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

Revision as of 12:53, 30 January 2021

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 12, so $d=12$

Region 2: The volume is equal to the surface area of $B$ times $r$ . The surface area can be computed to be $2(4*3 + 3*1 + 4*1) = 38$ , so $c=38$ .

Region 3: The volume of each quarter cylinder is equal to $(\pi*r^2*h)/4$ . The sum of all such cylinders must equal $(\pi*r^2)/4$ times the sum of the edge lengths. This can be computed as $4(4+3+1) = 32$ , so the sum of the volumes of the quarter cylinders is $8\pi*r^2$ , so $b=8\pi$

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 $\frac{4}{3}\pi*r^3$ , so $a=\frac{4\pi}{3}$ .

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

~DrJoyo

Video Solution 1

https://youtu.be/3BvJeZU3T-M

~IceMatrix

Video Solution 2

https://www.youtube.com/watch?v=NAZTdSecBvs ~ MathEx

@@ Line 19: / Line 19: @@
 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\cdot3 + 3\cdot1 + 4\cdot1) = 38</math>, so <math>c=38</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*3 + 3*1 + 4*1) = 38</math>, so <math>c=38</math>.
-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 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 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>
-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>.
+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>.
 Using these values, <math>\frac{(8\pi)(38)}{(4\pi/3)(12)} = \boxed{\textbf{(B) }19}</math>

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
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