Difference between revisions of "2021 AMC 12A Problems/Problem 10"

Revision as of 18:24, 11 February 2021

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 $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ 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?

$[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("$3$",(-0.9,h1*s),N,fontsize(10)); real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1); draw(ellipse((d,0),r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]$

$\textbf{(A) }1 \qquad \textbf{(B) }\frac{47}{43} \qquad \textbf{(C) }2 \qquad \textbf{(D) }\frac{40}{13} \qquad \textbf{(E) }4$

Solution

I will be referring to the areas filled with liquid as the cones.

The area of a cone is $\frac13\pi r^2h$ , where $r$ is the radius of the cone and $h$ 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 $\frac{4}{1} = 4$ . After marbles are dropped, the volumes are still equal, so the ratio of the heights is still $4$ . Therefore, the ratio of the liquid rise is $\boxed{\textbf{(E) }4}$ .

Video Solution by Hawk Math

https://www.youtube.com/watch?v=AjQARBvdZ20

Video Solution by pi_is_3.14 (Similar Triangles, 3D Geometry - Cones)

https://youtu.be/4Iuo7cvGJr8

~ pi_is_3.14

@@ Line 43: / Line 43: @@
 https://www.youtube.com/watch?v=AjQARBvdZ20
-== Video Solution by pi_is_3.14(Similar Triangles, 3D Geometry - Cones) ==
+== Video Solution by pi_is_3.14 (Similar Triangles, 3D Geometry - Cones) ==
 https://youtu.be/4Iuo7cvGJr8

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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