Triangular Pool Bounce

We can visualize the ball travelling in a straight line through a series of reflected isosceles right triangle pool tables with legs that are $1$ unit long, and place the result on a coordinate grid, as shown below:

Then the goal would be to draw a straight segment from $(0, 0)$ to one of the red points without going through a blue point. However, since all of the red points are on points in which both coordinates are even, its $x:y$ ratio can be simplified by a greatest common power of $2$ factor to give new point with the same $x:y$ ratio where at least one coordinate is odd (a blue point). In other words, it is not possible to draw a straight segment from $(0, 0)$ to one of the red points without going through at least one blue point. Therefore, it is not possible to hit a ball that reflects back to where it started.

We can think of the table borders as mirrors and the ball trajectory as the light. Then we project the everything that's in front of the border to behind the border. If we do this process of reflecting over and over and follow a straight line from the origin, we can start to see when the ball would hit the origin point. The origin point on this case is just where the yellow and blues lines meet. So we just follow the ball trajectory. We have points in the border wall itself, but we can't hit these because the rules don't allow (it is the same as throwing the ball parallel to the border). So we have left the green points in the picture. But notice that in order to hit them, we always pass through a encounter of two lines (the black 'x' in the pictures). But we can't do that too because it's the same as hitting a vertex! This is actually just a way to visualize a geometric pattern that will repeat itself if we expand the representation many times more. It's no hard proof. But it's a precise representation of the trajectory, and from it we can try to work on a prove. It would probably involve writing the functions of all the lines in the form of y=mx+b; and finding where they cross each other (y=y). Then we show that every vector connecting the origin to the points where the yellow and blue lines meet (the return point) always passes through a vertex first. I didn't have the time now to try to formalize a proof but I think this visualization might be useful.

Let the initial velocity vector be $(u, v)$ . In order for the ball's motion to be inside the triangle, necessarily $0 < u < v$ .

The left, top and hypotenuse cushions give the velocity mappings $(x, y) \rightarrow (-x, y)$ , $(x, y) \rightarrow (x, -y)$ and $(x, y) \rightarrow (y, x)$ respectively. Thus the only possible velocities at a given time are $(u, v)$ , $(-u, v)$ , $(u, -v)$ , $(-u, -v)$ , $(v, u)$ , $(-v, u)$ , $(v, -u)$ and $(-v, -u)$ . Of these, only $(-u, -v)$ is possible when the ball is returning to the vertex it started at.

As such, in order to return to the vertex the ball must travel back along its path. Since the physics are invariant under time reversal, the second half of the journey will be the time reversal of the first half. As such, at the halfway point the ball would have to reverse direction instantaneously.

However, this is impossible:

• To reverse direction by bouncing off the left cushion would require $y = 0$ . Since $0 < u < v$ , no such velocity is reachable.
• To reverse direction by bouncing off the top cushion would require $x = 0$ . Since $0 < u < v$ , no such velocity is reachable.
• To reverse direction by bouncing off the hypotenuse cushion would require $x = -y$ . This would have to come from one of the velocities where the components have opposite signs. But at a given time, $|x| = u$ and $|y| = v$ or vice versa, so we would require $u = v$ , which we know isn't the case.

Just draw the picture, assume a direction firstly, then you can draw each point where the ball changes its direction according to the pattern of reflection. Once you have done that for four times, you can find out that the direction of the third reflection parallel to the first direction you assume. And now, you are very close to the final answer, because what you should do now is only to regard each three reflection as a circulation. And the only difference beteewn two circulation is the length of the ball's routine, all the directions in each circulation are same. So the just go deeper and deeper into the right angle but will never achieve it.

Let's call $a$ the clockwise angle between the y axis and the current trajectory, then

$l(a) = -a$ mapping of the left mirror

$t(a) = 180°-a$ mapping of the top mirror

$d(a) = 90°-a$ mapping of the diagonal mirror

So the initial angle $a_0$ is only changed by compounding the above transformation into a few possible angles.

The trajectory cannot retrace itself back to the origin because $a=0° and 45°$ are not admitted for $a_0$ , so multiples of 45° will never be reached compounding the transformations, so the transformations above will never behave as a retracement $f(a) = a+180°$

The alternative is a trajectory that doesn't retrace itself but eventually ends up at the origin. This is absurd thought because will require a final angle $a*=180+a_0$ given the finite options of the compound transformations, but this will mean retracing the first segment.

You gotta Have a nice reasoning to solve it At any number of step ,(considering that the ball returns to the original position) the coordinates must change (x,y) |------>(y,x) for an odd number of times considering the two sides two be the x and y axes we see that such change is never possible(even once).

No matter how many times the triangle is folded, the ball always passes from one vertex of the big triangle to the other vertex, and the ball cannot move along the edge, so this is impossible.

Let us consider a light ray instead of the ball and the angle of release (angle made with the perpendicular of ∆) as ø. After 1st reflection, the angle made by the reflected ray with the Hypotenuse would be (3π/4-ø). For the ray to retrace it's path, this angle should be equal to π/2 which isn't possible (as ø cannot be equal to π/4). Therefore, the ball will never get back to it's starting pt.