Six vertical mirrors are arranged to form a regular hexagon with only a small gap (of negligible size) at each vertex. A beam of light is shone through one of these gaps toward the one directly opposite, through which it exits.

However, the beam can be aimed slightly off this direct route to the other side, so that it is reflected off each of the mirrors exactly once before finally exiting the arrangement through that same opening opposite its entrance.

What is the smallest angle off of straight across which will make this possible?

Let ABCDEF be the hexagon with A and D
the source and target vertices respectively.

Reflect the hexagon six times about edges

       DE, B_1C_1, A_2F_2, C_3D_3, E_4F_4, 
       and A_5B_5.

Where the vertices of the hexagon are 
labeled thus for each reflection

       X --> X_1 --> X_2 --> X_3 --> X_4 
         --> X_5 --> X_6.

Draw the line AD_6.

The angle we desire is

       Angle BAD_6 - Angle BAD.

These angles are easily determined from
the reflection diagram.

       arctan(3*sqrt(3)/2) - 60

       =~ 8.948275564627

 

Posted by Bractals on 2009-05-08 15:03:37