Difference between revisions of "2020 AMC 10B Problems/Problem 20"
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4. The one-eighth spheres at each corner of <math>B</math> | 4. The one-eighth spheres at each corner of <math>B</math> | ||
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− | Region | + | Region 1: The volume of <math>B</math> is <math>1 \cdot 3 \cdot 4 = 12</math>, so <math>d=12</math>. |
− | Region | + | Region 2: This volume is equal to the surface area of <math>B</math> times <math>r</math> (these "extensions" are just more boxes). The volume is then <math>\text{SA} \cdot r=2(1 \cdot 3 + 1 \cdot 4 + 3 \cdot 4)r=38r</math> to get <math>c=38</math>. |
− | Region 4: There is an eighth | + | Region 3: We see that there are 12 quarter-cylinders, 4 of each type. We have 4 quarter-cylinders of height 1, 4 quarter-cylinders of height 3, 4 quarter-cylinders of height 4. Since 4 quarter-cylinders make a full cylinder, the total volume is <math>4 \cdot \dfrac{1\pi r^2}{4} + 4 \cdot \dfrac{3\pi r^2}{4} + 4 \cdot \dfrac{4 \pi r^2}{4}=8 \pi r^2</math>. Therefore, <math>b=8\pi</math>. |
+ | |||
+ | Region 4: There is an eighth-sphere of radius <math>r</math> at each corner of <math>B</math>. Since there are 8 corners, these eighth-spheres add up to 1 full sphere of radius <math>r</math>. The volume of this sphere is then <math>\frac{4}{3}\pi \cdot r^3</math>, so <math>a=\frac{4\pi}{3}</math>. | ||
+ | |||
+ | Using these values, <math>\dfrac{bc}{ad}=\frac{(8\pi)(38)}{(4\pi/3)(12)} = \boxed{\textbf{(B) }19}</math>. | ||
+ | |||
+ | To see a diagram of <math>S(r)</math>, view TheBeautyofMath's explanation video (Video Solution 1). | ||
− | |||
~DrJoyo | ~DrJoyo | ||
− | ==Video Solution== | + | ~Edits by BakedPotato66 |
− | https://youtu.be/3BvJeZU3T-M | + | |
+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/SIwp-70zu7o | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | ===Video Solution=== | ||
+ | https://youtu.be/3BvJeZU3T-M?t=1351 | ||
− | + | ===Video Solution=== | |
+ | https://www.youtube.com/watch?v=NAZTdSecBvs | ||
==See Also== | ==See Also== |
Latest revision as of 14:53, 8 June 2023
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 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
, so
.
Region 2: This volume is equal to the surface area of times
(these "extensions" are just more boxes). The volume is then
to get
.
Region 3: We see that there are 12 quarter-cylinders, 4 of each type. We have 4 quarter-cylinders of height 1, 4 quarter-cylinders of height 3, 4 quarter-cylinders of height 4. Since 4 quarter-cylinders make a full cylinder, the total volume is . Therefore,
.
Region 4: There is an eighth-sphere of radius at each corner of
. Since there are 8 corners, these eighth-spheres add up to 1 full sphere of radius
. The volume of this sphere is then
, so
.
Using these values, .
To see a diagram of , view TheBeautyofMath's explanation video (Video Solution 1).
~DrJoyo
~Edits by BakedPotato66
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/3BvJeZU3T-M?t=1351
Video Solution
https://www.youtube.com/watch?v=NAZTdSecBvs
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.