Difference between revisions of "2020 AMC 10B Problems/Problem 20"
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Region 4: There is an eighth of a sphere of radius r at each corner. Since there are 8 corners, these add up to one full sphere of radius r. The volume of this sphere is <math>\frac{4}{3}\pi*r^3</math>. <math>a=\frac{4}{3}</math>. | Region 4: There is an eighth of a sphere of radius r at each corner. Since there are 8 corners, these add up to one full sphere of radius r. The volume of this sphere is <math>\frac{4}{3}\pi*r^3</math>. <math>a=\frac{4}{3}</math>. | ||
| − | Using these values, <math>\frac{(8\pi)(38)}{(4\pi/3)(12)} = 19</math> | + | Using these values, <math>\frac{(8\pi)(38)}{(4\pi/3)(12)} = 19</math> (D) |
~DrJoyo | ~DrJoyo | ||
Revision as of 03:05, 8 February 2020
Contents
Problem
Let 
be a right rectangular prism (box) with edges lengths 
and 
, together with its interior. For real 
, let 
be the set of points in 
-dimensional space that lie within a distance 
of some point . The volume of can be expressed as 
, where 
and 
are positive real numbers. What is 
Solution
Split the volume into 4 regions:
1. The rectangular prism itself
2. The extensions of the faces of B
3. The quarter cylinders at each edge of B
4. The one-eighth spheres at each corner of B.
Region 1: The volume of B is 12, so 
Region 2: The volume is equal to the surface area of B times r. The surface area can easily be computed to be 38, so 
.
Region 3: The volume of each quarter cylinder is equal to 
. The sum of all such cylinders must equal 
times the sum of the edge lengths. This can easily be computed as 32, so the sum of the volumes of the quarter cylinders is 
. 
Region 4: There is an eighth of a sphere of radius r at each corner. Since there are 8 corners, these add up to one full sphere of radius r. The volume of this sphere is 
. 
.
Using these values, 
(D)
~DrJoyo
Video Solution
~IceMatrix
See Also
| 2020 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
