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**Bouncing Barney**

**By Mary
Negley**

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For this assignment, I will explore the
following problem:

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.

To
better understand this problem, I constructed the following GSP sketch.

I want to assume that, so. First I will
identify all the parallelograms in the figure. By construction, the following quadrilaterals are
parallelograms: *AFIH*, *GFIC*, *EDCH*, *EBIH*, *AFLG*, *GFMH*,
and *GFEK*.

By definition of parallelogram, opposite sides of a
parallelogram are not only parallel; they are also equal. So the following congruent segment
pairs occur:, , , and .

Now I want to focus on the parallelogram *GFIC*.

It is a parallelogram by construction and by
definition and . Notice that
(1)and (2). By
construction, *GLIH* is also a
parallelogram and by definition, (3)and (4) .

So we have

By (1) and (2)

By (4), we can substitute *GH* for *LI*
and we get

Subtracting *GH* from both sides yields

In our congruent segment pairs above, we saw that and . Now that we
know that, we can substitute *FL* for *HC* in the second
congruent segment pair, so we get. So we see that
and. Thus, , which is a contradiction, so our assumption must be
false. Therefore, and Barney
started at the same place that he ended.

Another
question that I can answer is given above: How many times will Barney reach a wall before returning to
his starting point? In my GSP
sketch, it appears that he reaches five walls before returning to his starting
point (or six, if one counts the wall that contains his starting/ending
point). Through exploration with
the GSP sketch given here,
one sees that this is the case unless Barney starts at one of the vertices or
the midpoint. If he starts at one
of the vertices, then he walks along the walls the whole time, so he either
does not reach a wall at all or always reaches a wall, depending on how one
looks at it. However, if he starts
at the midpoint then the starting/ending point is on the third wall that he
meets. What is the mathematical
reason behind that?

Below is a
sketch of the situation where Barney starts at the midpoint.

Since we
know the midpoint of *AC*, . Since *GCMidpointH* was
constructed to be a parallelogram, . Since *BGMidpointH* was
constructed to be another parallelogram, . By
transitivity, since and , . So *G* is the
midpoint of *BC*. By a similar
argument, *H *is the midpoint of *AB*. So we know that Barney reaches a wall at each of the three
midpoints of the sides of the triangle.
Notice that the triangle formed by BarneyÕs path is by definition a
medial triangle. Every triangle
has a medial triangle, so it makes sense that BarneyÕs path would be along the
medial triangle. So that must be
why he reaches three walls when he starts at the midpoint.