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)