Difference between revisions of "2021 Fall AMC 10A Problems/Problem 3"
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~NH14 ~MRENTHUSIASM | ~NH14 ~MRENTHUSIASM | ||
− | == Solution 2 ( | + | == Solution 2 (Inequality) == |
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− | ~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> | ||
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+ | ~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== | ==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}} |
Revision as of 16:29, 23 November 2021
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 Clearly, we get
~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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.