## Another Obedient Ray

Two mirrors, A and B, are joined at a fixed angle at point O, and a ray of light is shone into the angle between them. It first strikes one mirror at a point X. When the ray re-emerges, the last point it strikes is point Y on one mirror.
1. If the incident ray is parallel to mirror A, and the exit ray is parallel to mirror B, find the possible angles for AOB and the possible relationships between OX and OY.
2. Find the requirements on the angle of incidence, with respect to angle AOB, such that OX=OY.

Source: Original. Similar to Ken's POTW January 2, 1997

Solutions were received from Ravi Subramanian, John Hewson, and Philippe Fondanaiche.
1. Angle AOB = PI / (2n + 1) for non negative integers n. OX = OY
2. [ PI - (2*n+1)*AOB ] / 2.
If measured with respect to normal, this would be (2*n+1)*AOB.
Philippe's explanation of these answers is below.
```Question 1

Let angle(AOB) = a.First of all,we can observe that if the incident ray is
parallel to mirror A and the exit ray is parallel to mirror B,then a is
always acute.Moreover a simple graph of the path of the ray shows that the
possible values of a are
a = pi/(2*n+1) with n any integer >=1,that is to say a=pi/3,
pi/5,pi/7,pi/9,etc.... Due to the symmetry, the incident ray can become the
exit ray and vice versa, we have obviously OX=OY with 2*n strikes, n on each
mirror such that the increasing abscissa of the n points of impact on the
mirror A are equal to the corresponding ones on the mirror B.

Question 2

Let alpha=angle of incidence. As mentioned above, due to the symmetry, OX=OY
implies that there is always an even number of strikes 2*n such that there
are n strikes on each mirror with identical abscissa.
The angles of incidence of the ray at the consecutive strikes are
respectively on the mirror A,then B,then A,etc...alpha,
alpha+a,alpha+2*a,alpha+3*a.....,alpha+k*a...if and only if alpha + k*a
remainspi/2.
OX=OY implies that we have an isosceles triangle defined by the points of
contact of the kth and the (k+1)th strikes and we have the equality of the
two angles of incidence : pi - a - (alpha +(k-1)*a)= alpha +(k-1)*a.
Therefore alpha  = [ pi - (2*k-1)*a ] / 2 which is valid for a <=pi/(2*k-1).
So the requirements on the angle of incidence alpha can be summarized as
follows:
- determine k such as k = integer of 1/2*(pi/a+1) (rounded down)
- calculate alpha = [ pi - (2*k-1)*a ] / 2
```

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