2001 AMC 12 Problem 15

Question

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.)

\dfrac{1}{2}\sqrt{3}

1

\sqrt{2}

\dfrac{3}{2}

2

Answer: B

Concepts:net (3D geometry)3D geometry rhombus

Difficulty rating: 1660

Small hint:

Unfold the two faces the path crosses into a single flat figure

Big hint:

The two faces form a rhombus of side

1,

and the midpoints become midpoints of opposite sides

Solution:

A shortest path leaves the starting edge through one of its two incident faces and reaches the opposite edge through one of its two incident faces. Any such pair of faces shares an edge. Unfolding that pair gives a rhombus of side

1

made of two equilateral triangles.

The two opposite-edge midpoints become the midpoints of opposite sides of this rhombus, which are exactly

1

unit apart along a straight segment. Folding back preserves the length, so the shortest trip is

1.

Thus, the correct answer is B.