2010 AMC 10A Problem 20

Attempt Problem 20 of the 2010 AMC 10A 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 2010 AMC 10A solutions, or check the answer key.

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20.

A fly trapped inside a cubical box with side length 11 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+424+4\sqrt{2}

2+42+232+4\sqrt{2}+2\sqrt{3}

2+32+332+3\sqrt{2}+3\sqrt{3}

42+434\sqrt{2}+4\sqrt{3}

32+533\sqrt{2}+5\sqrt{3}

Answer: D
Concepts:cube geometrygraph theoryoptimization
Difficulty rating: 2070
Solution:

Note that all the paths the fly can take have lengths of 1,2,1, \sqrt2, or 3.\sqrt3.

There are only 44 space diagonals in the cube, so at most 44 moves can have length 3.\sqrt3. The other 44 moves have length at most 2.\sqrt2.

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

The path has length 42+43. 4\sqrt2 + 4\sqrt3.

Thus, D is the correct answer.

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