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>(\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>\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}(\frac{1}{2})(\frac{1}{2}) = \frac{1}{12}</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>. |
== See also == | == See also == | ||
Revision as of 16:55, 28 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?
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 
. Then the square base of the pyramid has area 
.
We also know that the height of the pyramid is half the height of the cube, so it is 
. The volume of a pyramid with base area 
and height 
is 
so each of the pyramids has volume 
. The whole octahedron is twice this volume, so 
.
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 | ||
