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Difference between revisions of "2020 AMC 10B Problems/Problem 20"

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==Problem==
 
==Problem==
  
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Let <math>B</math> be a right rectangular prism (box) with edges lengths <math>1,</math> <math>3,</math> and <math>4</math>, together with its interior. For real <math>r\geq0</math>, let <math>S(r)</math> be the set of points in <math>3</math>-dimensional space that lie within a distance <math>r</math> of some point <math>B</math>. The volume of <math>S(r)</math> can be expressed as <math>ar^{3} + br^{2} + cr +d</math>, where <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> are positive real numbers. What is <math>\frac{bc}{ad}?</math>
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Let <math>B</math> be a right rectangular prism (box) with edges lengths <math>1,</math> <math>3,</math> and <math>4</math>, together with its interior. For real <math>r\geq0</math>, let <math>S(r)</math> be the set of points in <math>3</math>-dimensional space that lie within a distance <math>r</math> of some point in <math>B</math>. The volume of <math>S(r)</math> can be expressed as <math>ar^{3} + br^{2} + cr +d</math>, where <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> are positive real numbers. What is <math>\frac{bc}{ad}?</math>
  
 
<math>\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38</math>
 
<math>\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38</math>
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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 1: The volume of <math>B</math> is 12, so <math>d=12</math>
 
  
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Region 2: The volume is equal to the surface area of <math>B</math> times <math>r</math>. The surface area can easily be computed to be <math>2(4*3 + 3*1 + 4*1) = 38</math>, so <math>c=38</math>.
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Region 1: The volume of <math>B</math> is <math>1 \cdot 3 \cdot 4 = 12</math>, so <math>d=12</math>.
  
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Region 3: The volume of each quarter cylinder is equal to <math>(\pi*r^2*h)/4</math>. The sum of all such cylinders must equal <math>(\pi*r^2)/4</math> times the sum of the edge lengths. This can easily be computed as <math>4(4+3+1) = 32</math>, so the sum of the volumes of the quarter cylinders is <math>8\pi*r^2</math>, so <math>b=8\pi</math>
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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>.
  
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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*r^3</math>, so <math>a=\frac{4\pi}{3}</math>.
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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>.
 +
 
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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).  
  
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Using these values, <math>\frac{(8\pi)(38)}{(4\pi/3)(12)} = \boxed{\textbf{(B) }19}</math>
 
  
 
~DrJoyo
 
~DrJoyo
  
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==Video Solution==
+
~Edits by BakedPotato66
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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
  
−
~IceMatrix
+
===Video Solution===
 +
https://www.youtube.com/watch?v=NAZTdSecBvs
  
 
==See Also==
 
==See Also==

Latest revision as of 13:53, 8 June 2023

Problem

Let $B$ be a right rectangular prism (box) with edges lengths $1,$ $3,$ and $4$, together with its interior. For real $r\geq0$, let $S(r)$ be the set of points in $3$-dimensional space that lie within a distance $r$ of some point in $B$. The volume of $S(r)$ can be expressed as $ar^{3} + br^{2} + cr +d$, where $a,$ $b,$ $c,$ and $d$ are positive real numbers. What is $\frac{bc}{ad}?$

$\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38$

Solution

Split $S(r)$ 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 $1 \cdot 3 \cdot 4 = 12$, so $d=12$.

Region 2: This volume is equal to the surface area of $B$ times $r$ (these "extensions" are just more boxes). The volume is then $\text{SA} \cdot r=2(1 \cdot 3 + 1 \cdot 4 + 3 \cdot 4)r=38r$ to get $c=38$.

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 $4 \cdot \dfrac{1\pi r^2}{4} + 4 \cdot \dfrac{3\pi r^2}{4} + 4 \cdot \dfrac{4 \pi r^2}{4}=8 \pi r^2$. Therefore, $b=8\pi$.

Region 4: There is an eighth-sphere of radius $r$ at each corner of $B$. Since there are 8 corners, these eighth-spheres add up to 1 full sphere of radius $r$. The volume of this sphere is then $\frac{4}{3}\pi \cdot r^3$, so $a=\frac{4\pi}{3}$.

Using these values, $\dfrac{bc}{ad}=\frac{(8\pi)(38)}{(4\pi/3)(12)} = \boxed{\textbf{(B) }19}$.

To see a diagram of $S(r)$, view TheBeautyofMath's explanation video (Video Solution 1).


~DrJoyo

~Edits by BakedPotato66

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

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

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