Iran TST 2007 Day 2 P1 wrote: In an isosceles right-angled triangle shaped billiards table , a ball starts moving from one of the vertices adjacent to hypotenuse. When it reaches to one side then it will reflect its path. Prove that if we reach to a vertex then it is not the vertex at initial position By Sam Nariman Solution: Let the given triangle be $\Delta ABC$ right angled at $B$ & Assume, the ball starts from $A$, then it must hit & bounce off $\overline{BC}$ in the end, in order to reach $A$ back. Say, the ball hits the sidelines of $\Delta ABC$, $n$ times, at $X_1,X_2, \ldots , X_n$. We know, $X_1,X_n $ $\in$ $\overline{BC}$. Also, $AX_1,AX_n > AB$. When, $BX_1 < CX_1$, keep reflecting $\Delta ABC$ over $\overline{BC}$ such that the ball traces a straight line. Build a coordinate system with $AB$ as $X-$axis & $BC$ as $Y-$axis. Let, $A$ $\equiv (-1,0)$, $B$ $\equiv$ $(0,0)$ & $C$ $\equiv$ $(0,1)$. Suppose, the straightline- trace endpoint is $T$, Then, due to $BX_1 < CX_1$ $\implies$ $T \equiv (t,1)$. Join the line segments, $(2x-1,0),(2x-2,1)$ and $(2x-1,0),(2x,1)$ and $(x-1,0),(x-1,1)$ for $x$ $\in$ $\mathbb{N}$. Then, due to reflection argument, the number of times $AT$ hits these line segments is $n$. Let these new hit-points be $Y_1,$ $Y_2,$ $\ldots$ $,Y_n$. Also, it's clear, $X_1 \equiv Y_1$. Now Due to Congruence, $$Y_iY_{i+1}=X_iX_{i+1} \implies X_nA=Y_nT > AB$$Clearly, when $t$ is odd $\implies$ $Y_nT$ $<$ $AB$, because $Y_n$ $\in$ $\overline{(2x-1,0),(2x-2,1)}$. When $t$ is even $\implies$ $Y_nT$ $=$ $X_nA$ $=$ $AX_1$ $>$ $AB$ $\implies$ $X_1$ $\equiv$ $X_n$. This fact implies, $X_2 \equiv X_{n-1}$ $\implies$ $\ldots$ $\implies$ $X_k $ $\equiv$ $ X_{(n+1)-k}$. Regardless of parity of $n$, this coincidence of point eventually lead that there exists a perpendicular bounce $\implies$ Contradiction! When, $BX_1 > CX_1$, the only difference is that $T \equiv (t,u)$ for $t > u > 1$. Now, Erase all the existing lines, Draw the lines only in the form of $x=k$, $y=\ell$, $x+y=2r+1$ and $x-y$ $=$ $2r+1$ for $k, \ell ,r \in \mathbb{Z}$ and only focus in the region enclosed by $y-x=1$ and $y=0$. When, $u$ is odd, the method demonstrated in $(t,1)$ works similarly too. When $u$ is even & $t$ is odd, we can use the method demonstrated in "$(t,1)$ when $t$ is even" and When $u,t$ both are even, we can use the method in "$(t,1)$ when $t$ is odd". Summing up, all cases lead contradiction & hence, the ball cannot bounce back in point $A$ $\qquad \blacksquare$

consider this shape we reflect the shape every time by the hypotenuse or the perpendicular side to get this shape. it is obvious that if the ball starts its journey through any point from the initial orange point on the lower left. it doesn't reach any orange point. note that the orange points are the initial vertex where the ball started its movements.