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

(→‎Video Solution by Aaron He)
(→‎Solution 1 (Fraction Trick):)
Line 39: Line 39:
  
 
==Solution 1 (Fraction Trick):==
 
==Solution 1 (Fraction Trick):==
−
<b>Initially:</b>
+
<b>Initial Condition</b>
  
−
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>
+
<cmath>\begin{array}{cccc}
 +
& \text{Base} & \text{Height} & \text{Volume} \\
 +
\text{Narrow Cone} & 3 & h_1 & \frac13\pi(3)^2h_1 \\
 +
\text{Wide Cone} & 6 & h_2 & \frac13\pi(6)^2h_2
 +
\end{array}</cmath>
 +
By similar triangles:
  
−
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, the ratio of base radius to height is <math>\frac{3}{h_1},</math> which remains constant.
  
−
Equating initial volumes gives <math>3\pi h_1=12\pi h_2,</math> from which <math>\frac{h_1}{h_2}=4.</math>
+
For the wide cone, the ratio of base radius to height is <math>\frac{6}{h_2},</math> which remains constant.
  
−
<b>Finally:</b>
+
Equating the initial volumes gives <math>\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2,</math> which simplifies to <math>\frac{h_1}{h_2}=4.</math>
  
−
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>
 
  
−
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>
+
<b>Final Condition</b>
  
−
Equating final volumes simplifies to <math>x^3=y^3,</math> or <math>x=y.</math>
+
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:
 +
<cmath>\begin{array}{cccc}
 +
& \text{Base} & \text{Height} & \text{Volume} \\
 +
\text{Narrow Cone} & 3x & h_1x & \frac13\pi(3x)^2h_1 \\
 +
\text{Wide Cone} & 6y & h_2y & \frac13\pi(6y)^2h_2
 +
\end{array}</cmath>
 +
 
 +
Equating final volumes gives <math>\frac13\pi(3x)^2h_1=\frac13\pi(6y)^2h_2,</math> which simplifies to <math>x^3=y^3,</math> or <math>x=y.</math>
  
 
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>
 
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>
 +
  
 
<b>PS:</b>
 
<b>PS:</b>
Line 67: Line 79:
 
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>
 
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 information.
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 15:21, 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.

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 1 (Fraction Trick):

Initial Condition

\[\begin{array}{cccc} & \text{Base} & \text{Height} & \text{Volume} \\ \text{Narrow Cone} & 3 & h_1 & \frac13\pi(3)^2h_1 \\ \text{Wide Cone} & 6 & h_2 & \frac13\pi(6)^2h_2 \end{array}\] By similar triangles:

For the narrow cone, the ratio of base radius to height is $\frac{3}{h_1},$ which remains constant.

For the wide cone, the ratio of base radius to height is $\frac{6}{h_2},$ which remains constant.

Equating the initial volumes gives $\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2,$ which simplifies to $\frac{h_1}{h_2}=4.$


Final Condition

Let the base radii of the narrow cone and the wide cone be $3x$ and $6y,$ respectively. We have the following table: \[\begin{array}{cccc} & \text{Base} & \text{Height} & \text{Volume} \\ \text{Narrow Cone} & 3x & h_1x & \frac13\pi(3x)^2h_1 \\ \text{Wide Cone} & 6y & h_2y & \frac13\pi(6y)^2h_2 \end{array}\]

Equating final volumes gives $\frac13\pi(3x)^2h_1=\frac13\pi(6y)^2h_2,$ which 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 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 $\boxed{\textbf{(E) } 4}$ 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)

https://youtu.be/4Iuo7cvGJr8

~ 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

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