Help please with reflection problem!?
posted by Knights on .
A laser is shot from vertex A of square ABCD of side length 1, towards point P on BC so that BP = 3/4. The laser reflects off the sides of the square, until it hits another vertex, at which point it stops. What is the length of the path the laser takes?
Help?? Won't the laser bounce around until infinity?
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
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!!