Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1.1 (Properties of Fractions))
(→‎Solution 1.1 (Properties of Fractions))
Line 71: Line 71:
 
<u><b>Remarks</b></u>
 
<u><b>Remarks</b></u>
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
−
  <li>This solution uses the following property of fractions: <p>  
+
This solution uses the following property of fractions: <p>  
 
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}=\frac{bk\pm dk}{b\pm d}=\frac{(b\pm d)k}{b\pm d}=k.</math></li><p>
 
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}=\frac{bk\pm dk}{b\pm d}=\frac{(b\pm d)k}{b\pm d}=k.</math></li><p>
−
</ol>
+
 
 
~huashiliao2020
 
~huashiliao2020
  

Revision as of 21:43, 5 December 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]

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

Solution 1 (Algebra)

Initial Scenario

Let the heights of the narrow cone and the wide cone be $h_1$ and $h_2,$ respectively. We have the following table: \[\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & 3 & h_1 & & \frac13\pi(3)^2h_1=3\pi h_1 & \\ [2ex] \textbf{Wide Cone} & 6 & h_2 & & \hspace{2mm}\frac13\pi(6)^2h_2=12\pi h_2 & \end{array}\] Equating the volumes gives $3\pi h_1=12\pi h_2,$ which simplifies to $\frac{h_1}{h_2}=4.$

Furthermore, by similar triangles:

  • For the narrow cone, the ratio of the base radius to the height is $\frac{3}{h_1},$ which always remains constant.
  • For the wide cone, the ratio of the base radius to the height is $\frac{6}{h_2},$ which always remains constant.

Two solutions follow from here:

Solution 1.1 (Properties of Fractions)

Final Scenario

For the narrow cone and the wide cone, let their base radii be $3x$ and $6y$ (for some $x,y>1$), respectively. By the similar triangles discussed above, their heights must be $h_1x$ and $h_2y,$ respectively. We have the following table: \[\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & 3x & h_1x & & \frac13\pi(3x)^2(h_1x)=3\pi h_1 x^3 & \\ [2ex] \textbf{Wide Cone} & 6y & h_2y & & \hspace{2.0625mm}\frac13\pi(6y)^2(h_2y)=12\pi h_2 y^3 & \end{array}\] Recall that $\frac{h_1}{h_2}=4.$ Equating the volumes gives $3\pi h_1 x^3=12\pi h_2 y^3,$ which simplifies to $x^3=y^3,$ or $x=y.$

Finally, the requested ratio is \[\frac{h_1 x - h_1}{h_2 y - h_2}=\frac{h_1 (x-1)}{h_2 (y-1)}=\frac{h_1}{h_2}=\boxed{\textbf{(E) }4:1}.\] Remarks

    This solution uses the following property of fractions:

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

    ~huashiliao2020

    Solution 1.2 (Bash)

    Final Scenario

    For the narrow cone and the wide cone, let their base radii be $r_1$ and $r_2,$ respectively; let their rises of the liquid levels be $\Delta h_1$ and $\Delta h_2,$ respectively. We have the following table: \[\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & r_1 & h_1+\Delta h_1 & & \frac13\pi r_1^2(h_1+\Delta h_1) & \\ [2ex] \textbf{Wide Cone} & r_2 & h_2+\Delta h_2 & & \frac13\pi r_2^2(h_2+\Delta h_2) & \end{array}\] By the similar triangles discussed above, we get \begin{align*} \frac{3}{h_1}&=\frac{r_1}{h_1+\Delta h_1} &\implies \quad r_1&=\frac{3}{h_1}(h_1+\Delta h_1), & \hspace{10mm} (1) \\ \frac{6}{h_2}&=\frac{r_2}{h_2+\Delta h_2} &\implies \quad r_2&=\frac{6}{h_2}(h_2+\Delta h_2). & (2) \end{align*} The volume of the marble dropped into each cone is $\frac43\pi(1)^3=\frac43\pi.$

    Now, we set up an equation for the volume of the narrow cone, then express $\Delta h_1$ in terms of $h_1:$ \begin{align*} \frac13\pi r_1^2(h_1+\Delta h_1) &= 3\pi h_1+\frac43\pi \\ \frac13 r_1^2(h_1+\Delta h_1) &= 3h_1+\frac43 \\ \frac13\left(\frac{3}{h_1}(h_1+\Delta h_1)\right)^2(h_1+\Delta h_1) &= 3h_1+\frac43 &&\text{by }(1) \\ \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, then express $\Delta h_2$ in terms of $h_2:$ \[\frac13\pi r_2^2(h_2+\Delta h_2) = 12\pi h_2+\frac43\pi.\] Using a similar process from above, 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) }4:1}. \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

    Solution 3

    Since the radius of the narrow cone is 1/2 the radius of the wider cone, the ratio of their areas is $\dfrac{1}{4}$. Therefore, the ratio of the height of the narrow cone to the height of the wide cone must be $\dfrac{4}{1}$. Note that this ratio is constant, regardless of how much water is dropped as long as it is an equal amount for both cones. See Solution 2 for another explanation.

    Video Solution (Simple and Quick)

    https://youtu.be/TgjvviBALac

    ~ Education, the Study of Everything

    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 by TheBeautyofMath

    First-this is not the most efficient solution. I did not perceive the shortcut before filming though I suspected it.

    https://youtu.be/t-EEP2V4nAE?t=231 (for AMC 10A)

    https://youtu.be/cckGBU2x1zg?t=814 (for AMC 12A)

    ~IceMatrix

    Video Solution by WhyMath

    https://youtu.be/c-5-8PnCvCk

    ~savannahsolver

    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

    The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_10&oldid=167382"