Bouncing Barney

Barney is in the triangular room and 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.

We want to show that Barney will eventually return to his starting point.  We will consider the starting point as point S.

Since line segment ED is parallel to line segment BC, we can consider line segment SD as a transversal through two parallel lines.  Therefore, angle DSC is congruent to angle EDS because they are alternate interior angles.

Since line segment ED is parallel to line segment BC, we can consider line segment EF as a transversal through two parallel lines.  Therefore, angle BFE is congruent to angle DEF by alternate interior angles.

Since line segment FG is parallel to line segment AB, angle GFE is congruent to angle BEF by alternate interior angles.

Since line segment BC is parallel to line segment HG, angle GFC is congruent to angle FGH by alternate interior angles if line segment FG is the transversal.  Since line segment FG is parallel to line segment AB, angle FGH is congruent to angle AHG by alternate interior angles when line segment GH is the transversal.

Since line segment BC is parallel to line segment GH, angle ABC is congruent to angle AHG by corresponding angles.

Since line segment DS is parallel to line segment AB, angle ABC is congruent to angle DSC by corresponding angles.

Since line segment SH is parallel to line segment FE, angle EFB is congruent to angle HSB by corresponding angles if line segment BC is the transversal.

Since line segment SD is parallel to line segment AB, angle AHG is congruent to angle DIG by corresponding angles with line segment HG as the transversal.  Angle DIG is congruent to angle HIS by vertical angles.

Since line segment ED is parallel to line segment HG, angle DEF is congruent to angle EJH by alternate interior angles.  Angle EJH is congruent to angle GJF by alternate vertical angles.

Since the sum of the interior angles of a triangle is 180̊, angle EFG is congruent to angle SKF.

Since line segment EF is parallel to line segment AC and line segment HS is parallel to line segment AC, line segment EF is parallel to line segment HS.  Therefore, if we consider line segment DS as the transversal, angle SKF is congruent to angle HSD.

Since angle BFE, angle EFG, and angle GFC forms a straight line, the sum of their angles is 180̊.  Therefore, the sum of angle BSH, angle HSD, and angle DSC is also 180̊.  Since these angles form a straight line, they intersect at the common point S.  Thus, Barney will return to his starting point.

Does the path create a pattern of similar triangles?

If we finish labeling the angles of all of the triangles by using properties of parallel lines with transversals, we can see that all of the triangles have the same angle measures.  Therefore, any triangle formed by Barney’s path will produce similar triangle to the original triangle ABC.

What is the distance that Barney will travel?

First, let’s show that triangles BHS, EAD, and FGC are congruent.

Since line segment HS is parallel to line segment AC and line segment AH is parallel to line segment SD, they form parallelogram ADSH.  Therefore, line segment AD is congruent to line segment HS.

Since line segment BC is parallel to line segment GH and line segment AC is parallel to line segment HS, they form parallelogram HGCS.  Therefore, line segment HS is congruent to line segment GC.

Since triangles BHS, EAD, and FGC are similar and they have a side length in common, they are congruent by angle-side-angle.

Therefore, line segment AB=DS+GF, line segment BC=GH+DE, and line segment AC=EF+HS.

The distance that Barney travels is equal to

SD+DE+EF+FG+GH+HS

= (SD+GF)+(GH+DE)+(EF+HS)

=AB+BC+AC

Therefore, the distance that Barney travels is equal to the perimeter of the triangle.

What if Barney started at the midpoint of the triangle?

When Barney starts at the midpoint of the triangle, his path only forms four triangles with its path.

By labeling the angles using the properties of parallel lines with transversals, we see that all four of these triangles are similar.

By using a similar argument as the one used to show that triangles BHS, EAD, and FGC are congruent, all four of these triangles are also congruent.

Since these triangles are congruent, we can say that line segment DS=1/2(AB), DE=1/2(BC), and ES=1/2(AC).

Therefore, the distance that Barney travels is

DS+DE+ES

=1/2(AB)+1/2(BC)+1/2(AC)

=1/2(AB+BC+AC)

Thus, the distance that Barney travels is equal to ˝ of the perimeter of the triangle.

What happens when the starting point is on the centroid?

When Barney starts at the centroid of the triangle, his path forms nine triangles with its path.

By labeling the angels using the properties of parallel lines with transversals, we see that all nine of the triangles are similar.  In particular these triangles are congruent.  We know this by looking that the properties of the parallelograms formed by Barney’s path.

Since these triangles are congruent, we can say that line segment DK=1/3(AB), EI=1/3(AC), and ED=1/3(BC).

Therefore, the distance Barney travels is

DK+EI+ED

=1/3(AB)+1/3(AC)+1/3(BC)

=1/3(AB+AC+BC)

Thus, the distance that Barney traveled is equal to 1/3 of the perimeter.

What happens when Barney starts on the orthocenter of the circle?

If Barney starts at the orthocenter of the triangle, his path looks very similar to the path he took at his original starting point.  He produces a path that forms many similar triangles, and he has traveled the distance of the perimeter of the triangle.