Launching a particle to reflect off of the interior of a circle in exactly eight distinct points

Question

This question was recently asked on a National level Olympiad. It reads:

Consider a circle of radius $R$ centered at the origin. A particle is launched from the $x$-axis at a distance $d$ from the origin such that $0<d<R$ at an angle $\alpha$ with the $x$-axis. The particle is reflected from the boundary of the circle so that the angle of incidence equals the angle of reflection. Determine the angle $\alpha$ such that the particle strikes the interior of the circle at exactly eight distinct points. (The value of $\alpha$ is to be expressed in terms of $R$ and $d$)

My attempt:

I tried visualizing this problem in a trivial case where the particle strikes the interior exactly five times. Upon applying some elementary geometry, I came to know that the particle strikes the interior tracing congruent isosceles triangles $OA_i A_{i+1}$ where $A_i$ is the point of incidence. Since the particle traces the exact same path as it did in its first "launch" in its last strike (to strike at exactly 5 points), the particle traces a star like shape, where each "leg" of the star is congruent to another (since the triangles are congruent isosceles triangles). If we join the points of incidence, we obtain a regular pentagon and also see that the trajectory of the particle is the diagonals of the pentagon. Generalising this further, in case of 8 points, we obtain a regular octagon with the trajectory being the diagonals (excluding the main diagonals as they act as axes of symmetry). After this, the value of $\alpha$ can be computed easily by applying some elementary geometry and trigonometry.

Is this solution valid or does it need some modification?

Answer 1

There are two regular octagons that you can inscribe into a circle: an ordinary one, or a self-intersecting one (star-shaped, "octagram"). You can imagine that you draw one of those (or both - even if that would make a messy picture) and then rotate the octagon until one of its legs goes through the point $(r,0)$. At that point, you just need to let the particle go along that leg.

This implies that, depending on $r$, this may have $0$, $1$ or $2$ different solutions:

• If $r$ is "small", then you cannot position either of the octagons.
• If $r$ is big enough, you can position the star-shaped octagon, but maybe not the ordinary octagon.
• With $r$ quite close to the circle, you can position both the star-shaped octagon and the ordinary octagon.

Answer 2

use octagon $T_0T_1T_2T_3T_4T_5T_6T_7$ rotated by $\phi$: $PT_1$ is on the diagonal $T_5T_0$