## Posts Tagged ‘**angles of incidence**’

## Round 2: Problem 5 Solution

I spent a tiny bit of time on this problem, because the solution did not come immediately to mind.

Ultimately, realizing that being that the angle of incidence equals the angle of reflection helped to come up with a bit of magic that helps visualize the path. Because the angles are equal, we can unwind the path that the puck takes into a straight line by “mirroring” the rink. This will allow us to solve the problem with just a tiny bit of high school trigonometry.

Essentially what we have done is mapped the bouncing path into a straight one. For the sake of rigorousness we could show that this straight line is a bijective mapping of a bouncing path, but I will assume that it is trivial for our purposes and trust that if you don’t believe me that you trace it out on one of the unused boxes.

This path represents a trajectory that bounces off each of the four walls and arrives in the “goal.” To solve the problem, all we need is the angle . We know that (because both the puck and goal are 1m away from the walls) and we also know that , therefore , or since the problem asks for degrees, .

What is interesting is that we can modify this method to work with an arbitrary amount of bounces and end up at any point along the rink. Check out some further reading on the topic of Dynamical Billiards.

Further Reading:

The Geometry of Billiards (Pre-print version) (PDF, S. Tabachinkov [Penn State])

Running Header: The Diamond System (PDF, R. Miller [UNCC])