**Part A: Bouncing Barney**

*A. Bouncing Barney. We discussed this investigation in class. Your
challenge now is to prepare a write-up on it, exploring the underlying mathematics
ideas and conjectures.*

**Barney is in the triangular room shown here. He walks from a point
on BC parallel to AC. When he reaches AB, he turns and walks parallel to BC.
When he reaches AC, he turns and walks parallel to AB. Prove that Barney will
eventually return to his starting point. How many times will Barney reach a
wall before returning to his starting point? Explore and discuss for various
starting points on line BC, including points exterior to segment BC. Discuss
and prove any mathematical conjectures you find in the situation. **

Who's your favorite Barney?

To solve this problem, I first made this
GSP sketch which models poor Barney's situation. It seems that if Barney
(who starts at point **P1** and progresses to **P6**
before he returns to **P1** again) always touches the same six
points on the triangle before he reaches his start, no matter where he begins
(except for two special cases I will discuss below). Can we prove this? You
bet!

**P1P2** is parallel to **BC**, so by
the angle angle similarity theorem, triangle **AP2P1** is similar
to triangle **ABC**. Therefore **length(AP2)/length(AB) =
length(AP1)/length(AC)**. Similarly, since **P2P3** is parallel
to **AC**, triangle **BP2P3** is similar to triangle
**BAC**, and so **length(AP2)/length(AB) = length(CP3)/length(BC)**.
Following this pattern we find that:

**length(CP3)/length(BC) = length(CP4)/length(AC)**,

**length(CP4)/length(AC) = length(BP5)/length(AB)**,

**length(BP5)/length(AB) = length(BP6)/length(BC)**,

We are trying to prove that the next point, **P7**
is equal to **P1**. Still following the pattern, we get:

**length(BP6)/length(BC) = length(AP7)/length(AC)**.

Our long chain of equalities, however, tells us that **length(AP1)/length(AC)
= length(AP7)/length(AC)**. Since **P1** and **P7**
both lie on the segment **AC**, **P1 = P7**, and so
Barney is in a six stop cycle, unless he started on a point mentioned in the
two cases below.

**Case 1:** Barney starts on a midpoint of any side
of the triangle, say the midpoint of **AC**.

If Barney walks parallel to **BC**, he will reach
the midpoint of **AB**, because the line connecting the midpoints
of two edges of a triangle is parallel to the third edge. Similarly, when Barney
leaves the midpoint of **AB**, he will next reach the midpoint
of **BC**, then the midpoint of **AC**, where he started.
If Barney leaves from a midpoint, he thus only gets to see two points. He would
have a much more pleasant trip if he took a scenic route not beginning at a
midpoint. Poor Barney.

**Case 2:** Barney starts on a vertex of the triangle

In this case Barney walks around the edges of the triangle.

If Barney starts on a point not on a point on the line through
**AC** but not on the segment **AC** and moves parallel
to **BC**, he must go towards **AB** if he wishes
to change his direction again. If he does this, he will see six different point,
as in the usual case. The proof of this fact is similar to the proof above.