Difference between revisions of "2016 AMC 12B Problems/Problem 23"
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<math>\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ 1</math> | <math>\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ 1</math> | ||
| − | =Solution 1 (Non Calculus)= | + | ==Solution 1 (Non Calculus)== |
The first inequality refers to the interior of a regular octahedron with top and bottom vertices <math>(0,0,1),\ (0,0,-1)</math>. Its volume is <math>8\cdot\tfrac16=\tfrac43</math>. The second inequality describes an identical shape, shifted <math>+1</math> upwards along the <math>Z</math> axis. The intersection will be a similar octahedron, linearly scaled down by half. Thus the volume of the intersection is one-eighth of the volume of the first octahedron, giving an answer of <math>\textbf{(A) }\tfrac16</math>. | The first inequality refers to the interior of a regular octahedron with top and bottom vertices <math>(0,0,1),\ (0,0,-1)</math>. Its volume is <math>8\cdot\tfrac16=\tfrac43</math>. The second inequality describes an identical shape, shifted <math>+1</math> upwards along the <math>Z</math> axis. The intersection will be a similar octahedron, linearly scaled down by half. Thus the volume of the intersection is one-eighth of the volume of the first octahedron, giving an answer of <math>\textbf{(A) }\tfrac16</math>. | ||
Revision as of 20:16, 19 June 2017
Problem
What is the volume of the region in three-dimensional space defined by the inequalities 
and 
Solution 1 (Non Calculus)
The first inequality refers to the interior of a regular octahedron with top and bottom vertices 
. Its volume is 
. The second inequality describes an identical shape, shifted 
upwards along the 
axis. The intersection will be a similar octahedron, linearly scaled down by half. Thus the volume of the intersection is one-eighth of the volume of the first octahedron, giving an answer of 
.
Solution 2 (Calculus)
Let 
, then we can transform the two inequalities to 
and 
. Then it's clear that 
, consider 
, 
, then since the area of 
is 
, the volume is 
. By symmetry, the case when 
is the same. Thus the answer is 
.
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
