Difference between revisions of "2006 AMC 10A Problems/Problem 24"
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== Problem == | == Problem == | ||
− | + | Centers of adjacent faces of a unit cube are joined to form a regular [[octahedron]]. What is the volume of this octahedron? | |
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+ | <math>\textbf{(A) } \frac{1}{8}\qquad\textbf{(B) } \frac{1}{6}\qquad\textbf{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{1}{3}\qquad\textbf{(E) } \frac{1}{2}\qquad</math> | ||
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== 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. | We can break the octahedron into two [[square pyramid]]s by cutting it along a [[plane]] [[perpendicular]] to one of its internal [[diagonal]]s. |
Revision as of 10:18, 19 December 2021
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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