Difference between revisions of "2016 AMC 12B Problems/Problem 23"
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Let <math>z\rightarrow z-1/2</math>, then we can transform the two inequalities to <math>|x|+|y|+|z-1/2|\le1</math> and <math>|x|+|y|+|z+1/2|\le1</math>. Then it's clear that <math>-1/2\le z \le 1/2</math>, consider <math>0 \le z \le 1/2</math>, <math>|x|+|y|\le 1/2-z</math>, then since the area of <math>|x|+|y|\le k</math> is <math>2k^2</math>, the volume is <math>\int_{0}^{1/2}2k^2 \,dk=\frac{1}{12}</math>. By symmetry, the case when <math>\frac{-1}{2}\le z\le0</math> is the same. Thus the answer is <math>\frac{1}{6}</math>. | Let <math>z\rightarrow z-1/2</math>, then we can transform the two inequalities to <math>|x|+|y|+|z-1/2|\le1</math> and <math>|x|+|y|+|z+1/2|\le1</math>. Then it's clear that <math>-1/2\le z \le 1/2</math>, consider <math>0 \le z \le 1/2</math>, <math>|x|+|y|\le 1/2-z</math>, then since the area of <math>|x|+|y|\le k</math> is <math>2k^2</math>, the volume is <math>\int_{0}^{1/2}2k^2 \,dk=\frac{1}{12}</math>. By symmetry, the case when <math>\frac{-1}{2}\le z\le0</math> is the same. Thus the answer is <math>\frac{1}{6}</math>. | ||
| + | ==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 <math>1/4</math>th of the area of the tetrahedrons (since they are the same area). The area of a tetrahedron is bh/3, which gives us <math>1/6</math>, and then divide that by <math>4</math> to get the bounded area of <math>1/24</math> 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 <math>\textbf{(A) }\tfrac16</math>. | ||
| + | |||
| + | Sol by IronicNinja~ | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2016|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:36, 2 January 2019
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 |
| 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. 
