2001 AMC 12 Problem 15

Attempt Problem 15 of the 2001 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 12 solutions, or check the answer key.

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15.

An insect lives on the surface of a regular tetrahedron with edges of length 1.1. 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.)

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

11

2\sqrt{2}

32\dfrac{3}{2}

22

Answer: B
Concepts:net (3D geometry)3D geometryrhombus
Difficulty rating: 1660
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 11 made of two equilateral triangles.

The two opposite-edge midpoints become the midpoints of opposite sides of this rhombus, which are exactly 11 unit apart along a straight segment. Folding back preserves the length, so the shortest trip is 1.1.

Thus, the correct answer is B.

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