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Double Reflection (Posted on 2016-03-24) Difficulty: 3 of 5
Two flat mirrors are placed to form a wedge with the reflective sides facing inward. A laser is shined in from the end of one mirror. It hits each mirror once and leaves by the end of the other mirror. How large can the angle between the mirrors be?

Crude diagram: the laser starts at point 1 (at the end of one mirror), travels through 2, reflects off of 3, travels trough 4, reflects off of 5, travels through 6 and ends at point 7 (at the end of the other mirror).
     /\
    /  \
   /    \
  /3 4  5\
 /  6  2  \
7          1
What if I want the laser to bounce off of each side n times?

No Solution Yet Submitted by Brian Smith    
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Solution Alternate solution Comment 2 of 2 |
Charlie's is quicker than mine but I enjoyed following the angles around. There are some nice patterns.

Call the start D, the first bounce A, the second bounce B, and the end C.  (The order will allow for the extension to be more natural.)  The path is DABC.  Call the vertex O and the intersection of DA and BC is Z.

Call the measure of the angle sought AOB=θ.  There are then a few interesting angles:
ABZ=θ
BZD=2θ
ODA=90-3θ/2
This last angle tells you how to aim the laser based on the mirror angle.  Since it must be positive solving gives θ<60.

Extend the problem.  D and C are no longer the start and end but the first and last reflection.  Add points E and F so the path is now EDABCF.  The intersection of ED and CF is Y.
More interesting angles:
ZDY=3θ
DYF=4θ
OFY=90-5θ/2
Again this last angle is positive so θ<36

It's not hard to see how these internal angles keep increasing and how the formula becomes θ<180/(2n+1)

Remark:  we could also aim the laser in such a way as to have it end up back where we started.  It would bounce n times on one side but only n-1 on the other.  The formula for this is θ<180/(2n). 


  Posted by Jer on 2016-03-24 10:53:44
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