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 "2006 AMC 10A Problems/Problem 24"

(convert to wikisyntax, add box)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
[[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?  
+
[[Center]]s of adjacent [[face]]s 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 ==
−
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>.
+
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>(\frac{1}{2}\sqrt{2})^2 = \frac{1}{2}</math>.
−
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>.
+
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}(\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} \Longrightarrow \mathrm{(B)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 09:56, 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 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 in of length $\frac{\sqrt{2}}{2}$. Then the square base of the pyramid has area $(\frac{1}{2}\sqrt{2})^2 = \frac{1}{2}$. 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 $\frac{1}{3}(\frac{1}{2})(\frac{1}{2}) = \frac{1}{12}$. The whole octahedron is twice this volume, so $\frac{1}{12} \cdot 2 = \frac{1}{6} \Longrightarrow \mathrm{(B)}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_24&oldid=13216"