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

## 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 (Fraction Trick)

Initially:

For the narrow cone liquid, the base radius is $3.$ Let its height be $h_1.$ By similar triangles, the ratio of base radius to height is $\frac{3}{h_1}.$ The volume is $\frac13\pi(3)^2h_1=3\pi h_1.$

For the wide cone liquid, the base radius is $6.$ Let its height be $h_2.$ By similar triangles, the ratio of base radius to height is $\frac{6}{h_2}.$ The volume is $\frac13\pi(6)^2h_2=12\pi h_2.$

Equating initial volumes gives $3\pi h_1=12\pi h_2,$ from which $\frac{h_1}{h_2}=4.$

Finally:

For the narrow cone liquid, the base radius is $3x,$ where $x>1.$ By similar triangles, it follows that its height is $h_1x$ and its volume is $3\pi h_1 x^3.$

For the wide cone liquid, the base radius is $6y,$ where $y>1.$ By similar triangles, it follows that its height is $h_2y$ and its volume is $12\pi h_2 y^3.$

Equating final volumes simplifies to $x^3=y^3,$ or $x=y.$

Lastly, the fraction we seek is $$\frac{h_1 x - h_1}{h_2 y - h_2}=\frac{h_1 (x-1)}{h_2 (y-1)}=\frac{h_1}{h_2}=\boxed{\textbf{(E) }4}.$$

PS:

1. This problem uses the following fraction trick:

For unequal positive numbers $a,b,c$ and $d,$ if $\frac ab = \frac cd = k,$ then $\frac{a\pm c}{b\pm d}=k.$

Quick Proof

From $\frac ab = \frac cd = k,$ we know that $a=bk$ and $c=dk$. Therefore, $$\frac{a\pm c}{b\pm d}=\frac{bk\pm dk}{b\pm d}=\frac{(b\pm d)k}{b\pm d}=k.$$

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 work.

~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 $\boxed{\textbf{(E) } 4}$ times as much.

-scrabbler94

~ pi_is_3.14