Difference between revisions of "2012 AMC 12B Problems/Problem 15"
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==Problem== | ==Problem== | ||
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Jesse cuts a circular disk of radius 12, along 2 radii to form 2 sectors, one with a central angle of 120. He makes two circular cones using each sector to form the lateral surface of each cone. What is the ratio of the volume of the smaller cone to the larger cone? | Jesse cuts a circular disk of radius 12, along 2 radii to form 2 sectors, one with a central angle of 120. He makes two circular cones using each sector to form the lateral surface of each cone. What is the ratio of the volume of the smaller cone to the larger cone? | ||
| − | + | <math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{\sqrt{10}}{10}\qquad\textbf{(D)}\ \frac{\sqrt{5}}{6}\qquad\textbf{(E)}\ \frac{\sqrt{5}}{5}</math> | |
==Solution== | ==Solution== | ||
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If the original radius is <math>12</math>, then the circumference is <math>24\pi</math>; since arcs are defined by the central angles, the smaller arc, a <math>120</math> degree angle, is half the size of the larger sector. so the smaller arc is <math>8\pi</math>, and the larger is <math>16\pi</math>. Those two arc lengths become the two circumferences of the new cones; so the radius of the smaller cone is <math>4</math> and the larger cone is <math>8</math>. Using the Pythagorean theorem, the height of the larger cone is <math>4\cdot\sqrt{5}</math> and the smaller cone is <math>8\cdot\sqrt{2}</math>, and now for volume just square the radii and multiply by <math>\tfrac{1}{3}</math> of the height to get the volume of each cone: <math>128\cdot\sqrt{2}</math> and <math>256\cdot\sqrt{5}</math> [both multiplied by three as ratio come out the same. now divide the volumes by each other to get the final ratio of <math>\boxed{\textbf{(C) } \frac{\sqrt{10}}{10}}</math> | If the original radius is <math>12</math>, then the circumference is <math>24\pi</math>; since arcs are defined by the central angles, the smaller arc, a <math>120</math> degree angle, is half the size of the larger sector. so the smaller arc is <math>8\pi</math>, and the larger is <math>16\pi</math>. Those two arc lengths become the two circumferences of the new cones; so the radius of the smaller cone is <math>4</math> and the larger cone is <math>8</math>. Using the Pythagorean theorem, the height of the larger cone is <math>4\cdot\sqrt{5}</math> and the smaller cone is <math>8\cdot\sqrt{2}</math>, and now for volume just square the radii and multiply by <math>\tfrac{1}{3}</math> of the height to get the volume of each cone: <math>128\cdot\sqrt{2}</math> and <math>256\cdot\sqrt{5}</math> [both multiplied by three as ratio come out the same. now divide the volumes by each other to get the final ratio of <math>\boxed{\textbf{(C) } \frac{\sqrt{10}}{10}}</math> | ||
== See Also == | == See Also == | ||
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{{AMC12 box|year=2012|ab=B|num-b=14|num-a=16}} | {{AMC12 box|year=2012|ab=B|num-b=14|num-a=16}} | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:42, 2 September 2020
Problem
Jesse cuts a circular disk of radius 12, along 2 radii to form 2 sectors, one with a central angle of 120. He makes two circular cones using each sector to form the lateral surface of each cone. What is the ratio of the volume of the smaller cone to the larger cone?
Solution
If the original radius is 
, then the circumference is 
; since arcs are defined by the central angles, the smaller arc, a 
degree angle, is half the size of the larger sector. so the smaller arc is 
, and the larger is 
. Those two arc lengths become the two circumferences of the new cones; so the radius of the smaller cone is 
and the larger cone is 
. Using the Pythagorean theorem, the height of the larger cone is 
and the smaller cone is 
, and now for volume just square the radii and multiply by 
of the height to get the volume of each cone: 
and 
[both multiplied by three as ratio come out the same. now divide the volumes by each other to get the final ratio of 
See Also
| 2012 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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. 
