Difference between revisions of "2006 AMC 10A Problems/Problem 24"

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<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 ==
We can break the octahedron into two [[square pyramid]]s by cutting it along a [[plane]] [[perpendicular]] to one of its internal [[diagonal]]s.  The cube has [[edge]]s of [[length]] one so the edge of the octahedron in of length <math>\frac{\sqrt{2}}{2}</math>.  Then the [[square (geometry) | square]] base of the [[pyramid]] has [[area]] <math>\left(\frac{1}{2}\sqrt{2}\right)^2 = \frac{1}{2}</math>.
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We can break the octahedron into two [[square pyramid]]s by cutting it along a [[plane]] [[perpendicular]] to one of its internal [[diagonal]]s.  The cube has [[edge]]s of [[length]] one so the edge of the octahedron is of length <math>\frac{\sqrt{2}}{2}</math>.  Then the [[square (geometry) | square]] base of the [[pyramid]] has [[area]] <math>\left(\frac{1}{2}\sqrt{2}\right)^2 = \frac{1}{2}</math>.
 
We also know that the height of the pyramid is half the height of the cube, so it is <math>\frac{1}{2}</math>.  The volume of a pyramid with base area <math>B</math> and height <math>h</math> is <math>A=\frac{1}{3}Bh</math> so each of the pyramids has volume <math>\frac{1}{3}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{12}</math>. The whole octahedron is twice this volume, so <math>\frac{1}{12} \cdot 2 = \frac{1}{6} \Longrightarrow \mathrm{(B)}</math>.
 
We also know that the height of the pyramid is half the height of the cube, so it is <math>\frac{1}{2}</math>.  The volume of a pyramid with base area <math>B</math> and height <math>h</math> is <math>A=\frac{1}{3}Bh</math> so each of the pyramids has volume <math>\frac{1}{3}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{12}</math>. The whole octahedron is twice this volume, so <math>\frac{1}{12} \cdot 2 = \frac{1}{6} \Longrightarrow \mathrm{(B)}</math>.
  

Revision as of 12:41, 30 October 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 square pyramids by cutting it along a plane perpendicular to one of its internal diagonals. The cube has edges of length one so the edge of the octahedron is of length $\frac{\sqrt{2}}{2}$. Then the square base of the pyramid has area $\left(\frac{1}{2}\sqrt{2}\right)^2 = \frac{1}{2}$. We also know that the height of the pyramid is half the height of the cube, so it is $\frac{1}{2}$. The volume of a pyramid with base area $B$ and height $h$ is $A=\frac{1}{3}Bh$ so each of the pyramids has volume $\frac{1}{3}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{12}$. The whole octahedron is twice this volume, so $\frac{1}{12} \cdot 2 = \frac{1}{6} \Longrightarrow \mathrm{(B)}$.

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
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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