2020 AMC 10B Problems/Problem 20

Revision as of 13:42, 8 February 2020 by Drjoyo (talk | contribs) (Solution)

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 $B$. 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 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 $d=12$

Region 2: The volume is equal to the surface area of B times r. The surface area can easily 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 easily 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 $a=\frac{4}{3}$.

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

~DrJoyo

Video Solution

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

~IceMatrix

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
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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