Bouncing Barney

Given triangle ABC, Barney is standing against one wall in a
triangular room.

Barney walks to a 2^{nd} wall (wall AB) by moving
parallel to the third wall(wall AC).

Once Barney gets to this wall he turns and walks to the third
wall AC by moving parallel to wall BC. This process continues as Barney walks
from one wall to the next by moving parallel to the third wall.

Do you think that Barney will return to his starting point? If
so, how many times will he intersect each side?

What if Barney starts from another point on BC, will the result
remain true? LetŐs see.

Yes, Barney seems to always return to his starting point and
crosses each side twice in the process.
Now letŐs investigate some findings when Barney starts at various
points.

Here Barney starts at the midpoint of BC. Notice, that Barney
still returns to the starting point, but only intersects each side of the
triangle once. Also, it appears that Barney also intersects the other two sides
at their midpoint. Additionally, it appears that BarneyŐs path divides triangle
ABC into 4 congruent triangles. LetŐs see if this is true.

Yes, it
appears that my conjecture may be correct. But remember that GSP measurements
are not proofs.

What if
the starting point was one third the distance of BC, does anything noteworthy
happen?

It
appears that the paths create 9 congruent triangles, letŐs measure to see if my
conjecture is correct.

Yes, the path creates 9 triangles
of equal perimeter and area and also trisects each side of the triangle.
Lastly, what if the starting point was one fourth the distance of BC?

This
is very intersecting, the path divides the triangle into 3 congruent trapezoids
and 7 congruent triangles.

As I explore the different starting points
of Barney, I notice that when Barney starts inside the triangle (or on one of
the vertices), the distance he travels to reach his starting point is equal to
the perimeter of the original triangle. LetŐs prove that here:

Let the starting point be D and H, M, N, O and Q
represent the points where Barney reaches the different sides of the triangle.
So, the segment paths that Barney travels are DH, HO, OQ, OM, MN, and MD, in
that order. Let J, L and P be the intersections of BarneyŐs travel.

I know that AB = BM+MA. Since BMHD is a parallelogram,
BM=HD and since MAOQ is a parallelogram, MA=OQ. So, by substitution, AB=HD +
OQ.

Similarly, I know that AC= AN+NC. Since ANDH is a
parallelogram, AN=DH and since NCQM is a parallelogram, NC=QM. So, AC= DH + QM.

And finally, I know that BC = BQ+ QC. Since BQOH is a
parallelogram, BQ=OH and since QCNM is a parallelogram, QC=MN. So, BC=OH+MN.

Therefore, AB+BC+AC= HD + OQ+ DH + QM+ OH+MN.

Another exception to this proof, is when the starting
point is the midpoint of a side, in which case the sum of the paths is half of
the perimeter of the triangle. _{}

What happens if Barney starts on a point outside of the
triangle? LetŐs see:

The path
encompasses the triangle, but it does return to the starting point. By using
directed measuresments, it can be determined that the path of Barney outside of
the triangle, is also the perimeter of the original triangle.