Difference between revisions of "1989 AHSME Problems/Problem 26"

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== Problem ==
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A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is
 
A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is
  
 
<math> \mathrm{(A) \frac{\sqrt{3}}{12} } \qquad \mathrm{(B) \frac{\sqrt{6}}{16} } \qquad \mathrm{(C) \frac{1}{6} } \qquad \mathrm{(D) \frac{\sqrt{2}}{8} } \qquad \mathrm{(E) \frac{1}{4} }  </math>
 
<math> \mathrm{(A) \frac{\sqrt{3}}{12} } \qquad \mathrm{(B) \frac{\sqrt{6}}{16} } \qquad \mathrm{(C) \frac{1}{6} } \qquad \mathrm{(D) \frac{\sqrt{2}}{8} } \qquad \mathrm{(E) \frac{1}{4} }  </math>
  
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== Solution ==
  
 
Call the length of a side of the cube x. Thus, the volume of the cube is <math>x^3</math>. We can then find that a side of this regular octahedron is the square root of <math>(\frac{x}{2})^2</math>+<math>(\frac{x}{2})^2</math> which is equivalent to <math>\frac{x\sqrt{2}}{2}</math>. Using our general formula for the volume of a regular octahedron of side length a, which is <math>\frac{a^3\sqrt2}{3}</math>, we get that the volume of this octahedron is...
 
Call the length of a side of the cube x. Thus, the volume of the cube is <math>x^3</math>. We can then find that a side of this regular octahedron is the square root of <math>(\frac{x}{2})^2</math>+<math>(\frac{x}{2})^2</math> which is equivalent to <math>\frac{x\sqrt{2}}{2}</math>. Using our general formula for the volume of a regular octahedron of side length a, which is <math>\frac{a^3\sqrt2}{3}</math>, we get that the volume of this octahedron is...
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Comparing the ratio of the volume of the octahedron to the cube is…
 
Comparing the ratio of the volume of the octahedron to the cube is…
  
<math>\frac{\frac{x^3}{6}}{x^3} \rightarrow \framebox[1.1\width]{(C) \frac{1}{6} }</math>
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<math>\frac{\frac{x^3}{6}}{x^3} \rightarrow \frac{1}{6} </math> or <math>\fbox{C}</math>
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== See also ==
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{{AHSME box|year=1989|num-b=25|num-a=27}} 
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[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:08, 22 October 2014

Problem

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is

$\mathrm{(A) \frac{\sqrt{3}}{12} } \qquad \mathrm{(B) \frac{\sqrt{6}}{16} } \qquad \mathrm{(C) \frac{1}{6} } \qquad \mathrm{(D) \frac{\sqrt{2}}{8} } \qquad \mathrm{(E) \frac{1}{4} }$


Solution

Call the length of a side of the cube x. Thus, the volume of the cube is $x^3$. We can then find that a side of this regular octahedron is the square root of $(\frac{x}{2})^2$+$(\frac{x}{2})^2$ which is equivalent to $\frac{x\sqrt{2}}{2}$. Using our general formula for the volume of a regular octahedron of side length a, which is $\frac{a^3\sqrt2}{3}$, we get that the volume of this octahedron is...

$(\frac{x\sqrt{2}}{2})^3 \rightarrow \frac{x^3\sqrt{2}}{4} \rightarrow \frac{x^3\sqrt{2}}{4}*\frac{\sqrt{2}}{3} \rightarrow \frac{2x^3}{12}=\frac{x^3}{6}$

Comparing the ratio of the volume of the octahedron to the cube is…

$\frac{\frac{x^3}{6}}{x^3} \rightarrow \frac{1}{6}$ or $\fbox{C}$


See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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