INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More

Contents

[hide]

Try this beautiful problem from Geometry based on cubical box.

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?

- \(4\sqrt 5+3\sqrt 2\)
- \(4\sqrt 5+3\sqrt 2\)
- \(4\sqrt 3+4\sqrt 2\)
- \(7\sqrt 3+4\sqrt 2\)
- \(4\sqrt 3+7\sqrt 2\)

Geometry

Cubical

Pythagoras

But try the problem first...

Answer: \(4\sqrt 3+4\sqrt 2\)

Source

Suggested Reading

AMC-10A (2010) Problem 20

Pre College Mathematics

First hint

Given that "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".................................Therefore we may say that from a corner to any other corner the straight path will be \(A \to G \to B \to H \to C \to E \to D \to F \to A\)

can you finish the problem........

Second Hint

The distance of an interior diagonal in this cube is \(\sqrt 3\) ( i.e \(HB\)) and the distance of a diagonal on one of the square faces is \(\sqrt 2\) ( i.e \(HA\))

can you finish the problem........

Final Step

Now the fly visits each corner exactly once, it cannot traverse such a line segment twice. Also, the cube has exactly four such diagonals, so the path of the fly can contain at most four segments of length.Therefore the maximum distance traveled is \(4\sqrt 3+4\sqrt 2\)

- https://www.cheenta.com/area-of-trapezoid-amc-10a-2002-problem-25/
- https://www.youtube.com/watch?v=fvhDNPfri9w

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL
Google