Difference between revisions of "2001 AMC 12 Problems/Problem 15"
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== Solution == | == Solution == | ||
| − | Given any path on the surface, we can unfold the surface into a plane to get a path of the same length in the plane. Consider the net of a tetrahedron in the picture below. A pair of opposite points is marked by dots. It is obvious that in the plane the shortest path is just a segment that connects these two points. | + | Given any path on the surface, we can unfold the surface into a plane to get a path of the same length in the plane. Consider the net of a tetrahedron in the picture below. A pair of opposite points is marked by dots. It is obvious that in the plane the shortest path is just a segment that connects these two points. By symmetry (as the tetrahedron is regular, so all of its faces are equilateral triangles), its length is the same as the length of the tetrahedron's edge, i.e. <math>\boxed{\text{(B) }1}</math>. |
<asy> | <asy> | ||
Latest revision as of 06:03, 6 June 2025
Contents
Problem
An insect lives on the surface of a regular tetrahedron with edges of length 1. It wishes to travel on the surface of the tetrahedron from the midpoint of one edge to the midpoint of the opposite edge. What is the length of the shortest such trip? (Note: Two edges of a tetrahedron are opposite if they have no common endpoint.)
Solution
Given any path on the surface, we can unfold the surface into a plane to get a path of the same length in the plane. Consider the net of a tetrahedron in the picture below. A pair of opposite points is marked by dots. It is obvious that in the plane the shortest path is just a segment that connects these two points. By symmetry (as the tetrahedron is regular, so all of its faces are equilateral triangles), its length is the same as the length of the tetrahedron's edge, i.e. 
.
![[asy] unitsize(2cm); defaultpen(0.8); pair A=(0,0), B=(1,0), C=rotate(60)*B, D1=B+C-A, D2=A+C-B, D3=A+B-C; draw(A--B--C--cycle); draw(D1--B--C--cycle); draw(D2--A--C--cycle); draw(D3--B--A--cycle); dot( A+0.5*(C-A) ); dot( B+0.5*(C-A) ); draw( ( A+0.5*(C-A) ) -- ( B+0.5*(C-A) ), dashed ); [/asy]](http://latex.artofproblemsolving.com/f/2/d/f2d84548c68b8c26bea559248ac713844ce32e42.png)
Video Solution
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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. 
