Question 1

By Krista Floer

When I started working on this problem, I did not know which question to answer first. So I just moved the starting point around to see if I could notice anything. I found that no matter where the starting point on BC was, Barney would always return to the starting point. I also noticed that he had to touch each side twice. There was only one case that I could find where Barney would not have to touch each side twice. This is discussed later. Consider these pictures:

Random Point on BC Point on a Trisection Point Point on the Midpoint

We can see that Barney returned to the start no matter where the starting point was. But how can we prove this?

Let us prove that Barney will always return to his starting point just for the general case.

Proof:

 Consider the parallelogram Bfab. By definition of a parallelogram, Bf = ba. Now let M be the point where Barney finishes his path. This point has to be on the segment AC because of how Barney must travel. MbBf must be a parallelogram by construction. This gives us that Mb = Bf.

Since ba = Bf and Bf = Mb, by the transitive property we have ba = bM. Since the segments are equal, then the points a and M must be the same point. Hence, Barney always stops where he started from.