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.)

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

$1$

$\sqrt{2}$

$\dfrac{3}{2}$

$2$

Solution:

Unfold the two faces the insect crosses into the plane. They form 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.

2001 AMC 12 Problem 15

Solution:

Problem 15 in Other Years