2010 AMC 10A Problem 20

Question

A fly trapped inside a cubical box with side length

1

meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?

4+4\sqrt{2}

2+4\sqrt{2}+2\sqrt{3}

2+3\sqrt{2}+3\sqrt{3}

4\sqrt{2}+4\sqrt{3}

3\sqrt{2}+5\sqrt{3}

Answer: D

Concepts:cube geometry graph theory optimization

Difficulty rating: 2070

Small hint:

Each move has length

1,\sqrt2,

or

\sqrt3

Big hint:

There are only

4

space diagonals, so the remaining moves are at most face diagonals

Solution:

Note that all the paths the fly can take have lengths of

1, \sqrt2,

or

\sqrt3.

There are only

4

space diagonals in the cube, so at most

4

moves can have length

\sqrt3.

The other

4

moves have length at most

\sqrt2.

This upper bound is attainable, for example by alternating space diagonals and face diagonals around the vertices.

The path has length

4\sqrt2 + 4\sqrt3.

Thus, D is the correct answer.