Suppose you have an xy coordinate plane with two circles both with a radius of one, centered at (-2,2) and (2,2). You have a line segment with one endpoint at (0,0) and forms an angle (∠a) with the x-axis. The line segment continues until it intersects one of the circles and then “bounces off” -that is it forms a new line segment with an endpoint at the intersection, forming an equivalent angle with the tangent of the circle at that point to the angle formed by the original segment and that tangent (as if the segments show the path of light and the circles are mirrors). The new segment extends until it intersects a circle and “bounces off” to form a new segment and so on. For what measure of ∠a are there infinitely many intersections? (The light never stops bouncing).

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    $\begingroup$ Intuitively, there is no solution. The only way the light can bounce forever is if it starts at $(1,2)$ and bounces to $(-1,2)$ and back. But in order to start from there, it must have been able to get there. $\endgroup$ – Andrew Chin Dec 27 '19 at 23:03
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    $\begingroup$ @AndrweChin Not at all. It can bounce closer and closer to those two points, never quite getting there, never turning back. $\endgroup$ – Arthur Dec 27 '19 at 23:37
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    $\begingroup$ I guess you meant ask whether there are infinitely many intersections? Mathematics tends to want precision in the use of language . . . $\endgroup$ – Lubin Dec 28 '19 at 0:48
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    $\begingroup$ A pleasing question. Seems to me that $\forall N>0$, there is a closed interval $I_N$ consisting of rays $\alpha$ that reflect $\ge N$ times from the right-hand circle. So the number of possible reflections would be unbounded. Furthermore, you have $I_N\supset I_{N+1}$ for all $N$, so that there would be a nonempty intersection of all, presumably consisting of one critical $\alpha_\infty$.But I can’t see what happens to the ray emanating from the origin with angle $\alpha_\infty$ , although you’d think that there would be infinitely many reflections in this case. $\endgroup$ – Lubin Dec 28 '19 at 5:43
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    $\begingroup$ @toritoverdejo There are several ways to see how this works. Lubin above has one explanation. Alternately, consider the interval of angles between aiming at the center of the right circle, and aiming at $(1,2)$. For each beam, if it only reflects a finite number of times, consider whether it eventually disappears upwards or downwards. The interval of angles that result in the beam disappearing upwards has an infimum, and the interval of angles that result in the beam disappearing downwards has a supremum. Those two bounds are equal. What happens if you aim at that exact angle? $\endgroup$ – Arthur Dec 28 '19 at 18:54

I wrote a Mathematica routine to compute the number of bounces as a function of angle $\theta$, assuming the light beam first strikes the circle on the right, at point $(2-\cos\theta,2-\sin\theta)$.

Using machine precision I found a maximum of $27$ bounces for some values in the interval

$$ 0.40923372682221769905317<\theta< 0.40923372682221769905375 \quad\text{(in radians).}$$

The limiting value leading to infinitely many bounces must then lie within this interval.

I could improve this result (at the expense of longer execution time). Notice that angle $\angle a$ defined in the question can be obtained from $$ \angle a=\arctan{2-\sin\theta\over2-\cos\theta}. $$

  • $\begingroup$ @Robo300 $\arctan{\sqrt3\over4}\approx0.408637855097592422374188$, which is outside the interval I found. $\endgroup$ – Intelligenti pauca Jan 4 '20 at 21:08
  • $\begingroup$ How do you prove that such a limiting value exists? $\endgroup$ – Sophie Jan 5 '20 at 23:19
  • $\begingroup$ @Sophie See comments above by Lubin and Arthur. $\endgroup$ – Intelligenti pauca Jan 6 '20 at 7:47

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