Brenda King

Bouncing Barney

__Introduction:__

Since this problem is
often presented in middle school, I thought it would be nice to introduce
Barney as an actual character. Barney,
the bear, hears a noise outside his home which frightens him. He panics and starts running as fast as he
can go, without looking ahead, and finds himself bouncing off the walls. Barney is in the triangular room shown here.

Barney, a good math
student, wants to return to his starting point. 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. Barney will continue in this
manner for his entire journey.

Prove that Barney will
eventually return to his starting point.

How many times will
Barney reach a wall before returning to his starting point?

If we start Barney’s
journey on the line BC, at which point should the path begin?

__Starting point on a vertex__

If he starts at vertex B and
travels parallel to side AC, then the path would be outside the triangle and he
would not return to his starting point (see diagram 1). If he starts at vertex B and travels parallel
to side AB, then he must be traveling on side AB itself. He will stay on this path until he reaches
vertex A. At vertex A, he will turn
(#1), and continue down side AC. When he
reaches vertex C, he will turn (#2) and return to his starting point. In this case, Barney has returned to his
starting point and made two turns on the journey. The length of his path is the perimeter of ∆ABC.

In a likewise manner, if
he starts at vertex C and travels parallel to AB, then, again, the path will be
outside the triangle. (See dashed lines in diagram 1). When he travels parallel to AC, he will be on
the side AC. He will make two turns,
return to his starting point, and travel a distance equal to the perimeter of ∆ABC.

Diagram 1

__Starting point on the midpoint of side BC__

If Barney starts on the
midpoint of side BC, diagram 2, then_{}.

Diagram 2

Barney’s path has been
set up to always be parallel with one side of ∆ABC. If
side BC is a transversal (red line) crossing parallel lines _{} and _{} (blue lines), then _{}. Since ∆Bnm and ∆BAC both share angle B, then the triangles are
similar from AA (if two triangles have corresponding two pairs of angles with
the same measure then they are similar).
The points m, n, and o, are all midpoints (proportions all = ˝ from _{}).

Barney has returned to
his starting point and made two turns on the journey. If Barney starts at the midpoint of one of
the sides, he will follow a path consisting of mid-segments. His path length will be the sum of the three
mid-segments which is half of the perimeter of ∆ABC (see measurement in diagram 2). The assumption that Barney did not return to
his starting point would be a contradiction to the work shown.

__Starting point between vertex B and C__

Diagram 3 shows the path
Barney would take if a random starting point is selected on the segment between
vertex B and C. All of the triangles are
similar in this case since the sides are all parallel to one of the sides of
the original triangle. For example in Diagram 3, ∆DBE and ∆EFD are similar
triangles. With the added condition ED=ED,
∆DBE and ∆EFD are also congruent from AAS.

Diagram 3

Lots of parallelograms are
produced in Barney’s travels. All parallelograms have equal opposite
sides. EDKJ and DFHI (see yellow region)
are parallelograms, therefore AE=DL, EB=HI, ED=KJ, EH=BI. Using segment addition, AB=AE+EB. By substitution, AB=DL+ HI. This shows that the two paths parallel to
side AB are equivalent to side AB (see blue segments). In a likewise manner FH + KL = BC (red segments
parallel to BC) are equal to side BC and KI + ED = AC (green segment parallel
to AC) are equivalent to side AC.

The results, Barney
has returned to his starting point and made five turns on the journey (at
points E, H, I, K, and L). If Barney
starts at any random point of segment BC, he will follow a path equal to the
perimeter of ∆ABC.

__Starting at the Centroid of ∆ABC __

I constructed the
centroid (intersection of segments from vertex to midpoint) for ∆ABC. I moved the starting point for Barney until
it coincided with this point. While
doing this, I noticed that the diagram changed from having 10 similar triangles
(for random placement of point on side BC) to having only 9 triangles. The 9 triangles are all similar from the
parallel line construction and share a common side. With these conditions, the 9 triangles are
congruent. Barney returns to his
starting point after two turns.

One of the characteristics of the centroid is that
it divides the sides into ˝ and 2/3 consistent with the congruent triangles
already mentioned. Orthocenters can also
be located outside the triangle. The
case of starting outside the triangles is investigated next.

Diagram 4

__Starting at a point outside ∆ABC __

If Barney were traveling on a point outside the
triangle but still on paths that are parallel to the perimeter of the triangle
his path would be longer than the path on the inside of the triangle. The further out he starts from the triangle,
the longer will be his path.

In diagram 5, Barney starts outside the triangle at
point D. He then travels parallel to
each side until he comes to the extension of that side. He then turns and follows another parallel
path, turning at each extended side, until he returns to his starting
point. With Barney starting in the
middle of the side, he will have 6 turns before returning back to his starting
point.

It would be impossible for Barney not to return to
his starting point given the requirement of following parallel lines.

Diagram 5

__Discussion of Barney returning to his starting point__

If Barney did not return to
his starting point, then he must return to either the left or the right of that
location. Diagram 6 shows two possible finishing points for this situation.

Barney’s walk is defined to
be parallel to one of the sides of the triangle. If we assume he finishes at point F, then the
path used to get there was parallel to side AB (by definition). All of the triangles are similar in this case
because the sides are all parallel to one of the sides of the original
triangle. This means ∆NBE and
∆EFN are similar triangles (looking at the diagram, it can be seen that
triangles are not even constructed under this scenario). With the added condition EN=EN, the triangles
are also congruent by AAS. Corresponding parts of congruent triangles are
congruent, therefore EF = BN. This
condition can only happens when F and D coincide. With D being the starting point, this proves
that Barney must return to his starting point, and not to the left of this
location. In a likewise manner, it can
be shown that Barney can not finish to the right of this location either.

Another contradiction is side
BF is not on the path traveled by Barney.

Diagram 6

__Summary __

- Barney will returns to his starting point in all cases, inside,
outside, or on the edge of the triangle except when he is on a vertex
traveling parallel to the opposite
side (diagram 1).
- The number of turns Barney makes depends on where he starts his
journey.

Start on a vertex traveling parallel to
the opposite side: no turns.

Start on a vertex (along adjoining
side), midpoint, or centroid: two turns.

Start on a side between the vertices:
five turns.

Start outside the triangle in the middle
of the path: six turns.

Start
outside on an extended line (corner
point): five turns.

- The length of Barney’s walk also depends on where he begins the
journey:

Path starts on midpoint or centroid
< start path between vertices <
start path outside

To see an animated Barney walk click here.