Difference between revisions of "2006 AMC 10A Problems/Problem 24"
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== Problem == | == Problem == | ||
− | + | [[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 == | ||
− | == See | + | We can break the octahedron into two [[square pyramid]]s by cutting it along a [[plane]] [[perpendicular]] to one of its internal [[diagonal]]s. |
− | + | <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 [[edge]]s of [[length]] 1 so all edges of the regular octahedron have 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}\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 == | ||
+ | {{AMC10 box|year=2006|ab=A|num-b=23|num-a=25}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:32, 4 July 2013
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 1 so all edges of the regular octahedron have 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 |
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