2016 AMC 12B Problems/Problem 23
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 .
Solution 3
Do this problem first on the first quadrant. Graph this by using test points and you will see that you will have two tetrahedrons - each of them intersects at the middle, and thus its th of the area of the tetrahedrons (since they are the same area). The area of a tetrahedron is bh/3, which gives us , and then divide that by to get the bounded area of in the shaded region. Scale this up to the other quadrants now (since they are the same due to the abs value) and you get .
Sol by IronicNinja~
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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