2023 AMC 12B Problems/Problem 21
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 |
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