 BOUNCING BARNEY

Part (A) OF Final

By:  Gloria L. Jones

Barney is in the triangular room.  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 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 on line exterior to segment BC.  Discuss and prove any mathematical conjectures you find in the situation. In this write-up, we will explore some properties or postulates that will be identified and useful in proving conjectures in this exploration.  Look at the sketch above.  Starting at the left point on segment line BC, Barney will follow the red dashed lines as described above and will eventually return to where it started.

Barney accomplishes this return by moving parallel to one of the sides of the triangle every time it hits a wall.  In moving in this type of pattern, Barney will hit a wall of the triangle five times before it returns to its starting point.  In fact, its path will form a hexagon as shown below: Because this hexagon was formed by parallel paths cut by yet other parallel lines, there are some special properties or postulates in this sketch.  One of which is when two parallel lines are cut by a transversal line, then corresponding and alternate interior angles are equal.  The same-side interior angles are supplementary and the alternate exterior angles are equal.  So, in the diagram, we are looking at six equal triangles in the interior of ΔABC formed by Barneyπs parallel paths.

Notice that these six triangles are similar triangles to the original triangle.  These triangles share the same angle measurements.  Outside of the hexagon are three triangles in light blue left.  These triangles are equal in measurement as those found in the hexagon.  Also, when we extended the lines beyond the boundaries of the original triangle, more equal triangles are formed that are similar to ΔDEF.  It should not be a surprise then that the perimeter of ΔABC and the perimeter of TWXUVS are both equal to 19.92cm.  Because of this phenomenon, Barney will always return to the starting point.

Letπs take a look at this sketch from the area perspective: Notice that the hexagon shaded in light and dark green measure a total of 9.74cm² and the total area of ΔBAC is 14.61cm².   The ratio of the hexagon to the total triangle is 2/3 and the three equal triangles outside of the hexagon measures about 1/3 of the total area.  Barneyπs path in dark green makes up about a third of the total area as well.

Another method for showing the equality of the perimeters, is measuring the perimeter of the parallelograms found in the interior of ΔABC compared to the perimeter of ΔABC: When the interior triangles are aligned at the centroid, the parallelograms actually span the entire area of the original triangle.  For each segment on the perimeter of the original triangle, there is a segment along Barneyπs path with the same length.

Now, letπs contrast the path of Barney when the starting point is outside of segment BC: Again we can see that even starting from a point outside of the segment BC, Barney will return eventually to its start.  Also as the ΔABC collapses it appears as if it becomes the area of the path.  Nonetheless, because the triangles are similar, Barney will follow the same path along the segments which will bring it right back to wherever it started from.

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