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

(→‎Solution 2 (Fraction Trick))
(→‎Solution 2 (Fraction Trick))
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==Solution 2 (Fraction Trick)==
 
==Solution 2 (Fraction Trick)==
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Initially, for the narrow cone liquid, the base radius is <math>3.</math> Let its height be <math>h_1.</math> By similar triangles, the ratio of base radius to height is <math>\frac{3}{h_1}.</math> The volume is <math>\frac13\pi(3)^2h_1=3\pi h_1.</math>
+
<b>Initially:</b>
  
−
Initially, for the wide cone liquid, the base radius is <math>6.</math> Let its height be <math>h_2.</math> By similar triangles, the ratio of base radius to height is <math>\frac{6}{h_2}.</math> The volume is <math>\frac13\pi(6)^2h_2=12\pi h_2.</math>
+
For the narrow cone liquid, the base radius is <math>3.</math> Let its height be <math>h_1.</math> By similar triangles, the ratio of base radius to height is <math>\frac{3}{h_1}.</math> The volume is <math>\frac13\pi(3)^2h_1=3\pi h_1.</math>
 +
 
 +
For the wide cone liquid, the base radius is <math>6.</math> Let its height be <math>h_2.</math> By similar triangles, the ratio of base radius to height is <math>\frac{6}{h_2}.</math> The volume is <math>\frac13\pi(6)^2h_2=12\pi h_2.</math>
  
 
Equating initial volumes gives <math>3\pi h_1=12\pi h_2,</math> from which <math>\frac{h_1}{h_2}=4.</math>
 
Equating initial volumes gives <math>3\pi h_1=12\pi h_2,</math> from which <math>\frac{h_1}{h_2}=4.</math>
  
−
Finally, for the narrow cone liquid, the base radius is <math>3x,</math> where <math>x>1.</math> By similar triangles, it follows that its height is <math>h_1x</math> and its volume is <math>3\pi h_1 x^3.</math>
+
<b>Finally:</b>
 +
 
 +
For the narrow cone liquid, the base radius is <math>3x,</math> where <math>x>1.</math> By similar triangles, it follows that its height is <math>h_1x</math> and its volume is <math>3\pi h_1 x^3.</math>
  
−
Finally, for the wide cone liquid, the base radius is <math>6y,</math> where <math>y>1.</math> By similar triangles, it follows that its height is <math>h_2y</math> and its volume is <math>12\pi h_2 y^3.</math>
+
For the wide cone liquid, the base radius is <math>6y,</math> where <math>y>1.</math> By similar triangles, it follows that its height is <math>h_2y</math> and its volume is <math>12\pi h_2 y^3.</math>
  
 
Equating final volumes simplifies to <math>x^3=y^3,</math> or <math>x=y.</math>
 
Equating final volumes simplifies to <math>x^3=y^3,</math> or <math>x=y.</math>
  
−
Finally, the fraction we seek is <cmath>\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}.</cmath>
+
Lastly, the fraction we seek is <cmath>\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}.</cmath>
  
−
PS:
+
<b>PS:</b>
  
 
1. This problem uses the following fraction trick:
 
1. This problem uses the following fraction trick:
  
 
For unequal positive numbers <math>a,b,c</math> and <math>d,</math> if <math>\frac ab = \frac cd = k,</math> then <math>\frac{a\pm c}{b\pm d}=k.</math>
 
For unequal positive numbers <math>a,b,c</math> and <math>d,</math> if <math>\frac ab = \frac cd = k,</math> then <math>\frac{a\pm c}{b\pm d}=k.</math>
 +
 +
<u>Quick Proof</u>
 +
 +
From <math>\frac ab = \frac cd = k,</math> we know that <math>a=bk</math> and <math>c=dk</math>. Therefore, <cmath>\frac{a\pm c}{b\pm d}=\frac{bk\pm dk}{b\pm d}=\frac{(b\pm d)k}{b\pm d}=k.</cmath>
  
 
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.
 
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.

Revision as of 23:45, 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 $\boxed{\textbf{(E) }4}$.

Solution 2 (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

Video Solution by Hawk Math

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

Video Solution by OmegaLearn (Similar Triangles, 3D Geometry - Cones)

https://youtu.be/4Iuo7cvGJr8

~ 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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