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

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==Video Solution==
 
==Video Solution==
 +
https://www.youtube.com/watch?v=eDpxHt1LL7g
  
https://www.youtube.com/watch?v=eDpxHt1LL7g
+
~David
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2003|num-b=14|num-a=16}}
 
{{AMC8 box|year=2003|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:56, 15 April 2023

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

In order to minimize the amount of cubes needed, we must match up as many squares of our given figures with each other to make different sides of the same cube. One example of the solution with $\boxed{\textbf{(B)}\ 4}$ cubes. Notice the corner cube cannot be removed for a figure of 3 cubes because each face of a cube must be touching another face.

[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(1,0,0)*unitcube, white, thick(), nolight); draw(shift(1,0,1)*unitcube, white, thick(), nolight);[/asy]


Video Solution

https://www.youtube.com/watch?v=eDpxHt1LL7g

~David

See Also

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

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