Let B and C be the first and second bounce points.
The path will be ΔABC. We wish to find ∠OAB. Let
p = |OA|. We will assume that the angle of incidence
equals the angle of reflection at a bounce. Therefore,
∠ABO = ∠OBC and ∠BCO = ∠OCA.
Since ΔBOC is isosceles,
∠OBC = ∠BCO.
Let φ = ∠ABO = ∠OBC = ∠BCO = ∠OCA.
Using the law of sines on triangles OAB and OAC,
r p p
----------- = ------------ = ---------
sin(∠OAB) sin(∠OBA) sin(φ)
p r
= ----------- = ------------
sin(∠OCA) sin(∠OAC)
Let θ = ∠OAB = ∠OAC.
Summing the interior angles of ΔABC gives
θ = 90° - 2φ
Applying the law of sines again to ΔOAB gives
r p
-------- = -------
sin(θ) sin(φ)
or
r*sin(φ) = p*sin(θ)
= p*sin(90° - 2φ)
= p*cos(2φ)
= p*[1 - 2*sin2(φ)]
or
2*p*sin2(φ) + r*sin(&phi) - p = 0.
Therefore,
sin(φ) = [√(r2 + 8*p2)-r]/(4*p).
Thus,
∠OAB = 90° - 2*arcsin{ [√(r2 + 8*p2)-r]/(4*p) }.
QED
Note: If p=r, then ΔABC is equilateral;
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