Difference between revisions of "2003 AMC 8 Problems/Problem 15"

(Solution)
Line 24: Line 24:
 
draw(shift(1,0,1)*unitcube, white, thick(), nolight);</asy>
 
draw(shift(1,0,1)*unitcube, white, thick(), nolight);</asy>
  
<math> \textbf{(B)} 3</math>
+
<math> \textbf{(A)} 3</math>
  
 
{{AMC8 box|year=2003|num-b=14|num-a=16}}
 
{{AMC8 box|year=2003|num-b=14|num-a=16}}

Revision as of 12:02, 24 August 2012

Problem

A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?

[asy] defaultpen(linewidth(0.8)); path p=unitsquare; draw(p^^shift(0,1)*p^^shift(1,0)*p); draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p); label("FRONT", (1,0), S); label("SIDE", (5,0), S); [/asy]

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution

One example of the solution [asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,-1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,1)*unitcube, white, thick(), nolight);[/asy]

$\textbf{(A)} 3$

2003 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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