Difference between revisions of "2021 AMC 12A Problems/Problem 10"
Pi is 3.14 (talk | contribs) (→Video Solution by pi_is_3.14(Similar Triangles, 3D Geometry - Cones)) |
Pi is 3.14 (talk | contribs) (→Video Solution by pi_is_3.14 (Similar Triangles, 3D Geometry - Cones)) |
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− | == Video Solution by | + | == Video Solution by OmegaLearn (Similar Triangles, 3D Geometry - Cones) == |
https://youtu.be/4Iuo7cvGJr8 | https://youtu.be/4Iuo7cvGJr8 | ||
Revision as of 23:18, 11 February 2021
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
I will be referring to the areas filled with liquid as the cones.
The area of a cone is , where
is the radius of the cone and
is the height. Since the first cone has half the radius of the second cone, and both cones have the same volume, the ratio of the height of the first cone to the height of the second cone is
. After marbles are dropped, the volumes are still equal, so the ratio of the heights is still
. Therefore, the ratio of the liquid rise is
.
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
Video Solution by OmegaLearn (Similar Triangles, 3D Geometry - Cones)
~ pi_is_3.14
See also
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 |
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