2010 AMC 10A Problems/Problem 20

Problem

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?

$\textbf{(A)}\ 4+4\sqrt{2} \qquad \textbf{(B)}\ 2+4\sqrt{2}+2\sqrt{3} \qquad \textbf{(C)}\ 2+3\sqrt{2}+3\sqrt{3} \qquad \textbf{(D)}\ 4\sqrt{2}+4\sqrt{3} \qquad \textbf{(E)}\ 3\sqrt{2}+5\sqrt{3}$

Solution

The distance of an interior diagonal in this cube is $\sqrt{3}$ and the distance of a diagonal on one of the square faces is $\sqrt{2}$ . It would not make sense if the fly traveled an interior diagonal twice in a row, as it would return to the point it just came from, so at most the final sum can only have 4 as the coefficient of $\sqrt{3}$ . The other 4 paths taken can be across a diagonal on one of the faces, so the maximum distance traveled is $\textbf{(D)}\ 4\sqrt{2}+4\sqrt{3}$ .

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2010 AMC 10A Problems/Problem 20

Problem

Solution

See also