Hearing the shape of right triangle billiard tables

Figure 2.1.  After a reflection, $ P_{n} $ becomes $ P_{n+1} $ in Case ($ i $) and $ P_{n-1} $ in Case ($ ii $)

Figure 2.2.  From $ B(\gamma^+) = ({r}, {b}, {r}, {g}, {b}, {r}, {b}, {r}, {b}, {g}, {r}, \cdots) $ to get the return sequence $ R(\gamma^+) = (1, 2, 1, 2, \cdots) $ and then to the get the level sequence $ L(\gamma^+) = (0, 1, 0, -1, 0, \cdots) $

Figure 3.1.  The visual diagram of $ \theta_{n} $, $ \theta_{n+1} $ and $ \theta_{n+2} $

Figure 3.2.  Cases ($ i $) and ($ ii $) correspond to $ LM_n = ({r}, {g}, {r}) $ and $ \theta_{n}\in(\alpha, \pi-\alpha) $; Cases ($ iii $) and ($ iv $) correspond to $ LM_n = ({r}, {b}, {r}) $ and $ \theta_{n}\in[0, \alpha)\bigcup\, (\pi-\alpha, \pi] = (-\alpha, \alpha)\, (\bmod\, \pi) $

Figure 3.3.  In Case ($ i $), $ \theta_{n_0} = \pi-\alpha $. In Case ($ ii $), $ \theta_{n_0} = \alpha $

Figure 3.4.   

Figure 3.5.   

Figure 3.6.  The left-hand part is a right triangle billiard table $ Rt $ and a periodic orbit: $ p_{0} $ $ \longrightarrow $ $ p_{1} $ $ \longrightarrow $ $ p_{2} $ $ \longrightarrow $ $ \cdots $ $ \longrightarrow p_{9} $ $ \longrightarrow $ $ p_{10} $ $ \longrightarrow $ $ \cdots $ in $ Rt $. The right-hand part is the unfolding floors of the corresponding rhombus of $ Rt $ along the periodic orbit

Figure 3.7.  The left-hand part is a right triangle billiard table $ Rt $ and a periodic orbit: $ p_{0} $ $ \longrightarrow $ $ p_{1} $ $ \longrightarrow $ $ p_{2} $ $ \longrightarrow $ $ \cdots $ $ \longrightarrow $ $ p_{6} $ $ \longrightarrow $ $ \cdots $ in $ Rt $. The right-hand part is the unfolding floors of the corresponding rhombus of $ Rt $ along the periodic orbit