2009 AMC 8 Problems/Problem 4: Difference between revisions
No edit summary |
Basketball8 (talk | contribs) |
||
| Line 44: | Line 44: | ||
</asy> | </asy> | ||
<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> | ||
==Solution== | |||
The answer is $\boxed{\textbf{(B)}} because the longest piece can't fit in the figure. | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=3|num-a=5}} | {{AMC8 box|year=2009|num-b=3|num-a=5}} | ||
Revision as of 22:27, 12 November 2012
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]](http://latex.artofproblemsolving.com/4/1/a/41a5a2bf3bad2afc8d7f420343ef308f0a208aa3.png)
![[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]](http://latex.artofproblemsolving.com/f/b/4/fb48f64accc93ee27ac75f695ac7d512c9e12f7a.png)
![\[\textbf{(A)}\qquad\qquad\qquad\textbf{(B)}\quad\qquad\qquad\textbf{(C)}\:\qquad\qquad\qquad\textbf{(D)}\quad\qquad\qquad\textbf{(E)}\]](http://latex.artofproblemsolving.com/c/9/1/c91a399a954405c9abe5b1a2036d7e5d2917d99c.png)
Solution
The answer is $\boxed{\textbf{(B)}} because the longest piece can't fit in the figure.
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 | ||
