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

m (→‎Solution 1.2 (Bash))
m (→‎Solution 1 (Use Tables to Organize Information))
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<cmath>\begin{array}{cccc}
 
<cmath>\begin{array}{cccc}
−
& \text{Base} & \text{Height} & \text{Volume} \\ [2ex]
+
& \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex]
−
\text{Narrow Cone} & 3 & h_1 & \frac13\pi(3)^2h_1=3\pi h_1 \\ [2ex]
+
\textbf{Narrow Cone} & 3 & h_1 & \frac13\pi(3)^2h_1=3\pi h_1 \\ [2ex]
−
\text{Wide Cone} & 6 & h_2 & \frac13\pi(6)^2h_2=12\pi h_2
+
\textbf{Wide Cone} & 6 & h_2 & \frac13\pi(6)^2h_2=12\pi h_2
 
\end{array}</cmath>
 
\end{array}</cmath>
 
By similar triangles:
 
By similar triangles:
Line 58: Line 58:
 
Let the base radii of the narrow cone and the wide cone be <math>3x</math> and <math>6y,</math> respectively, where <math>x,y>1.</math> We have the following table:
 
Let the base radii of the narrow cone and the wide cone be <math>3x</math> and <math>6y,</math> respectively, where <math>x,y>1.</math> We have the following table:
 
<cmath>\begin{array}{cccc}
 
<cmath>\begin{array}{cccc}
−
& \text{Base} & \text{Height} & \text{Volume} \\ [2ex]
+
& \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex]
−
\text{Narrow Cone} & 3x & h_1x & \frac13\pi(3x)^2h_1=3\pi h_1 x^3 \\ [2ex]
+
\textbf{Narrow Cone} & 3x & h_1x & \frac13\pi(3x)^2h_1=3\pi h_1 x^3 \\ [2ex]
−
\text{Wide Cone} & 6y & h_2y & \frac13\pi(6y)^2h_2=12\pi h_2 y^3
+
\textbf{Wide Cone} & 6y & h_2y & \frac13\pi(6y)^2h_2=12\pi h_2 y^3
 
\end{array}</cmath>
 
\end{array}</cmath>
  
Line 86: Line 86:
 
Let the rises of the liquid levels of the narrow cone and the wide cone be <math>\Delta h_1</math> and <math>\Delta h_2,</math> respectively. We have the following table:
 
Let the rises of the liquid levels of the narrow cone and the wide cone be <math>\Delta h_1</math> and <math>\Delta h_2,</math> respectively. We have the following table:
 
<cmath>\begin{array}{cccc}  
 
<cmath>\begin{array}{cccc}  
−
& \text{Base} & \text{Height} & \text{Volume} \\ [2ex]  
+
& \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex]  
−
\text{Narrow Cone} & r_1 & h_1+\Delta h_1 & \frac13\pi r_1^2(h_1+\Delta h_1) \\ [2ex]  
+
\textbf{Narrow Cone} & r_1 & h_1+\Delta h_1 & \frac13\pi r_1^2(h_1+\Delta h_1) \\ [2ex]  
−
\text{Wide Cone} & r_2 & h_2+\Delta h_2 & \frac13\pi r_2^2(h_2+\Delta h_2)
+
\textbf{Wide Cone} & r_2 & h_2+\Delta h_2 & \frac13\pi r_2^2(h_2+\Delta h_2)
 
\end{array}</cmath>
 
\end{array}</cmath>
  

Revision as of 16:31, 17 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 (Use Tables to Organize Information)

Initial Scenario

\[\begin{array}{cccc} & \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex] \textbf{Narrow Cone} & 3 & h_1 & \frac13\pi(3)^2h_1=3\pi h_1 \\ [2ex] \textbf{Wide Cone} & 6 & h_2 & \frac13\pi(6)^2h_2=12\pi h_2 \end{array}\] 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 $3\pi h_1=12\pi h_2,$ which simplifies to $\frac{h_1}{h_2}=4.$

Final Scenario (Two solutions follow from here.)

Solution 1.1 (Fraction Trick)

Let the base radii of the narrow cone and the wide cone be $3x$ and $6y,$ respectively, where $x,y>1.$ We have the following table: \[\begin{array}{cccc} & \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex] \textbf{Narrow Cone} & 3x & h_1x & \frac13\pi(3x)^2h_1=3\pi h_1 x^3 \\ [2ex] \textbf{Wide Cone} & 6y & h_2y & \frac13\pi(6y)^2h_2=12\pi h_2 y^3 \end{array}\]

Equating the final volumes gives $3\pi h_1 x^3=12\pi h_2 y^3,$ which simplifies to $x^3=y^3,$ or $x=y.$

Lastly, the requested ratio 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. The work above shows that, regardless of the shape or the volume of the solid dropped in, as long as the solid sinks to the bottom and is completely submerged without spilling any liquid, the answer will remain unchanged.

~MRENTHUSIASM

Solution 1.2 (Bash)

Let the base radii of the narrow cone and the wide cone be $r_1$ and $r_2,$ respectively.

Let the rises of the liquid levels of the narrow cone and the wide cone be $\Delta h_1$ and $\Delta h_2,$ respectively. We have the following table: \[\begin{array}{cccc} & \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex] \textbf{Narrow Cone} & r_1 & h_1+\Delta h_1 & \frac13\pi r_1^2(h_1+\Delta h_1) \\ [2ex] \textbf{Wide Cone} & r_2 & h_2+\Delta h_2 & \frac13\pi r_2^2(h_2+\Delta h_2) \end{array}\]

By similar triangles discussed above, we have \[\begin{array}{cccc} \frac{3}{h_1}=\frac{r_1}{h_1+\Delta h_1} &\Rightarrow &r_1=\frac{3}{h_1}(h_1+\Delta h_1) & \ \ \ \ \ \ \ (1) \\ [2ex] \frac{6}{h_2}=\frac{r_2}{h_2+\Delta h_2} &\Rightarrow &r_2=\frac{6}{h_2}(h_2+\Delta h_2) & \ \ \ \ \ \ \ (2) \end{array}\]

The volume of the marble dropped in is $\frac43\pi(1)^3=\frac43\pi.$

Now, we set up an equation for the volume of the narrow cone and solve for $\Delta h_1:$ \begin{align*} \frac13\pi r_1^2(h_1+\Delta h_1) &= 3\pi h_1+\frac43\pi \\ \frac13\pi{\underbrace{\left(\frac{3}{h_1}(h_1+\Delta h_1)\right)}_{\text{by (1)}}}^2(h_1+\Delta h_1) &= 3\pi h_1+\frac43\pi \\ \frac{3}{h_1^2}(h_1+\Delta h_1)^3 &= 3h_1+\frac43 \\ (h_1+\Delta h_1)^3 &= h_1^3 + \frac{4h_1^2}{9} \\ \Delta h_1 &= \sqrt[3]{h_1^3 + \frac{4h_1^2}{9}}-h_1. \end{align*}

Next, we set up an equation for the volume of the wide cone $\Delta h_2:$ \[\frac13\pi r_2^2(h_2+\Delta h_2) = 12\pi h_2+\frac43\pi.\] Using the exact same process from above (but with different numbers), we get \[\Delta h_2 = \sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2.\] Recall that $\frac{h_1}{h_2}=4.$ Therefore, the requested ratio is \begin{align*} \frac{\Delta h_1}{\Delta h_2}&=\frac{\sqrt[3]{h_1^3 + \frac{4h_1^2}{9}}-h_1}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{\sqrt[3]{(4h_2)^3 + \frac{4(4h_2)^2}{9}}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{\sqrt[3]{4^3\left(h_2^3 + \frac{h_2^2}{9}\right)}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{4\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\boxed{\textbf{(E) }4}. \end{align*}

~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

Video Solution (Simple and Quick)

https://youtu.be/TgjvviBALac

~ Education, the Study of Everything

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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