The 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, parallel to mirror A. It bounces a number of times on mirrors B and A, strikes one mirror at right angles at a point X on the mirror, and then re-emerges along its original path. What is the distance between the original ray and mirror A, in terms of the angle between the mirrors and the distance OX? (Solve for X on either A or B).

Source: The Penguin Book of Curious and Interesting Puzzles, David Wells, 1992, #378. Used as Ken's POTD 3/29/95.


Solutions:
These are from Ken Burres, Wilfred Theunissen and Evert Offereins, and myself.

Ken Burres sent a graphical solution with several similar triangles for an angle of 30 degrees, and two bounces on the mirrors, which showed that the distance for that special case is OX.


Wilfred Theunissen and Evert Offereins sent the following solution:
The two mirrors A and B are joint in point O with an angle alpha. As you already
know, alpha has to be 90/n where n is the number of bounces the ray takes to end
in a right angle. The angle at first bounce is alpha, at second bounce 2 * alpha
etc. The last bounce the angle is n * alpha = 90 which implies that alpha is 
90/n.

The first bounce on mirror B occurs at point X(1). Drop a perpendicular from 
X(1) on  mirror A and call the intersection point Y. We are going to prove that:

(1)     YX(1) = OX, where X is the point where the Ray makes a right angle with
                  one of the mirrors.

We define the i-th bounce in point X(i), so X = X(n). 
The second bounce is in point X(2) with an angle of 2 * alpha. We are going to 
calculate the length of X(1)X(2) by using the sine-rule in triangle X(1)X(2)Y. 
This leads to the formula:

(2)     X(1)X(2) = X(1)Y / sin(2*alpha)

Because the triangle OX(1)X(2) is an isosceles triangle we know that :

(3)     X(1)X(2) = OX(2)

Using (2) and (3), we have:
 
(4)     OX(2) = X(1)Y/sin(2*alpha)

We now look at triangle OX(2)X(3) to calculate the length of OX(3). 
With the sine-rule we get: 

(5)     OX(2) / sin(180 - 3*aplha) = OX(3)/ sin(2*alpha)

Since sin(180 - x) = sin x (for all x) and using (4) we get:

(6)     ( X(1)Y / sin(2*alpha) ) / sin(3*alpha) = OX(3) / sin(2*alpha)

equals:

(7)     OX(3) = X(1)Y / sin(3*alpha)


Suppose 3 * alpha = 90. This results with (7) in:

        OX = OX(3) = X(1)Y/sin(90) = X(1)Y

In general this means that:

(8)     OX(n) = X(1)Y / sin(n*alpha)

knowing that alpha = 90/n we got what we wanted: 

(9)     OX = OX(n) = X(1)Y/ sin(90) = X(1)Y.

        QED (Quod erat demonstrantem)

Ken's solution and comments:

The above solution is a good rigorous proof that the original distance is OX. I have a more esoteric description, that is basically an "unfolding" of the ray's path to show that it is always at a perpendicular distance OX from O.

Label backwards from X the bounces of the ray (W, V, U, T, etc.). Now reflect X over OW, by finding X' such that OWX' is congruent to OWX. It's easy to show that X', W, and V are colinear (angle VWU = angle OWX'), so we now have a right triangle OX'V showing that the perpendicular distance from ray VW to O is OX.

Now reflect X' over OV to find X", such that OVX" is congruent to OVX'. Again, X", V and U are colinear, so right triangle OX"U shows that the perpendicular distance from ray UV to O is OX.

By reflecting X across the two mirrors in this way, the angle at O continually increases by alpha. After one less than n reflections, the angle at O will be 90 degrees, and other leg of the "triangle" will actually line up with the original ray, parallel to the other mirror, showing that it's original distance was also OX.

Any comments on this solution? Feel free to write.


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