Difference between revisions of "1994 AJHSME Problems/Problem 12"
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| − | ==Problem== | + | == Problem 12 == |
| − | + | Each of the three large squares shown below is the same size. Segments that intersect the sides of the squares intersect at the midpoints of the sides. How do the shaded areas of these squares compare? | |
| − | < | + | <center> |
| + | <asy> | ||
| + | unitsize(36); | ||
| + | fill((0,0)--(1,0)--(1,1)--cycle,gray); | ||
| + | fill((1,1)--(1,2)--(2,2)--cycle,gray); | ||
| + | draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); | ||
| + | draw((1,0)--(1,2)); | ||
| + | draw((0,0)--(2,2)); | ||
| − | + | fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,gray); | |
| − | <math> | + | draw((3,0)--(5,0)--(5,2)--(3,2)--cycle); |
| + | draw((4,0)--(4,2)); | ||
| + | draw((3,1)--(5,1)); | ||
| + | |||
| + | fill((6,1)--(6.5,0.5)--(7,1)--(7.5,0.5)--(8,1)--(7.5,1.5)--(7,1)--(6.5,1.5)--cycle,gray); | ||
| + | draw((6,0)--(8,0)--(8,2)--(6,2)--cycle); | ||
| + | draw((6,0)--(8,2)); | ||
| + | draw((6,2)--(8,0)); | ||
| + | draw((7,0)--(6,1)--(7,2)--(8,1)--cycle); | ||
| + | |||
| + | label("$I$",(1,2),N); | ||
| + | label("$II$",(4,2),N); | ||
| + | label("$III$",(7,2),N); | ||
| + | </asy> | ||
| + | </center> | ||
| + | |||
| + | <math>\text{(A)}\ \text{The shaded areas in all three are equal.}</math> | ||
| + | |||
| + | <math>\text{(B)}\ \text{Only the shaded areas of }I\text{ and }II\text{ are equal.}</math> | ||
| + | |||
| + | <math>\text{(C)}\ \text{Only the shaded areas of }I\text{ and }III\text{ are equal.}</math> | ||
| − | <math> | + | <math>\text{(D)}\ \text{Only the shaded areas of }II\text{ and }III\text{ are equal.}</math> |
| − | <math> | + | <math>\text{(E)}\ \text{The shaded areas of }I, II\text{ and }III\text{ are all different.}</math> |
| − | <math> | + | ==Solution== |
| + | Square II clearly has <math>1/4</math> shaded. Partitioning square I into eight right triangles also shows <math>1/4</math> of it is shaded. Lastly, square III can be partitioned into sixteen triangles, and because four are shaded, <math>1/4</math> of the total square is shaded. <math>\boxed{\text{(A)}}</math>. | ||
| − | |||
==See Also== | ==See Also== | ||
{{AJHSME box|year=1994|num-b=11|num-a=13}} | {{AJHSME box|year=1994|num-b=11|num-a=13}} | ||
Revision as of 23:47, 22 December 2012
Problem 12
Each of the three large squares shown below is the same size. Segments that intersect the sides of the squares intersect at the midpoints of the sides. How do the shaded areas of these squares compare?
![[asy] unitsize(36); fill((0,0)--(1,0)--(1,1)--cycle,gray); fill((1,1)--(1,2)--(2,2)--cycle,gray); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((1,0)--(1,2)); draw((0,0)--(2,2)); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,gray); draw((3,0)--(5,0)--(5,2)--(3,2)--cycle); draw((4,0)--(4,2)); draw((3,1)--(5,1)); fill((6,1)--(6.5,0.5)--(7,1)--(7.5,0.5)--(8,1)--(7.5,1.5)--(7,1)--(6.5,1.5)--cycle,gray); draw((6,0)--(8,0)--(8,2)--(6,2)--cycle); draw((6,0)--(8,2)); draw((6,2)--(8,0)); draw((7,0)--(6,1)--(7,2)--(8,1)--cycle); label("$I$",(1,2),N); label("$II$",(4,2),N); label("$III$",(7,2),N); [/asy]](http://latex.artofproblemsolving.com/d/9/3/d93e618cec4517a590f4d808d9c80dc5979437a7.png)
Solution
Square II clearly has 
shaded. Partitioning square I into eight right triangles also shows of it is shaded. Lastly, square III can be partitioned into sixteen triangles, and because four are shaded, of the total square is shaded. 
.
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
