1981 AHSME Problems/Problem 20
Problem
A ray of light originates from point and travels in a plane, being reflected
times between lines
and
before striking a point
(which may be on
or
) perpendicularly and retracing its path back to
(At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for
). If
, what is the largest value
can have?
Solution
Notice that when we start, we want the smallest angle possible of reflection. The ideal reflection would be , but that would be impossible. Therefore we start by working backwards. Since angle
is
, the reflection would give us a triangle with angles
, and
. Then, when we reflect again, we will have
=
. Since the other side of the reflection when we had the
degrees had carried over to the other side, we have a
triangle.
Notice that we keep decreasing by increments of . This is because the starting angle was
and since we always have to decrease
every time and that every triangle has every increasing angles of
, we must decrease by
every time. This is the most optimal path of the light beam.
The pattern of light will be . When we get to the angle of
degrees, we have reached angle
. Therefore, we don't count the
, so our total number of reflections between
and
is
~Arcticturn