Difference between revisions of "2020 AMC 10B Problems/Problem 20"
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Region 1: The volume of <math>B</math> is 12, so <math>d=12</math> | Region 1: The volume of <math>B</math> is 12, so <math>d=12</math> | ||
− | Region 2: The volume is equal to the surface area of <math>B</math> times <math>r</math>. The surface area can be computed to be <math>2(4 | + | Region 2: The volume is equal to the surface area of <math>B</math> times <math>r</math>. The surface area can be computed to be <math>2(4\cdot3 + 3\cdot1 + 4\cdot1) = 38</math>, so <math>c=38</math>. |
− | Region 3: The volume of each quarter cylinder is equal to <math>(\pi | + | Region 3: The volume of each quarter cylinder is equal to <math>(\pi\cdotr^2\cdoth)/4</math>. The sum of all such cylinders must equal <math>(\pi\cdotr^2)/4</math> times the sum of the edge lengths. This can be computed as <math>4(4+3+1) = 32</math>, so the sum of the volumes of the quarter cylinders is <math>8\pi\cdotr^2</math>, so <math>b=8\pi</math> |
− | Region 4: There is an eighth of a sphere of radius <math>r</math> at each corner. Since there are 8 corners, these add up to one full sphere of radius <math>r</math>. The volume of this sphere is <math>\frac{4}{3}\pi | + | Region 4: There is an eighth of a sphere of radius <math>r</math> at each corner. Since there are 8 corners, these add up to one full sphere of radius <math>r</math>. The volume of this sphere is <math>\frac{4}{3}\pi\cdotr^3</math>, so <math>a=\frac{4\pi}{3}</math>. |
Using these values, <math>\frac{(8\pi)(38)}{(4\pi/3)(12)} = \boxed{\textbf{(B) }19}</math> | Using these values, <math>\frac{(8\pi)(38)}{(4\pi/3)(12)} = \boxed{\textbf{(B) }19}</math> |
Revision as of 12:52, 30 January 2021
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 in . The volume of can be expressed as , where and are positive real numbers. What is
Solution
Split into 4 regions:
1. The rectangular prism itself
2. The extensions of the faces of
3. The quarter cylinders at each edge of
4. The one-eighth spheres at each corner of
Region 1: The volume of is 12, so
Region 2: The volume is equal to the surface area of times . The surface area can be computed to be , so .
Region 3: The volume of each quarter cylinder is equal to $(\pi\cdotr^2\cdoth)/4$ (Error compiling LaTeX. Unknown error_msg). The sum of all such cylinders must equal $(\pi\cdotr^2)/4$ (Error compiling LaTeX. Unknown error_msg) times the sum of the edge lengths. This can be computed as , so the sum of the volumes of the quarter cylinders is $8\pi\cdotr^2$ (Error compiling LaTeX. Unknown error_msg), so
Region 4: There is an eighth of a sphere of radius at each corner. Since there are 8 corners, these add up to one full sphere of radius . The volume of this sphere is $\frac{4}{3}\pi\cdotr^3$ (Error compiling LaTeX. Unknown error_msg), so .
Using these values,
~DrJoyo
Video Solution 1
~IceMatrix
Video Solution 2
https://www.youtube.com/watch?v=NAZTdSecBvs ~ MathEx
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.