Difference between revisions of "2015 AMC 10B Problems/Problem 17"

Latest revision as of 19:57, 25 August 2023

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?

$[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)--(5,0,3)--(5,0,0)--(5,4,0)--(0,4,0)--(0,4,3)--(0,0,3)--(5,0,3)); draw((0,4,3)--(5,4,3)--(5,4,0)); label("3",(5,0,3)--(5,0,0),W); label("4",(5,0,0)--(5,4,0),S); label("5",(5,4,0)--(0,4,0),SE); [/asy]$

$\textbf{(A) } \dfrac{75}{12} \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 10\sqrt2 \qquad\textbf{(E) } 15$

Solution 1

The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals $4$ and $5$ , for an area $A=10$ . The height $h$ , of one pyramid, is $\frac{3}{2}$ , so the volume of one pyramid is $\frac{Ah}{3}=5$ . Thus, the octahedron has volume $2\cdot5=\boxed{\textbf{(B) }10}$

Solution 2 (Scaling)

The "base" of the octahedron is half the base of the rectangular prism because it is connected by the midpoints. Additionally, the volume of an octahedron is $\dfrac{1}{3}$ of its respective prism. Thus, the octahedron's volume is $\dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6}$ of the rectangular prism's volume, meaning that the answer is $3 \cdot 4 \cdot 5 \cdot \dfrac{1}{6} = \boxed{\\10}$

Video Solution

https://www.youtube.com/watch?v=NEQNfr_T8OA

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 30: / Line 30: @@
 ==Video Solution==
-https://youtu.be/0uCMSH7-Ubk
+https://www.youtube.com/watch?v=NEQNfr_T8OA
 ==See Also==

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