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==
-The blue that's touching the center red 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 <math>\boxed{B=W\Rightarrow \text{(A)}}</math>.
+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==
@@ Line 41: / Line 41: @@
 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==

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