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2023 AMC 12B Problems/Problem 21: Difference between revisions

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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
==See Also==
{{AMC12 box|year=2023|ab=B|num-b=20|num-a=22}}
{{MAA Notice}}

Revision as of 19:34, 15 November 2023

Solution

We augment the frustum to a circular cone. Denote by $O$ the apex of the cone. Denote by $A$ the bug and $B$ the honey.

By using the numbers given in this problem, the height of the cone is $6 \sqrt{3}$. Thus, $OA = 12$ and $OB = 6$.

We unfold the lateral face. So we get a circular sector. The radius is 12 and the length of the arc is $12\pi$. Thus, the central angle of this circular sector is $180^\circ$.

Because $A$ and $B$ are opposite in the original frustum, in the unfolded circular cone, $\angle AOB = \frac{180^\circ}{2} = 90^\circ$.

Notice that a feasible path between $A$ and $B$ can only fall into the region with the range of radii between $OB = 6$ and $OA = 12$. Therefore, we cannot directly connect $A$ and $B$ and must make a detour. Denote by $AC$ a tangent to the circular sector with radius 6 that meets it at point $C$. Therefore, the shortest path between $A$ and $B$ consists of a segment $AC$ and an arc from $C$ to $B$.

Because $OA = 12$, $OC = 6$ and $\angle OCA = 90^\circ$, we have $AC = \sqrt{OA^2 - OC^2} = 6 \sqrt{3}$ and $\angle AOC = 60^\circ$. This implies $\angle COB = \angle AOB - \angle AOC = 30^\circ$. Therefore, the length of the arc between $C$ and $B$ is $OB \cdot \pi \cdot \frac{\angle COB}{180^\circ} = \pi$. Therefore, the shortest distance between $A$ and $B$ 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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