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

Revision as of 16:29, 23 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 $\left\lfloor\frac{81}{13}\right\rfloor \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq \left\lfloor\frac{81}{12}\right\rfloor,$ or $6 \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq 6.$ Clearly, we get $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.$

~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 $6\frac{39}{88} < \frac{81}{4\pi} < 6\frac34,$ 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

@@ Line 14: / Line 14: @@
 ~NH14 ~MRENTHUSIASM
-== Solution 2 (Arithmetic) ==
+== Solution 2 (Inequality) ==
-A sphere with radius <math>2</math> has volume <math>\frac {32\pi}{3}</math>. A cube with side length <math>6</math> has volume <math>216</math>. If <math>\pi</math> was <math>3</math>, it would fit 6.75 times inside. Since <math>\pi</math> is approximately <math>5</math>% larger than <math>3</math>, it is safe to assume that the <math>3</math> balls of clay can fit <math>6</math> times inside. Therefore, our answer is <math>\boxed {(D)6}</math>.
-~Arcticturn
+As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is <math>\left\lfloor\frac{81}{4\pi}\right\rfloor.</math>
+By an underestimation <math>\pi\approx3,</math> we have <math>4\pi>12,</math> or <math>\frac{81}{4\pi}<6\frac34.</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> from which <math>\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.</math>
+~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 <math>\left\lfloor\frac{81}{4\pi}\right\rfloor.</math>
+Approximating with <math>\pi\approx3,</math> we have <math>\frac{81}{4\pi}\approx6\frac34.</math> Since <math>\pi</math> is about <math>5\%</math> greater than <math>3,</math> it is safe to claim that <math>\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.</math>
+~Arcticturn ~MRENTHUSIASM
 ==See Also==
 {{AMC10 box|year=2021 Fall|ab=A|num-b=2|num-a=4}}
 {{MAA Notice}}

2021 Fall AMC 10A (Problems • Answer Key • Resources)
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