Difference between revisions of "2021 AMC 12A Problems/Problem 10"
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<math>\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1</math> | <math>\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1</math> | ||
− | ==Solution 1 ( | + | ==Solution 1 (Algebra)== |
<b><u>Initial Scenario</u></b> | <b><u>Initial Scenario</u></b> | ||
+ | |||
+ | Let the heights of the narrow cone and the wide cone be <math>h_1</math> and <math>h_2,</math> respectively. We have the following table: | ||
<cmath>\begin{array}{cccccc} | <cmath>\begin{array}{cccccc} | ||
& \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] | & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] | ||
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Equating the volumes gives <math>3\pi h_1=12\pi h_2,</math> which simplifies to <math>\frac{h_1}{h_2}=4.</math> | Equating the volumes gives <math>3\pi h_1=12\pi h_2,</math> which simplifies to <math>\frac{h_1}{h_2}=4.</math> | ||
− | + | Furthermore, by similar triangles: | |
− | For the narrow cone, the ratio of base radius to height is <math>\frac{3}{h_1},</math> which remains constant. | + | * For the narrow cone, the ratio of the base radius to the height is <math>\frac{3}{h_1},</math> which always remains constant. |
− | For the wide cone, the ratio of base radius to height is <math>\frac{6}{h_2},</math> which remains constant. | + | * For the wide cone, the ratio of the base radius to the height is <math>\frac{6}{h_2},</math> which always remains constant. |
Two solutions follow from here: | Two solutions follow from here: | ||
− | ===Solution 1.1 ( | + | ===Solution 1.1 (Properties of Fractions)=== |
<b><u>Final Scenario</u></b> | <b><u>Final Scenario</u></b> | ||
− | + | For the narrow cone and the wide cone, let their base radii be <math>3x</math> and <math>6y</math> (for some <math>x,y>1</math>), respectively. By the similar triangles discussed above, their heights must be <math>h_1x</math> and <math>h_2y,</math> respectively. We have the following table: | |
<cmath>\begin{array}{cccccc} | <cmath>\begin{array}{cccccc} | ||
& \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] | & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] | ||
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\textbf{Wide Cone} & 6y & h_2y & & \hspace{2.25mm}\frac13\pi(6y)^2\left(h_2y\right)=12\pi h_2 y^3 & | \textbf{Wide Cone} & 6y & h_2y & & \hspace{2.25mm}\frac13\pi(6y)^2\left(h_2y\right)=12\pi h_2 y^3 & | ||
\end{array}</cmath> | \end{array}</cmath> | ||
+ | Recall that <math>\frac{h_1}{h_2}=4.</math> Equating the volumes gives <math>3\pi h_1 x^3=12\pi h_2 y^3,</math> which simplifies to <math>x^3=y^3,</math> or <math>x=y.</math> | ||
− | + | Finally, the requested ratio 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) }4:1}.</cmath> | |
− | |||
− | |||
− | |||
<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 | + | <li>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}=k.</math> <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}=k.</math> <p> | ||
− | We can prove this | + | We can prove this property quickly: <p>From <math>\frac ab = \frac cd = k,</math> we know that <math>a=bk</math> and <math>c=dk</math>. Therefore, we conclude that <cmath>\frac{a\pm c}{b\pm d}=\frac{bk\pm dk}{b\pm d}=\frac{\left(b\pm d\right)k}{b\pm d}=k.</cmath></li><p> |
− | <li> | + | <li>This solution shows that, regardless of the shape or the volume of the solid dropped into each cone, the requested ratio stays the same as long as the solid sinks to the bottom and is completely submerged without spilling any liquid.</li><p> |
</ol> | </ol> | ||
− | |||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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<b><u>Final Scenario</u></b> | <b><u>Final Scenario</u></b> | ||
− | + | For the narrow cone and the wide cone, let their base radii be <math>r_1</math> and <math>r_2,</math> respectively; let their rises of the liquid levels be <math>\Delta h_1</math> and <math>\Delta h_2,</math> respectively. We have the following table: | |
<cmath>\begin{array}{cccccc} | <cmath>\begin{array}{cccccc} | ||
& \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] | & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] | ||
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\textbf{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> | ||
− | + | By the similar triangles discussed above, we get | |
− | By similar triangles discussed above, we | ||
<cmath>\begin{align*} | <cmath>\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{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) | \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*}</cmath> | \end{align*}</cmath> | ||
− | The volume of the marble dropped | + | The volume of the marble dropped into each cone is <math>\frac43\pi(1)^3=\frac43\pi.</math> |
− | Now, we set up an equation for the volume of the narrow cone | + | Now, we set up an equation for the volume of the narrow cone, then express <math>\Delta h_1</math> in terms of <math>h_1:</math> |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\frac13\pi r_1^2(h_1+\Delta h_1) &= 3\pi h_1+\frac43\pi \\ | \frac13\pi r_1^2(h_1+\Delta h_1) &= 3\pi h_1+\frac43\pi \\ | ||
− | \frac13\ | + | \frac13 r_1^2(h_1+\Delta h_1) &= 3h_1+\frac43 \\ |
+ | \frac13{\biggl(\phantom{ }\underbrace{\frac{3}{h_1}(h_1+\Delta h_1)}_{\text{by }(1)}\phantom{ }\biggr)}^2(h_1+\Delta h_1) &= 3h_1+\frac43 \\ | ||
\frac{3}{h_1^2}(h_1+\Delta h_1)^3 &= 3h_1+\frac43 \\ | \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} \\ | (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. | \Delta h_1 &= \sqrt[3]{h_1^3 + \frac{4h_1^2}{9}}-h_1. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | Next, we set up an equation for the volume of the wide cone, then express <math>\Delta h_2</math> in terms of <math>h_2:</math> | |
− | Next, we set up an equation for the volume of the wide cone | ||
<cmath>\frac13\pi r_2^2(h_2+\Delta h_2) = 12\pi h_2+\frac43\pi.</cmath> | <cmath>\frac13\pi r_2^2(h_2+\Delta h_2) = 12\pi h_2+\frac43\pi.</cmath> | ||
− | Using | + | Using a similar process from above, we get <cmath>\Delta h_2 = \sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2.</cmath> |
Recall that <math>\frac{h_1}{h_2}=4.</math> Therefore, the requested ratio is | Recall that <math>\frac{h_1}{h_2}=4.</math> Therefore, the requested ratio is | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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&=\boxed{\textbf{(E) }4:1}. | &=\boxed{\textbf{(E) }4:1}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | |||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 2 (Quick and | + | ==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 <math>\boxed{\textbf{(E) } 4}</math> times as much. | 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 <math>\boxed{\textbf{(E) } 4}</math> times as much. | ||
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~ Education, the Study of Everything | ~ Education, the Study of Everything | ||
− | |||
==Video Solution by Aaron He (Algebra)== | ==Video Solution by Aaron He (Algebra)== |
Latest revision as of 19:16, 19 June 2021
- The following problem is from both the 2021 AMC 10A #12 and 2021 AMC 12A #10, so both problems redirect to this page.
Contents
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 cm and cm. Into each cone is dropped a spherical marble of radius 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?
Solution 1 (Algebra)
Initial Scenario
Let the heights of the narrow cone and the wide cone be and respectively. We have the following table: Equating the volumes gives which simplifies to
Furthermore, by similar triangles:
- For the narrow cone, the ratio of the base radius to the height is which always remains constant.
- For the wide cone, the ratio of the base radius to the height is 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 and (for some ), respectively. By the similar triangles discussed above, their heights must be and respectively. We have the following table: Recall that Equating the volumes gives which simplifies to or
Finally, the requested ratio is Remarks
- This solution uses the following property of fractions:
For unequal positive numbers and if then
We can prove this property quickly:
From we know that and . Therefore, we conclude that
- This solution shows that, regardless of the shape or the volume of the solid dropped into each cone, the requested ratio stays the same as long as the solid sinks to the bottom and is completely submerged without spilling any liquid.
~MRENTHUSIASM
Solution 1.2 (Bash)
Final Scenario
For the narrow cone and the wide cone, let their base radii be and respectively; let their rises of the liquid levels be and respectively. We have the following table: By the similar triangles discussed above, we get The volume of the marble dropped into each cone is
Now, we set up an equation for the volume of the narrow cone, then express in terms of Next, we set up an equation for the volume of the wide cone, then express in terms of Using a similar process from above, we get Recall that Therefore, the requested ratio is ~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 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 . Therefore, the ratio of the height of the narrow cone to the height of the wide cone must be . 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)
~ 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)
~ 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
~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.