Final Project

In this final project, technology is used to investigate a problem regarding a special path defined within a triangle, by searching for patterns and looking for explanations of these patterns. The central problem is about a specially defined path of Barney within a triangle, which always leads him back to the starting point. It will be proved that the path indeed leads one back to the starting point regardless of the starting point. Also, the created path have various properties which will be investigated and explained. In the Further Exploration section, which is linked to another page, another similar but different path of Barney will be investigated.

Introduction of the Problem

Everyone's favorite character Barney was kidnapped by some evil gang members who were jealous of Barney's fame. The evil gang members took Barney to their hideout and cast Barney into an outer confinement of a triangular shape. Barney is very anxious and distressed within the confines so that he cannot relax, but keeps on walking. Without much thought, he started from one of the walls and decided to walk parallel to one of the walls. Whenever he reached the opposite wall, he decided to bounce off the wall and again walk parallel to one of the walls. Barney continued to anxiously walk in such a manner, and after a long time, Barney realized that he kept returning to the same starting point!

The next day, Barney's good friends Godzilla and Long Foot came and rescued Barney away to a safe place. After eating some food and taking a rest, Barney shared the strange pattern he observed within the triangular confinements. The three of them put their heads together, but they could not explain why this is so.

Outline

Further Extension - In this new page, the path of Barney is redefined, and this new path is investigated.

Observation of Barney's Path

We will clarify the path that Barney takes within the triangular confinements by walking through the path step by step. Let us say that Barney is initially against the left wall as in the left figure below. At first, Barney has two choices of direction that are parallel to the sides of the triangle, the right wall or the bottom wall. Let us assume that Barney chooses the bottom wall and walks toward the right wall as in the right figure below. He will reach the right wall at point #1. When he reaches the wall, then there is only one direction that is parallel to the sides of the triangle, which is parallel to the left wall. When he reaches point #2, he will bounce off parallel to the right wall and get to point #3, and so on.

Barney initially has two choices

Barney bounces off the walls at points 1, 2, 3, and so on

Here is a GSP file that has an animation of Barney's path within the triangular confinement: CLICK HERE

As you can see from the animation, Barney always returns to the same starting point no matter where the starting point is located along the side of the triangle!

Try moving Barney's starting point along the side and observe the pattern of the path for various starting points. There seems to be three cases of distinct patterns of the path that Barney takes. The first case is when Barney starts off from a point that is located within a distance of third of the side from a vertex. The next case is when Barney's starting point is between the trisect point and the midpoint of the side. Lastly, when the starting point is the midpoint of the side, there is a special case of Barney's path, where he only bounces off twice to get back to the starting point.

Case i) Distance from vertex < 1/3

Case ii) Distance from vertex < 1/2

Case iii) Distance from vertex = 1/2

Why does Barney Return to the Starting spot?

In this section, it will be proved that Barney always returns to his starting spot, using geometry of parallel lines. The proof will be conducted for two cases - the first when Barney is not at the midpoint of the segment, and second when Barney is at the half mark.

Let us observe the path of Barney when Barney bounces off three times and is at the fourth wall. The triangle confinement is labeled FGH, and the sequence of Barney's path is B - A - R - N - E - Y (hmm..). By the definition of Barney's path, the line segment numbered 1 is parallel to the bottom side of the triangle GH, the segment numbered 2 is parallel to the left side of the triangle FG, and so on.

Let us look at quadrilateral BARG. By definition of Barney's path, we know that

. Hence, quadrilateral BARG is a parallelogram, and from the properties of parallelogram, the opposing sides have the same length. That is,

Now let us consider quadrilateral FARN. Since Barney travelled in his second direction (#2) parallel to FG and his third direction (#3) parallel to FH, it follows that

, which means that quadrilateral FARN is a parallelogram. Hence,

Now let us consider when Barney makes it further to the next wall at his last point of redirection Y.

If we consider quadrilateral NEYG, we know that

, which implies that quadrilateral NEYG is a parallelogram. As the opposing sides of a parallelogram are equal in length,

Let us compare the two lengths of segments NG and FB. We can express NG and FB as,

From equation (1),

, which means that

. However,

from equation (2), and therefore

Now, for Barney's sixth and last change of direction at point Y, he will travel parallel to side FH from point Y. Since

by definition of Barney's path and equation (3), it follows that if Barney bounces off the wall at point Y parallel to FH, then he will return back to the starting point B.

When Barney starts off from the midpoint of the left side of the triangle, he moves toward point A parallel to GH. In this case, remember that he will only bounce off twice before returning to the starting point.

If we consider quadrilateral BARG, since

by definition of Barney's path, quadrilateral BARG is a parallelogram. Therefore,

Now consider the four points B, A, H, R. We know that

by definition. However, R is the midpoint of GH, which implies that GR = RH, and using the above equation,

Because

and

, the four points B, A, H, R form a parallelogram, and when Barney bounces off from point R parallel to the right side of the triangle, he must come back to the starting point B.

Similar and Congruent Triangles

One thing we can notice from our previous section is that there are many parallel lines of the path that Barney takes. These parallel lines create many angle relationships that generate sets of similar and congruent triangles within the original triangle. Let us investigate the similar and congruent triangles for each of the three cases categorized in the Observation of Barney's Path section.

Case 1. When Barney's starting point is within a third of the side's length and the vertex

Let us draw a complete path of Barney until he returns to the same starting spot. Let us color code the parallel lines so that it is easier to observe the angle relationships, and let us label the three angles of triangle FGH as f, g, and h.

Similarly, using parallel relationships of

and

, we can get the following angle relationships.

If we continue the process, using angle relationships of parallel lines and the fact that f + g + h =

, we can obtain the following angle relationships.

There are many triangles that have the same three angles as the main triangle FGH! Now we need to determine which ones are similar and which ones are congruent.

Using parallelograms NEYG, BGRA, FNRA, and etc., and the property that the opposing sides of the parallelogram are equal, we can find six congruent triangles, which are shaded light purple in the following figure.

Can you count how many similar triangles are there in total? There seems to be really many...

Case 2. When Barney's starting point is between the trisect point and midpoint of the side FG

The angle relationships are basically the same as in case 1. There are also six congruent triangles, which are the ones shaded.

When Barney's starting point is at the midpoint of the side of the triangle, there are now only two bounce off points, and we get four triangles.

If we consider quadrilateral BARG, since

, quadrilateral BARG is a parallelogram. As the two opposing sides of the parallelogram is equal in length, and also because R is the midpoint of GH, it follows that

Therefore, the four triangles in this case are all congruent, and we have four congruent triangles within the original triangle.

Distance of travel

While Barney was continuing to walk in this pattern in anguish, what was the distance that Barney travelled for each cycle of returning to the starting point?

Because of the way Barney chooses to travel, especially the parallel nature of the lines, we know that

. Therefore, Quadrilaterals GNEY and BAYH are parallelogram (The shaded regions in the figure below). Since the opposing sides of the parallelogram are equal in length,

Similarly, using the pairs of parallelograms BAHY and FARN, and FARN and GNEY, it follows that

Thus, lines numbered 1 and 4 is equal in length to the bottom side of the triangle, lines 6 and 3 is equal to the right side of the triangle, and lines 2 and 5 add up to the left segment of the triangle. We see that all the lines are exhausted, and the total length that Barney travels is equal to the perimeter of the triangle!

However, there is a special case, when Barney starts at the midpoint of the side of the triangle. In this case, we observed that Barney bounces off two times before reaching back to his original starting point.

Here, because the method Barney chooses to travel, we know that

. Using these parallel relationships, we can conclude that quadrilaterals BARG, FARB, and BARH are parallelograms. As opposing sides of the parallelogram are equal in length,

Also, as the starting point B is the midpoint of FG, and

, by the property of parallel lines in a triangle, points A and R are the midpoints of the respective sides of the triangle.

Therefore, the segment numbered 1 is half the length of side GH, 2 is half the length of FG, and 3 is half the distance of FH.

So when Barney starts off from the midpoint of the segment, the total length is half the perimeter of the triangle.

Special Starting Points

We have noticed how the path of Barney was special when he was starting out from the midpoint of the side of the triangle. In this case, Barney only bounced two times and the three paths connected the midpoints of each side. For a more detailed explanation go to the second part of the section Distance of Travel.

Are there other special starting points where the path of Barney has a distinct property? Let us investigate the path of Barney for various starting points, specifically the special points of the triangle - centroid, orthocenter, etc.

First, lets investigate the centroid of the triangle. The centroid of the triangle is the intersection of the three medians of the triangle, where a median is the line segment from the vertex to the opposing side's midpoint. One special property about the centroid is that the centroid divides each median in the ratio 2:1.

In the above diagram, let us start at the centroid located at the middle of the triangle. Let us choose to move right first, parallel to the bottom side of the triangle, and repeat the process bouncing off the wall parallel to each of the sides.

Now compare the pattern of Barney's path to the three general cases noted in the "Observation of Barney's Path" section. By comparison, we note that this is indeed a special case. In particular, we notice that the extra triangle usually found in the center of the triangle has disappeared. We also note that the original triangle FGH is dissected evenly by three rows of triangles. It seems that all these triangles are congruent, which would make this case the champion for the number of congruent triangles.

Consider the two triangles FBC and FGM'. As BC and GM' are parallel, the two triangles are clearly similar. However, we know from the property of a centroid that FC:CM' = 2:3, which means that the two triangles are similar in the ratio of lengths 2 : 3. Hence,

By the same procedure, it can be shown that triangles FCA and FM'H are also similar in the ratio 2 : 3, and

If we continue the process using similar triangles that involve the centroid and the parallel lines that go through the centroid, we can find that the points on the side are trisecting each side of the triangle.

Now consider the three quadrilaterals shown in the diagram. Since all the horizontal sides and the vertical sides of the quadrilaterals are parallel to one another, it follows that there are parallelograms. However, as NB = BG and GY = YR (the three points trisect each side), all the horizontal length and vertical length of these parallelograms are equal.

Using the same reasoning of using parallel lines and equal lengths, all these sub triangles can be shown to have the same dimension of lengths. Therefore they are all congruent.

Now, how about other special points of the triangle? I could not find anything special about other starting points.

When Barney's starting point is the Orthocenter (left), Incenter (middle), and Circumcenter (right)

Further Extension

In this extension of the problem of Barney, we will define another path of Barney where Barney will around outside the triangle. Let us extend the lines of the triangle and now Barney will start outside the triangle on this extended line. When Barney meets an extended line of the triangle, now instead of bouncing off the line, Barney will go through the line parallel to one of the sides of the triangle . . .

Bouncing Barney

by

David S. Park