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