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Difference between revisions of "2001 AMC 10 Problems/Problem 5"

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== Problem ==
 
== Problem ==
  
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A <math>5\times 5 \times 5</math> cube has all its faces painted, and then is cut into <math>125</math> small cubes, each <math>1\times 1\times 1</math>. How many of these small cubes have exactly <math>2</math> of their <math>6</math> faces painted?
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How many of the twelve pentominoes pictured below have at least one line of symmetry?
  
 
<asy>
 
<asy>
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draw(shift(18,2)*unitsquare);</asy>
 
draw(shift(18,2)*unitsquare);</asy>
  
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<math> \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7 </math>
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<math>\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7</math>
  
 
== Solution ==
 
== Solution ==
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The ones with lines over the shapes have at least one line of symmetry. Counting the number of shapes that have line(s) on them,
 
The ones with lines over the shapes have at least one line of symmetry. Counting the number of shapes that have line(s) on them,
 
we find <math> \boxed{\textbf{(D)}\ 6} </math> pentominoes.
 
we find <math> \boxed{\textbf{(D)}\ 6} </math> pentominoes.
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 +
==Video Solution by Daily Dose of Math==
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 +
https://youtu.be/svFpNvUUY7E?si=CloMWtqbbhBNgWy_
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 +
~Thesmartgreekmathdude
  
 
== See Also ==
 
== See Also ==
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{{AMC10 box|year=2001|num-b=4|num-a=6}}
 
{{AMC10 box|year=2001|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Geometry Problems]]

Latest revision as of 16:13, 18 October 2025

Problem

How many of the twelve pentominoes pictured below have at least one line of symmetry?

[asy] unitsize(5mm); defaultpen(linewidth(1pt)); draw(shift(2,0)*unitsquare); draw(shift(2,1)*unitsquare); draw(shift(2,2)*unitsquare); draw(shift(1,2)*unitsquare); draw(shift(0,2)*unitsquare); draw(shift(2,4)*unitsquare); draw(shift(2,5)*unitsquare); draw(shift(2,6)*unitsquare); draw(shift(1,5)*unitsquare); draw(shift(0,5)*unitsquare); draw(shift(4,8)*unitsquare); draw(shift(3,8)*unitsquare); draw(shift(2,8)*unitsquare); draw(shift(1,8)*unitsquare); draw(shift(0,8)*unitsquare); draw(shift(6,8)*unitsquare); draw(shift(7,8)*unitsquare); draw(shift(8,8)*unitsquare); draw(shift(9,8)*unitsquare); draw(shift(9,9)*unitsquare); draw(shift(6,5)*unitsquare); draw(shift(7,5)*unitsquare); draw(shift(8,5)*unitsquare); draw(shift(7,6)*unitsquare); draw(shift(7,4)*unitsquare); draw(shift(6,1)*unitsquare); draw(shift(7,1)*unitsquare); draw(shift(8,1)*unitsquare); draw(shift(6,0)*unitsquare); draw(shift(7,2)*unitsquare); draw(shift(11,8)*unitsquare); draw(shift(12,8)*unitsquare); draw(shift(13,8)*unitsquare); draw(shift(14,8)*unitsquare); draw(shift(13,9)*unitsquare); draw(shift(11,5)*unitsquare); draw(shift(12,5)*unitsquare); draw(shift(13,5)*unitsquare); draw(shift(11,6)*unitsquare); draw(shift(13,4)*unitsquare); draw(shift(11,1)*unitsquare); draw(shift(12,1)*unitsquare); draw(shift(13,1)*unitsquare); draw(shift(13,2)*unitsquare); draw(shift(14,2)*unitsquare); draw(shift(16,8)*unitsquare); draw(shift(17,8)*unitsquare); draw(shift(18,8)*unitsquare); draw(shift(17,9)*unitsquare); draw(shift(18,9)*unitsquare); draw(shift(16,5)*unitsquare); draw(shift(17,6)*unitsquare); draw(shift(18,5)*unitsquare); draw(shift(16,6)*unitsquare); draw(shift(18,6)*unitsquare); draw(shift(16,0)*unitsquare); draw(shift(17,0)*unitsquare); draw(shift(17,1)*unitsquare); draw(shift(18,1)*unitsquare); draw(shift(18,2)*unitsquare);[/asy]

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

Solution

Pentonimoes.gif

The ones with lines over the shapes have at least one line of symmetry. Counting the number of shapes that have line(s) on them, we find $\boxed{\textbf{(D)}\ 6}$ pentominoes.

Video Solution by Daily Dose of Math

https://youtu.be/svFpNvUUY7E?si=CloMWtqbbhBNgWy_

~Thesmartgreekmathdude

See Also

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 AMC 10 Problems and Solutions

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

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