Bouncing
Barney
By
Taylor Adams
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.
To
explore Bouncing Barney’s path, click here.
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.
