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

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</asy>
 
The cube has edges of length 1 so all edges of the regular octahedron have length <math>\frac{\sqrt{2}}{2}</math>.  Then the square base of the [[pyramid]] has area <math>\left(\frac{1}{2}\sqrt{2}\right)^2 = \frac{1}{2}</math>.
 
The cube has edges of length 1 so all edges of the regular octahedron have length <math>\frac{\sqrt{2}}{2}</math>.  Then the 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 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 = \boxed{\textbf{(B) }{\frac{1}{6}}</math>.
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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 = \boxed{\textbf{(B) }\frac{1}{6}}</math>.
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== See also ==
 
== See also ==
 
{{AMC10 box|year=2006|ab=A|num-b=23|num-a=25}}
 
{{AMC10 box|year=2006|ab=A|num-b=23|num-a=25}}

Latest revision as of 19:53, 1 October 2024

Problem

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

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

Solution

We can break the octahedron into two square pyramids by cutting it along a plane perpendicular to one of its internal diagonals. [asy] import three; real r = 1/2; triple A = (-0.5,1.5,0); size(400); currentprojection=orthographic(1,1/4,1/2); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)^^(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--(0,0,1)^^(0,0,0)--(0,0,1)^^(1,0,0)--(1,0,1)^^(0,1,0)--(0,1,1)^^(1,1,0)--(1,1,1),gray(0.8)); draw((0,r,r)--(r,1,r)--(1,r,r)--(r,0,r)--cycle^^(r,r,0)--(0,r,r)--(r,r,1)--(r,1,r)--(r,r,0)--(1,r,r)--(r,r,1)--(r,0,r)--(r,r,0)); draw((0,r,r)+A--(r,1,r)+A--(1,r,r)+A--(r,0,r)+A--cycle^^(0,r,r)+A--(r,r,1)+A--(1,r,r)+A^^(r,1,r)+A--(r,r,1)+A--(r,0,r)+A); [/asy] The cube has edges of length 1 so all edges of the regular octahedron have 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 $\frac{1}{12} \cdot 2 = \boxed{\textbf{(B) }\frac{1}{6}}$.

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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