Summer of Math Exposition

I guess exammple isnt very fortunete or maybe the rules at the begining are presented not right, because there are ways or ball to bounce odd amount of times and end up in the same place

I really liked the formulation of this problem as two friends who are playing billiards and wondering about cycles. It makes for a nice way to start simple and ramp up complexity. I also liked the way you introduced the final solution on a unit square and then adapted to fit the original problem. My favorite part was wondering if there is a cycle that hits only three edges of the pool. I found the question well motivated and the proof was very elegant. A close second was the synchronous animation of the straight line extending over the entire plane and the reflected line in the singular pool table. If I were to offer any criticism, I would say that the jump from equivalence to the generic case of any rational slope could have been more well motivated.

This is a really fantastic video. Beautiful to watch, great pacing, clearly explained, and nicely motivated. Well done!

This video was very creative, enjoyable and entertaining. It addressed a simple problem through create design and animations. I wonder if there is any further readings you could have shared with the audience at the end. Very well done team.

Goal Orientation: 2/10 Novelty: 1/10 Thought-Provoking: 0/10 Comprehensibility: 1/10 Technological: 3/10 Overall Average: 1.4/10

It seemed like it explained the topic well. It would have been nice to see some cooler or more advanced applications of the topic, though: it's not really explained as being that important. That said, it is of course billiard balls.

Nice! I like the tessellation, trick, that's a great visual way to think about this. I think there is too much emphasis on notation and formal proofs though. For example, explicitly mentioning the rectangular table case - it's intuitively obvious that you can just stretch the table and things are going to work the same way, I don't think the viewer needs to be convinced of that. And for the infinitely many periodic trajectories – that seems intuitively true already from the three examples you give at 0:50. It seems logical that you can just increase the number of vertical bounces per horizontal bounces arbitrarily (or use any rational number). I'd lean more into the visual aspect of the problem. On the infinite tessellation plane, my understanding is that for every square with the same orientation as the starting one (i.e. not flipped), there is a corresponding periodic trajectory. But if you count every square, you get duplicates because e.g. the square at (8, 6) gives the same trajectory as (4, 3) - that's because the fraction 8/6 can be reduced to 4/3. That's a lot more intuitive than working with formulas explicitly!