Difference between revisions of "2021 Fall AMC 10A Problems/Problem 3"
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<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7</math> | <math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7</math> | ||
− | == Solution == | + | == Solution 1 (Inequality) == |
− | |||
− | ~Arcticturn | + | The volume of the cube is <math>V_{\text{cube}}=6^3=216,</math> and the volume of a clay ball is <math>V_{\text{ball}}=\frac43\cdot\pi\cdot2^3=\frac{32}{3}\pi.</math> |
+ | |||
+ | Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is <cmath>\left\lfloor\frac{V_{\text{cube}}}{V_{\text{ball}}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.</cmath> | ||
+ | Approximating with <math>\pi\approx3.14,</math> we have <math>12<4\pi<13,</math> or <math>\left\lfloor\frac{81}{13}\right\rfloor \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq \left\lfloor\frac{81}{12}\right\rfloor.</math> We simplify to get <cmath>6 \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq 6,</cmath> | ||
+ | from which <math>\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.</math> | ||
+ | |||
+ | ~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 <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 < 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> | ||
+ | |||
+ | ~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 | ||
+ | |||
+ | ==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== | ==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
Contents
Problem
What is the maximum number of balls of clay of radius that can completely fit inside a cube of side length assuming the balls can be reshaped but not compressed before they are packed in the cube?
Solution 1 (Inequality)
The volume of the cube is and the volume of a clay ball is
Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is Approximating with we have or We simplify to get from which
~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
By an underestimation we have or
By an overestimation we have or
Together, we get from which
~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
Approximating with we have Since is about greater than it is safe to claim that
~Arcticturn ~MRENTHUSIASM
Video Solution
~savannahsolver
Video Solution
~Charles3829
Video Solution by TheBeautyofMath
https://youtu.be/o98vGHAUYjM?t=158
~IceMatrix
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
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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