Bouncing Barney

By: Kassie Smith


Barney is in the triangular room shown here. 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. 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 exterior to segment BC. Discuss and prove any mathematical conjectures you find in the situation.



To began this problem, I created a GSP file and continued my exploration within that file. Here is the file with a variety of explorations on different tabs within the file. The tabs are located along the bottom of the document, beginning on the left side. To get a better understanding of the problem, go to page 7 of the document. There is a sequence of buttons that you can press to show his path step by step. Below, I will compile and summarize my explorations.

The first thing I looked at was the length that Barney travels as he completes his journey. This is on page 1 of my GSP file. In total, I found that that length of his path is equal to the perimeter of the larger triangle that he is traveling within. Granted, this assertion assumes that he is starting somewhere along the sides or within the triangle (more on that later). The was found simply with the measuring utilities of the software. Further exploration showed that if the starting point is the midpoint of a side of the triangle, then the length of his path is half of the perimeter of the larger blue triangle. When Barney starts at the midpoint of a side of the larger triangle, then Barney's path actually creates two overlapping green triangle. The total of the two overlapping green triangles is equal to the perimeter of the triangle; however, Barney will make it back to the starting point by just traveling one of the triangles. Thus, this path is half the distance of any other path.

Now, what about when Barney begins outside of the triangle, along one of the lines that the triangle sides (segments) lie on. Obviously, if we just add the lengths of his path, that sum would be more than the perimeter of the original blue triangle. Nonetheless, we can introduce directional distances which would change this discussion. By directional distances, I mean that the distances can be both positive (green) and negative (red). I am considering the positive distances to be the parts of his path which are directly beside a side of the triangle; the negative distances would be those distances nearest to the vertices of the triangle. To further illustrate, the image below shows the positive distances in green and the negative distances in red. However, the interesting observation here is that the distance of the path that Barney travels when he begins on the exterior of the triangle is still equal to the perimeter of the original triangle when directional distances are considered. More clearly, you can find the sum of the red distances and subtract that from the sum of the green distances to get the perimeter of the original triangle. Alternatively, you could find the sum of the entire outer path (the green and the red distances) and then subtract the red distances twice.

Why does this work? I made a conjecture as to why this holds and again used the measuring capabilities of GSP to confirm. Now, this is by no means a proof, but it did help me to see why it works. I saw three sets of three parallel line segments in the figures. In the above figure, each set is composed of one red, one green, and one blue line segment. Using the same directional line segments, I took the longest green line segment and subtracted the shortest red segment. This resulted in the blue line segment, a side of the triangle for each of the three sets.

And what if Barney begins somewhere inside the triangle? The reader can drag around the starting point on page 4 of my GSP file to investigate this. Really, nothing too special happens if he starts inside the triangle. In fact, pretty much the same thing happens as if he were to start along the perimeter of the triangle. All of the same formations of the inner, smaller, green triangles appear as we drag the starting point around in the interior of the triangle as when we did the same along the perimeter of the triangle.

The only noticeable difference when the starting point is not limited to a side or extended side of the triangle occurs when the starting point is moved outside of the triangle. In this instance, it is possible for Barney to return to his beginning location but it is, by no means, always the case. I moved the point around for a while and made a variety of conjectures that turned out to be false when I was investigating this. In the end, I was not able to discover a solid pattern of behavior when Barney begins outside of the triangle and not on the extended sides of the triangle. Again, use page 4 of the file to move the starting point around, both inside and outside of the triangle.


While I was playing around with the different starting positions, I took a keen eye to the pattern of green triangles that I was creating in the interior of the larger, blue triangle. For most starting positions, there were four green triangles, but I noticed that at one point, I could create just three green triangles that all meet at one vertex. Actually, I noticed that the pattern of the green triangles was symmetric with respect to the midpoint of the side of the triangle. So, there were actually two starting points along one side of the triangle that I could create just three green triangle. These two points actually turned out to be the two points of the side of the triangle that trisected that side. So, the were evenly spaced. Even more, the place of vertex where each of the three triangles met turned out to be the centroid of the triangle.

Moreover, when the starting point is on one of the trisection points, the three green triangles are congruent and similar to the larger triangle (more on that later). Even more impressive is the fact that this division actually creates nine congruent triangles within the larger triangle and not just three. If we focus on the parallel lines, we can deduce congruence of many of the angles. Using that, the fact that the Barney's run in's with the walls trisect the sides of the triangle, and triangle congruences theorems, we can gather that all nine of the triangles are congruent.

But why does it make sense that the intersection of his paths in the middle of the triangle is the centroid? For one, we just said that we have created congruent triangles. Therefore, the centroid, being the center of gravity of a triangle, should be the balancing point. It is since there are equal numbers of triangles surrounding that point. Secondly, we know that the centroid is located at the intersection of the medians of the triangle and that these medians intersect 2/3 of the way from the vertex to the opposite side of the triangle. We can easily visualize this 2/3 distance because we have the congruent triangles there to help us.

So, regardless of whether Barney starts at a trisection point of the side of the triangle or the centroid actually inside the triangle, the overall path that he would take would remain unchanged. Barney could travel any direction from the centroid (granted that it is parallel to one of the sides) and he would begin, always follow, and end on the green path. And, again, the path would partition the triangle into nine smaller, congruent triangles.


Now, how did I know that these triangles were congruent? First, let's look at this image and I will use it to explain why.


Using the parallel lines from Barney's path, we can easily deduce that ∠DAI ≅ ∠EFB ≅ ∠CGH by corresponding angles. Similarly, ∠ADI ≅ ∠GCH ≅ ∠FEB and ∠AID ≅ ∠GHC ≅ ∠FBE.

Now, we look at side lengths rather than angles. In the mist of all these triangles, we can identify parallelograms formed by the parallel lines. First, look at ADEF, a parallelogram inside the triangle. Since it is a parallelogram, we know that opposite sides are equal. Thus, AD ≅ FE. We could continue this process to prove that the triangles are congruent. But we could also use the angle-side-angle theorem after this first step. So, by ASA we have △ADI ≅ △FEB. Similarly, we can apply this reasoning to show △ADI ≅ △GCH. By transitivity, △ADI ≅ △GCH ≅ △FEB.

Even more, we can connect these congruent triangles to the original triangle, △ABC. △ADI shares ∠CAB with △ACB. And △FEB shares ∠ABC with △ACB. And △GCH shares ∠ACB with △ACB. Therefore, we know that △ADI, △GCH, and △FEB are similar to △ACB.

Moreover, I can visualize a different set of triangles, △DJI, △GHL, and △KEF. Using the same ideas as above, we can show that this set is congruent to △ADI, △GCH, and △FEB and similar to △BCA as well.


Now that we know that these triangles are similar, we can actually show why the length of Barney's path is equal to the perimeter. Let's look at the diagram below. This is on page 9 of the GSP file.


Let's label the figure to make this more clear.

So, the length of the path is

d + [f + (c-3f) + f] + e + [d + (a-3d) +d] + f + [e + (b-3e) + e]

= d + (c-f) + e + (a-d) + f + (b-e)

= c + a + b

= perimeter of △ACB

Therefore, Barney travels and returns back to his starting position and he travels the same distance as the perimeter of △ACB



Now, let's see if we can extend Bouncing Barney to other n-gons.

First, I will try a rectangle.

My observations do not hold for a rectangle (or really any parallelogram). In the diagram above, Barney's path is the green dashed line. With the exception of the first step of the path, Barney travels along the sides of the rectangle. It can also easily be observed that the length of Barney's path is not equal to the perimeter of the rectangle. It would be less when Barney starts on a side of the rectangle or within the rectangle and more if he starts outside the rectangle.


What about a regular pentagon?


From these diagrams, we gather that Barney will return to his starting location. Also, it will two trips around unless he begins at the midpoint of the side of the pentagon in which case just one trip will take him back to the starting position. So that is parallel to the triangle. I kept searching for the parallels but it seemed to stop around there. I checked to see if the distance that Barney would travel would be the same as the perimeter of the pentagon and that is not the case. I decided to look at the ratio between the length of the path and the perimeter of the pentagon to see if that would give me any insight. For the diagram on the left, I got a ratio of 1.61782, something really close to the golden ratio. For the one on the right, I got 2.61812.


I also tried it with a hexagon.

This case was very similar to the pentagon. Again, I calculated the ratio and this time I got 1.5.


From my investigations with other n-gons, I was not able to deduce anything too formal. Nonetheless, I did find it very interesting to investigate. I was skeptical that it would work with both pentagons and hexagons but my explorations proved me wrong.


Return to EMAT 6680 Page