Difference between revisions of "2021 AMC 12A Problems/Problem 10"
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− | Let the base radii of the narrow cone and the wide cone be <math>3x</math> and <math>6y,</math> respectively. We have the following table: | + | Let the base radii of the narrow cone and the wide cone be <math>3x</math> and <math>6y,</math> respectively, where <math>x,y>1.</math> We have the following table: |
<cmath>\begin{array}{cccc} | <cmath>\begin{array}{cccc} | ||
& \text{Base} & \text{Height} & \text{Volume} \\ | & \text{Base} & \text{Height} & \text{Volume} \\ |
Revision as of 21:18, 13 February 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
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 (Fraction Trick):
Initial Condition
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 Condition
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 fraction we seek 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. Most of the steps can be done through mental math. Also, drawing a table showing the initial and final measurements can effectively organize the information.
~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
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.