Try this beautiful Geometry Problem based on a fly trapped inside cubical box from AMC 10 A, 2010. You may use sequential hints to solve the problem.

**Fly trapped inside cubical box **– AMC-10A, 2010- Problem 20

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?

,

- $4+4 \sqrt{2}$
- $2+4 \sqrt{2}+2 \sqrt{3}$
- $2+3 \sqrt{2}+3 \sqrt{3}$
- $4 \sqrt{2}+4 \sqrt{3}$
- $3 \sqrt{2}+5 \sqrt{3}$

**Key Concepts**

Geometry

Cube

Diagonal

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

AMC-10A, 2010 Problem-20

#### Check the answer here, but try the problem first

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

## Try with Hints

#### First Hint

Suppose the fly starts from the point \(A\).we have to find out the maximum possible length. The maximum possible length will be from one corner to another corner such as …..

$A \rightarrow G \rightarrow B \rightarrow H \rightarrow C \rightarrow E \rightarrow D \rightarrow F \rightarrow A$

Now can you find out this maximum path?

Now can you finish the problem?

#### Second Hint

Given that the side length of the cube is $1$. Therefore the diagonal \(AC\)=\(\sqrt 2\) and the diagonal \(AG=\sqrt 3\). Now we have to find out the path \(AG+GB+BH+HC+CE+ED+DF+FA\).Can you find it ?

Now can you finish the problem?

#### Third Hint

\(AG+GB+BH+HC+CE+ED+DF+FA\)=\(\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2\)=(\(4\sqrt3+4\sqrt2\))

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=OvduZbqenWU