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Consider a semicircle with diameter $AB$. A beam of light exits from $A$ at a $58^{\circ}$ to the horizontal $AB$, reflects off the arc $AB$ and continues reflecting off the "walls" of the semicircle until it returns to point $A$.

How many times does the beam of light reflect of the walls of the semicircle (not including when it hits $A$ at the end)?

Note: "walls" of the semicircle refer to the diameter $AB$ unioned with the arc $AB$.

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  • $\begingroup$ Are we sure it will ever come back to $A$? $\endgroup$ – imranfat Mar 14 '16 at 3:56
  • $\begingroup$ Yes the angle is chosen so that it does. $\endgroup$ – Joshua Benabou Mar 14 '16 at 3:57
  • $\begingroup$ Can the laser bounce of segment $AB$ should it hit the diameter? $\endgroup$ – imranfat Mar 14 '16 at 3:58
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    $\begingroup$ Yes $AB$ is considered a wall of the semicircle. $\endgroup$ – Joshua Benabou Mar 14 '16 at 3:58
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    $\begingroup$ Well I find that, if we replace the semicircle with a circle, we get that the beam hits the circle at points which forms 64 degree arcs. $\endgroup$ – Joshua Benabou Mar 14 '16 at 4:22
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I cheated and used GeoGebra. It appears to reflect 29 times (unless my old eyes miscounted) until it comes down vertically to point C. At that point it will reverse path for a total of 59 reflections until it returns to A.

Image from GeoGebra construction of problem

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By reflection across the diameter we can map what happens to the beam inside inside the semicircle to a full circle; them beam contacting the diameter and bounces off in the semicircle is equivalent to the beam passing through the diameter of the full circle.

We know that when the beam hits a wall, it reflects so that the angle of incidence equals the angle of reflection. By a simple angle-chase, we get that the beam hits the circle at points which forms 64 degree arcs.

After hitting the wall $n$ times the beam is at point $P$ on the circumference with $∠AOP=64n$. The beam first returns to $A$ when $360|64n$ or $n=45$. Thus the beam contacts the arc $AB$ 44 times, because we don't count when it hits $A$ at the end. The beam crosses from the upper half-circle to the lower half-circle or vice verca $64*45/180=16$ times, however this counts includes when the beam hits $A$ at the end, so it actually crosses $15$ times. The total numbner of contacts with the walls is then $44+15=59$.

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