Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AMC 8 Problems/Problem 4"

(Deleted repetitive solution.)
(→‎Solution)
Line 132: Line 132:
 
<cmath>\textbf{(A)}\qquad\qquad\qquad\textbf{(B)}\quad\qquad\qquad\textbf{(C)}\qquad\qquad\qquad\textbf{(D)}\quad\qquad\qquad\textbf{(E)}</cmath>
 
<cmath>\textbf{(A)}\qquad\qquad\qquad\textbf{(B)}\quad\qquad\qquad\textbf{(C)}\qquad\qquad\qquad\textbf{(D)}\quad\qquad\qquad\textbf{(E)}</cmath>
 
~Basketball8 ~MRENTHUSIASM
 
~Basketball8 ~MRENTHUSIASM
 +
==Solution==
 +
Note that in diagram two, there are no 5 squares horizontally or vertically in a row, therefore the piece that is five long cannot fit there, therefore the answer is <math>\boxed{\textbf{(B)}}</math>
 +
~LIUGRA001
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 17:45, 25 November 2024

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? [asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p); [/asy]

[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy] \[\textbf{(A)}\qquad\qquad\qquad\textbf{(B)}\quad\qquad\qquad\textbf{(C)}\:\qquad\qquad\qquad\textbf{(D)}\quad\qquad\qquad\textbf{(E)}\]

Solution

The answer is $\boxed{\textbf{(B)}}$ because the longest piece cannot fit into the figure.

Note that the five pieces can be arranged to form the figures in $\textbf{(A)},\textbf{(C)},\textbf{(D)},$ and $\textbf{(E)},$ as shown below:

[asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; fill(p,red); fill(shift(s,r)*p,yellow); fill(shift(s,-r)*p,yellow); fill(shift(2s,2r)*p,green); fill(shift(2s,0)*p,green); fill(shift(2s,-2r)*p,green); fill(shift(3s,3r)*p,cyan); fill(shift(3s,-3r)*p,cyan); fill(shift(3s,r)*p,cyan); fill(shift(3s,-r)*p,cyan); fill(shift(4s,-4r)*p,pink); fill(shift(4s,-2r)*p,pink); fill(shift(4s,0)*p,pink); fill(shift(4s,2r)*p,pink); fill(shift(4s,4r)*p,pink); draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p); [/asy]

[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; fill((0,0)--(0,5)--(1,5)--(1,0)--cycle,pink); fill((1,0)--(1,3)--(2,3)--(2,0)--cycle,green); fill((2,0)--(2,2)--(3,2)--(3,0)--cycle,yellow); fill((3,4)--(3,5)--(4,5)--(4,4)--cycle,red); fill((3,0)--(3,4)--(4,4)--(4,0)--cycle,cyan); fill((11,0)--(11,5)--(12,5)--(12,0)--cycle,pink); fill((12,1)--(12,5)--(13,5)--(13,1)--cycle,cyan); fill((13,2)--(13,5)--(14,5)--(14,2)--cycle,green); fill((14,3)--(14,5)--(15,5)--(15,3)--cycle,yellow); fill((15,4)--(15,5)--(16,5)--(16,4)--cycle,red); fill((17,0)--(17,5)--(18,5)--(18,0)--cycle,pink); fill((18,0)--(18,4)--(19,4)--(19,0)--cycle,cyan); fill((18,4)--(18,5)--(19,5)--(19,4)--cycle,red); fill((19,0)--(19,3)--(20,3)--(20,0)--cycle,green); fill((19,3)--(19,5)--(20,5)--(20,3)--cycle,yellow); fill((21,4)--(21,5)--(22,5)--(22,4)--cycle,red); fill((22,1)--(22,5)--(23,5)--(23,1)--cycle,cyan); fill((23,0)--(23,5)--(24,5)--(24,0)--cycle,pink); fill((24,1)--(24,4)--(25,4)--(25,1)--cycle,green); fill((24,4)--(24,5)--(26,5)--(26,4)--cycle,yellow); int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy] \[\textbf{(A)}\qquad\qquad\qquad\textbf{(B)}\quad\qquad\qquad\textbf{(C)}\qquad\qquad\qquad\textbf{(D)}\quad\qquad\qquad\textbf{(E)}\] ~Basketball8 ~MRENTHUSIASM

Solution

Note that in diagram two, there are no 5 squares horizontally or vertically in a row, therefore the piece that is five long cannot fit there, therefore the answer is $\boxed{\textbf{(B)}}$ ~LIUGRA001

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_4&oldid=235583"