Carisa Lindsay
Final Project
Bouncing Barney
For any triangle ABC,
the problem arises when trying to figure out how many times Barney will touch
the sides of the triangle as he travels parallel to each side.
Using Geometer’s
Sketchpad, we can construct an arbitrary triangle with Barney beginning at some
arbitrary point D. We can see through
constructing parallel lines and intersections that Barney will hit the wall six
times.
However, this is not
the case when Barney begins at any vertex,
or when Barney begins
at the midpoint of any side.
What if Barney begins
at the centroid?
If Barney begins at
the centroid of triangle ABC, he will reach the wall only reaches twice before
returns back to his starting place.
However, he will not actually STOP at his at his starting place, but
will instead pass through the starting place.
As he passes through his original starting place, he will maintain a
similar path as before.
What is the fewest
number of times Barney can reach the wall and still return to his original
starting place? Let’s consider triangle
ABC as a right triangle.
What if Barney begins
at one of the vertices? If Barney begins
at a vertex, he will not move since he will be unable to move parallel to the
opposite side due to the boundary of the room.
Suppose Barney begins
his journey outside of ABC.
Regardless of where
Barney begins, we see a similar pattern with how he “bounces” around the
triangle. Again, he will hit the wall
six times before returning to a location he has already been to. However, Barney will never reach the point in
which he began because of being constrained within the boundaries of the
triangle. Much of this problem depends
on the “rules” for which directions Barney is allowed to travel.
This would be an
appropriate activity in a secondary geometry class studying parallel lines and
their transversals, as well as interior angles resulting from them. Furthermore, a physics class could use a
problem like this when studying angles of deflection of light.
So how far is Barney
bouncing?
The answer is not
obvious. First, we need to divide
Barney’s paths into various line segments so we can create parallelograms. Because these sections are parallelograms and
using the definition of parallelograms, we can see that various sides are
congruent.
Since these sides are
congruent, they remain congruent as Barney travels. As barney approaches a vertex, it is more
obvious that Barney is actually traveling the same distance as the perimeter of
the “room”.