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Difference between revisions of "2005 AMC 12A Problems/Problem 17"

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</math>
 
</math>
  
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[[Image:2005 AMC 12A Problem 17.png]]
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<asy>
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path a=(0,0)--(10,0)--(10,10)--(0,10)--cycle;
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path b = (0,10)--(6,16)--(16,16)--(16,6)--(10,0);
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path c= (10,10)--(16,16);
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path d= (0,0)--(3,13)--(13,13)--(10,0);
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path e= (13,13)--(16,6);
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draw(a,linewidth(0.7));
 +
draw(b,linewidth(0.7));
 +
draw(c,linewidth(0.7));
 +
draw(d,linewidth(0.7));
 +
draw(e,linewidth(0.7));
 +
draw(shift((20,0))*a,linewidth(0.7));
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draw(shift((20,0))*b,linewidth(0.7));
 +
draw(shift((20,0))*c,linewidth(0.7));
 +
draw(shift((20,0))*d,linewidth(0.7));
 +
draw(shift((20,0))*e,linewidth(0.7));
 +
draw((20,0)--(25,10)--(30,0),dashed);
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draw((25,10)--(31,16)--(36,6),dashed);
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draw((15,0)--(10,10),Arrow);
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draw((15.5,0)--(30,10),Arrow);
 +
label("$W$",(15.2,0),S);
 +
label("Figure 1",(5,0),S);
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label("Figure 2",(25,0),S);
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</asy>
  
 
== Solution ==
 
== Solution ==

Revision as of 16:58, 17 June 2016

Problem

A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the same manner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volume of the piece that contains vertex $W$?

$(\mathrm {A}) \ \frac{1}{12} \qquad (\mathrm {B}) \ \frac{1}{9} \qquad (\mathrm {C})\ \frac{1}{8} \qquad (\mathrm {D}) \ \frac{1}{6} \qquad (\mathrm {E})\ \frac{1}{4}$

[asy] path a=(0,0)--(10,0)--(10,10)--(0,10)--cycle; path b = (0,10)--(6,16)--(16,16)--(16,6)--(10,0); path c= (10,10)--(16,16); path d= (0,0)--(3,13)--(13,13)--(10,0); path e= (13,13)--(16,6); draw(a,linewidth(0.7)); draw(b,linewidth(0.7)); draw(c,linewidth(0.7)); draw(d,linewidth(0.7)); draw(e,linewidth(0.7)); draw(shift((20,0))*a,linewidth(0.7)); draw(shift((20,0))*b,linewidth(0.7)); draw(shift((20,0))*c,linewidth(0.7)); draw(shift((20,0))*d,linewidth(0.7)); draw(shift((20,0))*e,linewidth(0.7)); draw((20,0)--(25,10)--(30,0),dashed); draw((25,10)--(31,16)--(36,6),dashed); draw((15,0)--(10,10),Arrow); draw((15.5,0)--(30,10),Arrow); label("$W$",(15.2,0),S); label("Figure 1",(5,0),S); label("Figure 2",(25,0),S); [/asy]

Solution

It is a pyramid, so $\frac{1}{3} \cdot \left(\frac{1}{4}\right) \cdot (1) = \frac {1}{12} \Rightarrow \boxed{(\mathrm {A})}$.

See also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 12 Problems and Solutions

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