Difference between revisions of "2021 AMC 12A Problems/Problem 10"
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==Solution 1 (Use Tables to Organize Information)== | ==Solution 1 (Use Tables to Organize Information)== |
Revision as of 20:35, 3 March 2021
- The following problem is from both the 2021 AMC 10A #12 and 2021 AMC 12A #10, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Use Tables to Organize Information)
- 3 Solution 2 (Quick and dirty)
- 4 Video Solution by Aaron He (Algebra)
- 5 Video Solution by OmegaLearn (Similar Triangles, 3D Geometry - Cones)
- 6 Video Solution (Simple and Quick)
- 7 Video Solution by TheBeautyofMath
- 8 Video Solution by WhyMath
- 9 See also
Problem
Two right circular cones with vertices facing down as shown in the figure below contains the same amount of liquid. The radii of the tops of the liquid surfaces are cm and cm. Into each cone is dropped a spherical marble of radius cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
Solution 1 (Use Tables to Organize Information)
Initial Scenario
By similar triangles:
For the narrow cone, the ratio of base radius to height is which remains constant.
For the wide cone, the ratio of base radius to height is which remains constant.
Equating the initial volumes gives which simplifies to
Final Scenario (Two solutions follow from here.)
Solution 1.1 (Fraction Trick)
Let the base radii of the narrow cone and the wide cone be and respectively, where We have the following table:
Equating the final volumes gives which simplifies to or
Lastly, the requested ratio is
PS:
1. This problem uses the following fraction trick:
For unequal positive numbers and if then
Quick Proof
From we know that and . Therefore,
2. The work above shows that, regardless of the shape or the volume of the solid dropped in, as long as the solid sinks to the bottom and is completely submerged without spilling any liquid, the answer will remain unchanged.
~MRENTHUSIASM
Solution 1.2 (Bash)
Let the base radii of the narrow cone and the wide cone be and respectively.
Let the rises of the liquid levels of the narrow cone and the wide cone be and respectively. We have the following table:
By similar triangles discussed above, we have
The volume of the marble dropped in is
Now, we set up an equation for the volume of the narrow cone and solve for
Next, we set up an equation for the volume of the wide cone Using the exact same process from above (but with different numbers), we get Recall that Therefore, the requested ratio is
~MRENTHUSIASM
Solution 2 (Quick and dirty)
The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii 3 and 6 and infinitely large height. Then the base area of the wide cylinder is 4 times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise times as much.
-scrabbler94
Video Solution by Aaron He (Algebra)
https://www.youtube.com/watch?v=xTGDKBthWsw&t=10m20s
Video Solution by OmegaLearn (Similar Triangles, 3D Geometry - Cones)
~ pi_is_3.14
Video Solution (Simple and Quick)
~ Education, the Study of Everything
Video Solution by TheBeautyofMath
First-this is not the most efficient solution. I did not perceive the shortcut before filming though I suspected it.
https://youtu.be/t-EEP2V4nAE?t=231 (for AMC 10A)
https://youtu.be/cckGBU2x1zg?t=814 (for AMC 12A)
~IceMatrix
Video Solution by WhyMath
~savannahsolver
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.