Difference between revisions of "2021 Fall AMC 10A Problems/Problem 3"

Revision as of 18:37, 7 April 2022

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 by TheBeautyofMath

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

~IceMatrix

@@ Line 39: / Line 39: @@
 ~savannahsolver
+==Video Solution by TheBeautyofMath==
+https://youtu.be/o98vGHAUYjM?t=158
+~IceMatrix
 ==See Also==
 {{AMC10 box|year=2021 Fall|ab=A|num-b=2|num-a=4}}
 {{MAA Notice}}

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