Bouncing Barney.


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.



The following figure shows Barney's path around a random triangular room ABC.   Notice that in 5 turns (6 lengths) he arrives back at his original starting point. 





Is this true for all starting points on segment BC, or does the figure above merely represent a special case?  Notice in the next figure, that as the starting point moves along BC, this observation still holds.





However, when the starting point lies coincident with vertex B, it only takes Barney 2 turns to arrive at his original starting point. 





The same is true when the starting point is the midpoint of segment BC.  It only takes Barney 2 turns (3 lengths) to arrive back at his original starting point.






Click here for an interactive java sketchpad exploration and examine this problem more.  Move the starting point along segment BC and observe the change in the parallel paths.  What happens when the starting point lies on a vertex of the triangle?  What about when the starting point coincides with the midpoint?

Now let's observe what happens when Barney's starting point lies exterior to segment BC.


Our conjecture seems to still hold.  Barney makes 5 turns before returning to his original starting point.

Use this interactive java sketchpad exploration to observe what happens when the starting point lies on the ray opposite BC.  What happens as you move this starting point?

When Barney walks on a path parallel to the sides of the room, he will always return to his starting point.  It also appears that as Barney walks his path creates a similar triangle at each vertex.

Let's look at ways to prove our conjecture holds true.  One way is by examining the parallelograms created by Barney's path.


The perimeter of these parallelograms, when the segments that are part of the triangle are excluded, are equivalent to the perimeter of the triangle.  Barney will always return to his starting point since his path is the same measure as the triangle.


If you would like to explore this proof more on your own, click on this interactive java sketchpad exploration.  Drag the starting point and notice what happens the measurements of the perimeters of the triangle and the parallelogram.

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