Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1989 AHSME Problems/Problem 26"
Line 3: Line 3:
 
<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>
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
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...
 +
 +
<math>(\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}</math>
 +
 +
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>

Revision as of 21:25, 7 June 2014

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

$\mathrm{(A) \frac{\sqrt{3}}{12} } \qquad \mathrm{(B) \frac{\sqrt{6}}{16} } \qquad \mathrm{(C) \frac{1}{6} } \qquad \mathrm{(D) \frac{\sqrt{2}}{8} } \qquad \mathrm{(E) \frac{1}{4} }$ These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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…

$\frac{\frac{x^3}{6}}{x^3} \rightarrow \framebox[1.1\width]{(C) \frac{1}{6} }$ (Error compiling LaTeX. Unknown error_msg)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_26&oldid=62228"