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Difference between revisions of "2006 AMC 10A Problems/Problem 24"
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== Problem ==
 
== Problem ==
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Centers of adjacent faces of a unit [[cube (geometry) | cube]] are joined to form a regular [[octahedron]]. What is the volume of this octahedron?  
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[[Center]]s of adjacent faces of a unit [[cube (geometry) | cube]] are joined to form a regular [[octahedron]]. What is the [[volume]] of this octahedron?  
  
 
<math>\mathrm{(A) \ } \frac{1}{8}\qquad\mathrm{(B) \ } \frac{1}{6}\qquad\mathrm{(C) \ } \frac{1}{4}\qquad\mathrm{(D) \ } \frac{1}{3}\qquad\mathrm{(E) \ } \frac{1}{2}\qquad</math>  
 
<math>\mathrm{(A) \ } \frac{1}{8}\qquad\mathrm{(B) \ } \frac{1}{6}\qquad\mathrm{(C) \ } \frac{1}{4}\qquad\mathrm{(D) \ } \frac{1}{3}\qquad\mathrm{(E) \ } \frac{1}{2}\qquad</math>  
 
== Solution ==
 
== Solution ==
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{{solution}}
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We can break the octahedron into two [[tetrahedron]]s. The cube has sides of length one, so we can solve for the area of the tetrahedron's base: <math>(\frac{1}{2}\sqrt{2})^2 = \frac{1}{2}</math>.
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*not confirmed. My own solution.
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We also know that the height of a tetrahedron, which would be half the height of the cube, is <math>\frac{1}{2}</math>. Using the volume formula of a tetrahedron: <math>A=\frac{1}{3}Bh</math> where B is the base side length, we find that the volume of one of the tetrahedrons is <math>\frac{1}{3}(\frac{1}{2})(\frac{1}{2}) = \frac{1}{12}</math>. The whole octahedron is twice this volume, so <math>\frac{1}{12} \cdot 2 = \frac{1}{6} \Rightarrow \mathrm{B}</math>.
  
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We can first break the octahedron into two tetrahedrons. since we know the cube has sides of length one, we can solve for the side length of the tetrahedron's base: (1/2)(sqrt(2))=(sqrt(2)/2)
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== See also ==
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We also know that the height of a tetrahedron is (1/2) from the fact that this is a unit cube. Using the area formula of a tetrahdron: A=(1/3)(b)(h) where b is the base side length, we find that the area of one tetrahedron is (1/12). The whole octahedron is twice this area. (1/12)(2)=(1/6)
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{{AMC10 box|year=2006|ab=A|num-b=23|num-a=25}}
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(B) 1/6
 
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== See Also ==
 
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*[[2006 AMC 10A Problems]]
 
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*[[2006 AMC 10A Problems/Problem 23|Previous Problem]]
 
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*[[2006 AMC 10A Problems/Problem 25|Next Problem]]
 
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 08:27, 15 February 2007

Problem

Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?

$\mathrm{(A) \ } \frac{1}{8}\qquad\mathrm{(B) \ } \frac{1}{6}\qquad\mathrm{(C) \ } \frac{1}{4}\qquad\mathrm{(D) \ } \frac{1}{3}\qquad\mathrm{(E) \ } \frac{1}{2}\qquad$

Solution

We can break the octahedron into two tetrahedrons. The cube has sides of length one, so we can solve for the area of the tetrahedron's base: $(\frac{1}{2}\sqrt{2})^2 = \frac{1}{2}$. We also know that the height of a tetrahedron, which would be half the height of the cube, is $\frac{1}{2}$. Using the volume formula of a tetrahedron: $A=\frac{1}{3}Bh$ where B is the base side length, we find that the volume of one of the tetrahedrons is $\frac{1}{3}(\frac{1}{2})(\frac{1}{2}) = \frac{1}{12}$. The whole octahedron is twice this volume, so $\frac{1}{12} \cdot 2 = \frac{1}{6} \Rightarrow \mathrm{B}$.

See also

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
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