2015 AMC 10B Problems/Problem 17
Contents
Problem
The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?
![[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)--(5,0,3)--(5,0,0)--(5,4,0)--(0,4,0)--(0,4,3)--(0,0,3)--(5,0,3)); draw((0,4,3)--(5,4,3)--(5,4,0)); label("3",(5,0,3)--(5,0,0),W); label("4",(5,0,0)--(5,4,0),S); label("5",(5,4,0)--(0,4,0),SE); [/asy]](http://latex.artofproblemsolving.com/0/f/e/0fecfb360827e29c3e9ad69126c0215a16cfd2ab.png)
Solution 1
The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals 
and 
, for an area 
. The height 
, of one pyramid, is 
, so the volume of one pyramid is 
. Thus, the octahedron has volume 
Solution 2
Scale the prism down by two, but let one vertice coincide with the big prism (The small prism is still inside the big one). Three of the vertices are the centers of faces. Since the vertices of the octahedron are made by the centers of the faces of the prism, connect those three vertices. Then draw a plane using those 3 points, and "slice" the prism. Notice how they are congruent. You do this for all of the eight prisms in the big prism. Since all of them are half the area, the area of the octahedron is 3 ⋅4 ⋅5=\boxed{{(B)}\\10}$
See Also
| 2015 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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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