Difference between revisions of "2009 AMC 8 Problems/Problem 4"
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fill((1,0)--(1,3)--(2,3)--(2,0)--cycle,green); | fill((1,0)--(1,3)--(2,3)--(2,0)--cycle,green); | ||
fill((2,0)--(2,2)--(3,2)--(3,0)--cycle,yellow); | fill((2,0)--(2,2)--(3,2)--(3,0)--cycle,yellow); | ||
− | fill((3, | + | fill((3,4)--(3,5)--(4,5)--(4,4)--cycle,red); |
− | fill((3, | + | fill((3,0)--(3,4)--(4,4)--(4,0)--cycle,cyan); |
fill((11,0)--(11,5)--(12,5)--(12,0)--cycle,pink); | fill((11,0)--(11,5)--(12,5)--(12,0)--cycle,pink); |
Latest revision as of 01:56, 26 November 2023
Contents
Problem
The five pieces shown below can be arranged to form four of the five figures shown in the choices. Which figure cannot be formed?
Solution
The answer is because the longest piece cannot fit into the figure.
Note that the five pieces can be arranged to form the figures in and
as shown below:
~Basketball8 ~MRENTHUSIASM
Video Solution
https://youtu.be/USVVURBLaAc?t=171
See Also
2009 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.