Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12B Problems/Problem 21"

m
m (→‎Graph)
 
(2 intermediate revisions by 2 users not shown)
Line 3: Line 3:
  
 
<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==
Line 27: Line 56:
 
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 <math>\boxed{\textbf{(E) } 6 \sqrt{3} + \pi}</math>.
 
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

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 $3\sqrt3$ inches, its top diameter is $6$ inches, and its bottom diameter is $12$ 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?

$\textbf{(A) } 6 + 3\pi\qquad \textbf{(B) }6 + 6\pi\qquad \textbf{(C) } 6\sqrt3 \qquad \textbf{(D) } 6\sqrt5 \qquad \textbf{(E) } 6\sqrt3 + \pi$

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

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)

Video Solution

https://youtu.be/43TAxlETKnA


~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_21&oldid=229054"