Difference between revisions of "2023 AMC 12B Problems/Problem 21"
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Revision as of 19:34, 15 November 2023
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 \boxed{\textbf{(E) } 6 \sqrt{3} + \pi}.
~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.