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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>
  
−
Together, we get <cmath>6\frac{39}{88} < \frac{81}{4\pi} < 6\frac34,</cmath>
+
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>
  

Revision as of 14:57, 28 November 2021

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

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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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