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

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== Problem ==
 
== Problem ==
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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 <math>W</math>?
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A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure <math>1</math>. The cube is then cut in the same manner along the dashed lines shown in Figure <math>2</math>. This creates nine pieces. What is the volume of the piece that contains vertex <math>W</math>?
  
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<asy>
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[[Image:2005 AMC 12A Problem 17.png]]
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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));
 
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draw(b,linewidth(0.7));
 
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draw(c,linewidth(0.7));
 
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draw(d,linewidth(0.7));
 
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draw(e,linewidth(0.7));
 
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draw(shift((20,0))*a,linewidth(0.7));
 
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draw(shift((20,0))*b,linewidth(0.7));
 
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draw(shift((20,0))*c,linewidth(0.7));
 
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draw(shift((20,0))*d,linewidth(0.7));
 
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draw(shift((20,0))*e,linewidth(0.7));
 
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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);
 
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label("$W$",(15.2,0),S);
 
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label("Figure 1",(5,0),S);
 
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label("Figure 2",(25,0),S);
 
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</asy>
 
  
 
<math>
 
<math>

Latest revision as of 14:58, 1 July 2025

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 $W$?

2005 AMC 12A Problem 17.png

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

Solution

It is a pyramid with height $1$ and base area $\frac{1}{4}$, so using the formula for the volume of a pyramid, $\frac{1}{3} \cdot \left(\frac{1}{4}\right) \cdot (1) = \frac {1}{12} \Rightarrow \boxed{(\mathrm {A})}$.

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

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

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