|
|
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #redirect [[2002 AMC 12A Problems/Problem 8]] |
| − | 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>R</math> 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>
| |
| − |
| |
| − | <math>\text{(A)}\ B = W \qquad \text{(B)}\ W = R \qquad \text{(C)}\ B = R \qquad \text{(D)}\ 3B = 2R \qquad \text{(E)}\ 2R = 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>.
| |
| − | | |
| − | ==See Also==
| |
| − | {{AMC10 box|year=2002|ab=A|num-b=7|num-a=9}}
| |
| − | | |
| − | [[Category:Introductory Geometry Problems]] | |
