Difference between revisions of "2023 AMC 12B Problems/Problem 21"
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==Problem== | ==Problem== | ||
− | A lampshade is made in the form of the lateral surface of the frustum of a right circular cone. The height of the frustum is | + | A lampshade is made in the form of the lateral surface of the frustum of a right circular cone. The height of the frustum is <math>3\sqrt3</math> inches, its top diameter is <math>6</math> inches, and its bottom diameter is <math>12</math> inches. A bug is at the bottom of the lampshade and there is a glob of honey on the top edge of the lampshade at the spot farthest from the bug. The bug wants to crawl to the honey, but it must stay on the surface of the lampshade. What is the length in inches of its shortest path to the honey? |
<math>\textbf{(A) } 6 + 3\pi\qquad \textbf{(B) }6 + 6\pi\qquad \textbf{(C) } 6\sqrt3 \qquad \textbf{(D) } 6\sqrt5 \qquad \textbf{(E) } 6\sqrt3 + \pi</math> | <math>\textbf{(A) } 6 + 3\pi\qquad \textbf{(B) }6 + 6\pi\qquad \textbf{(C) } 6\sqrt3 \qquad \textbf{(D) } 6\sqrt5 \qquad \textbf{(E) } 6\sqrt3 + \pi</math> | ||
+ | |||
+ | ==Graph== | ||
+ | |||
+ | |||
+ | <asy> | ||
+ | // Please make the graph better if you have time. | ||
+ | import olympiad; | ||
+ | size(10cm); | ||
+ | draw((-1,0)..(0,-1)..(1,0)); | ||
+ | import olympiad; | ||
+ | pair A = (-1,0); | ||
+ | import olympiad; | ||
+ | pair B = (-0.5,0); | ||
+ | import olympiad; | ||
+ | pair C = (0.5,0); | ||
+ | import olympiad; | ||
+ | pair D = (1,0), E = (0,0), F = (-0.25,-0.43301270189); | ||
+ | draw(A--B); | ||
+ | draw(C--D); | ||
+ | draw((-0.5,0)..(0,-0.5)..(0.5,0)); | ||
+ | draw(E--A); | ||
+ | label("$12$",(-0.5,0.15)); | ||
+ | draw(E--(0,-0.5)); | ||
+ | label("$6$",(0.1,-0.25)); | ||
+ | draw(A--F); | ||
+ | draw(F--E); | ||
+ | draw(anglemark(E,F,A), p=black); | ||
+ | label("$90^\circ$", anglemark(E,F,A), NE); | ||
+ | </asy> | ||
==Solution== | ==Solution== | ||
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This implies <math>\angle COB = \angle AOB - \angle AOC = 30^\circ</math>. | This implies <math>\angle COB = \angle AOB - \angle AOC = 30^\circ</math>. | ||
Therefore, the length of the arc between <math>C</math> and <math>B</math> is <math>OB \cdot \pi \cdot \frac{\angle COB}{180^\circ} = \pi</math>. | Therefore, the length of the arc between <math>C</math> and <math>B</math> is <math>OB \cdot \pi \cdot \frac{\angle COB}{180^\circ} = \pi</math>. | ||
− | Therefore, the shortest distance between <math>A</math> and <math>B</math> is \boxed{\textbf{(E) } 6 \sqrt{3} + \pi}. | + | Therefore, the shortest distance between <math>A</math> and <math>B</math> is <math>\boxed{\textbf{(E) } 6 \sqrt{3} + \pi}</math>. |
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/43TAxlETKnA | ||
+ | |||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Latest revision as of 19:38, 5 October 2024
Contents
Problem
A lampshade is made in the form of the lateral surface of the frustum of a right circular cone. The height of the frustum is inches, its top diameter is inches, and its bottom diameter is inches. A bug is at the bottom of the lampshade and there is a glob of honey on the top edge of the lampshade at the spot farthest from the bug. The bug wants to crawl to the honey, but it must stay on the surface of the lampshade. What is the length in inches of its shortest path to the honey?
Graph
Solution
We augment the frustum to a circular cone. Denote by the apex of the cone. Denote by the bug and the honey.
By using the numbers given in this problem, the height of the cone is . Thus, and .
We unfold the lateral face. So we get a circular sector. The radius is 12 and the length of the arc is . Thus, the central angle of this circular sector is .
Because and are opposite in the original frustum, in the unfolded circular cone, .
Notice that a feasible path between and can only fall into the region with the range of radii between and . Therefore, we cannot directly connect and and must make a detour. Denote by a tangent to the circular sector with radius 6 that meets it at point . Therefore, the shortest path between and consists of a segment and an arc from to .
Because , and , we have and . This implies . Therefore, the length of the arc between and is . Therefore, the shortest distance between and is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.