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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 3"
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By an overestimation <math>\pi\approx\frac{22}{7},</math> we have <math>4\pi<\frac{88}{7},</math> or <math>\frac{81}{4\pi}>6\frac{39}{88}.</math>
 
By an overestimation <math>\pi\approx\frac{22}{7},</math> we have <math>4\pi<\frac{88}{7},</math> or <math>\frac{81}{4\pi}>6\frac{39}{88}.</math>
  
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Together, we get <cmath>6\frac{39}{88} < \frac{81}{4\pi} < 6\frac34,</cmath>
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Together, we get <cmath>6 < 6\frac{39}{88} < \frac{81}{4\pi} < 6\frac34 < 7,</cmath>
 
from which <math>\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.</math>
 
from which <math>\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.</math>
  
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~Arcticturn ~MRENTHUSIASM
 
~Arcticturn ~MRENTHUSIASM
  
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== Solution 4 ==
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==Video Solution==
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We have
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https://youtu.be/0WrcZI_z9IE
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<cmath>
 
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\begin{align*}
 
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\left\lfloor \frac{6^3}{\frac{4}{3} \pi 2^3} \right\rfloor
 
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& = \left\lfloor \frac{81}{4 \pi } \right\rfloor \\
 
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& = 6 .
 
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\end{align*}
 
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</cmath>
 
  
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Therefore, the answer is \boxed{\textbf{(D) }6}.
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~savannahsolver
  
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~Steven Chen (www.professorchenedu.com)
+
==Video Solution==
 +
https://youtu.be/qOiO45IsE54
 +
 
 +
~Charles3829
 +
 
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/o98vGHAUYjM?t=158
 +
 
 +
~IceMatrix
 +
 
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/VcGR0GVqKBI
 +
 
 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=2|num-a=4}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:14, 19 October 2023

Problem

What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution 1 (Inequality)

The volume of the cube is $V_{\text{cube}}=6^3=216,$ and the volume of a clay ball is $V_{\text{ball}}=\frac43\cdot\pi\cdot2^3=\frac{32}{3}\pi.$

Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is \[\left\lfloor\frac{V_{\text{cube}}}{V_{\text{ball}}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.\] Approximating with $\pi\approx3.14,$ we have $12<4\pi<13,$ or $\left\lfloor\frac{81}{13}\right\rfloor \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq \left\lfloor\frac{81}{12}\right\rfloor.$ We simplify to get \[6 \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq 6,\] from which $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.$

~NH14 ~MRENTHUSIASM

Solution 2 (Inequality)

As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is $\left\lfloor\frac{81}{4\pi}\right\rfloor.$

By an underestimation $\pi\approx3,$ we have $4\pi>12,$ or $\frac{81}{4\pi}<6\frac34.$

By an overestimation $\pi\approx\frac{22}{7},$ we have $4\pi<\frac{88}{7},$ or $\frac{81}{4\pi}>6\frac{39}{88}.$

Together, we get \[6 < 6\frac{39}{88} < \frac{81}{4\pi} < 6\frac34 < 7,\] from which $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.$

~MRENTHUSIASM

Solution 3 (Approximation)

As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is $\left\lfloor\frac{81}{4\pi}\right\rfloor.$

Approximating with $\pi\approx3,$ we have $\frac{81}{4\pi}\approx6\frac34.$ Since $\pi$ is about $5\%$ greater than $3,$ it is safe to claim that $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.$

~Arcticturn ~MRENTHUSIASM

Video Solution

https://youtu.be/0WrcZI_z9IE

~savannahsolver

Video Solution

https://youtu.be/qOiO45IsE54

~Charles3829

Video Solution by TheBeautyofMath

https://youtu.be/o98vGHAUYjM?t=158

~IceMatrix

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/VcGR0GVqKBI

~Education, the Study of Everything

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
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Problem 4
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All AMC 10 Problems and Solutions

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