Posted by Knights on Friday, February 22, 2013 at 2:29pm.
No, the laser follows the old billiard table rule, that the angle of incidence is the angle of reflection.
I am going to enlarge your square to a 4 by 4, thus making your first bounce BP = 3
make a reasonable sketch, graph paper might be a good idea.
From P it will bounce and hit CD at Q
look at triangles ABP and PCQ, they are both right-angles and similar.
BP^2 + AB^2 = AP^2
9 + 16 = AP^2
AP = √25 = 5
So each triangle formed by a bounce will have the ratio
3:4:5
We are interested in the sum of the paths formed by the hypotenuses
in the second triangle PC, the short side is 1
so by ratios,
PQ/5 = 1/3
PP = 5/3
find CQ by ratios, then you can find QD
On my diagram, I have the following paths:
From A to P ---- 5 units
from P to Q on CD --- done : 5/3 units
from Q to R on AD , R is close to D,
from R to S on BC
from S to T on AB
from T to U on AD and ahhhh
from U to C , which is a vertex.
At this point you should notice that there is a lot of symmetry
and AP = RS = UC
PQ = TU
QR = TS
so once you have found QR, again by using the ratios 3:4:5
you have found the 3 different path lengths
Add up the 7 lengths, don't forget to divide by 4 , my original step to avoid some initial fractions.
Thank you very much it helps a lot!!