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

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.

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

 

To see an animated Barney walk click here.

 

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