Exam Question 1

Bouncing Barney

 

Problem statement

Barney is in the triangular room shown in Figure 14.1. 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.

Figure 14.1

 

Barney can start at any point along the segment BC. We denote the starting point by.

Figure 14.2 shows the positions of Barney as he moves from one side of the triangle to the other according to the rule specified in the problem until he reaches the point where he started. The different positions are denoted by  and . Thus, we observe that Barney makes 5 bounces (6 segments) to come to its starting point.

Figure 14.2

 

We consider two particular cases:

(i)

(ii)

 

Case (i)

Figure 14.3 shows the scenario when the starting point of Barney is at the first third of BC. We observe that there are nine congruent triangles. This can easily be verified by letting angle  to be x and  to be y and applying corresponding and alternate angle properties on the parallel lines.

Figure 14.3

 

The perimeter of the path  to  is equal to the perimeter of the triangle ABC. We shall prove this at a later point. Further, , and  lie on the vertices of a hexagon whose centre is the centroid of triangle ABC. Moreover, the area of the hexagon  is 2/3 the area of the triangle ABC.

Case (ii)

Now we consider the case when Barney is midway along BC. The path is illustrated in Figure 14.4. Here, the trajectory of Barney is the medial triangle. We have 4 congruent triangles and the path consists of only three segments. Therefore, the ratio of the area of the medial triangle to the original triangle is   1: 4.

Figure 14.4

 

We show that the six triangles  in Figure 14.4b are congruent.

From the parallelograms  and , we have .

From the parallelogram  and , we have .

Also, all the six triangles have same angles x, y, and z. Hence, applying the SAS property, the 6 triangles are congruent. 

 

Figure 14.4b

 

In fact, Barney returns to his initial point after . This can be seen from Figure 14.4b. Since is a parallelogram, the length of  is equal to .  Also,  is parallel to AB. If Barney does not return to his initial position after  then this will violate the condition that motion is parallel to side of triangle.

 

The scenario when the starting point of Barney is exterior to the triangle ABC is shown in Figure 14.5. 

Figure 14.5

 

Once again the points,,,,, and  lie on a hexagon but here it is outside the triangle.

Click HERE for animation

 

All the triangles that are generated as P1 moves exterior to BC are similar to the original triangle ABC. As P1 moves to the left of BC, the triangles are inverted and as P1 moves to the right of BC, the orientation is maintained.

Let the heights of similar triangles ABC and  be h1 and h2 respectively in Figure 14.6.

Figure 14.6

 

 or

Note  is a constant as it is the ratio of the height to the base of triangle ABC.

In Figure 14.6,  represents the distance between the starting point  and vertex B. A plot of against  is shown in Figure 14.7, where .  The origin has been made to coincide with the point B. When Barney moves to the left of B, this is considered as negative direction and when he moves to the right this is regarded as positive direction. Moreover, when Barney is at the midpoint of BC, .  

 

 

Figure 14.7

 

 

 

It should be noted that if Barney starts at one of the vertices of the triangle then the path that it traces is the triangle itself.

An interesting ratio involving ,,,,, and  in the Bouncing Barney problem is

 

or .

This ratio has been derived on the basis of numerical experiments and is yet to be proved. When Barney is at B, this ratio is 0 and when he is at the midpoint of BC it is 1.

 

Some results and proofs

Result 1:

We prove that the perimeter of the path is equal to the perimeter of the triangle ABC. Consider Figure 14.8 with the dimensions shown.

Figure 14.8

 

 

Perimeter of path =

                              =

                     =

                      =

 

Perimeter of  =

                                 =

                                 =

When  is at the midpoint,, , and ,

, , and .

Perimeter of  =

               =

              = 2 x length of path

The calculation of the perimeter of the path when the starting point is outside triangle ABC is shown in Figure 14.9.

Figure 14.9

 

Perimeter of path =

                             = 

                     =

 

Sum of long segments =

                                     =  

Sum of short segments =

Perimeter of  =

Result 2

1.   The path of Barney corresponds to the medial triangle not only when  is at the midpoint of  BC but for any point on the line which passes through the midpoint of BC and AB (Figure 14.10).

Figure 14.10

 

2.   Similarly, nine congruent triangles are obtained not only when  is at the points of trisection of BC or when  is at the centroid. Any points on the lines which passes through the points of trisection as shown in Figure 14.10b equally yield 9 congruent triangles.

Figure 14.10b

 

3.   Now we prove that if the center triangle is congruent to the triangles created by the path, then the sides are met at quarter points.

  Let the height of the congruent triangles be  in Figure 14.11. Therefore, the height of triangle ABC is 4h as it consists of 4 congruent triangles arranged vertically.

Triangle ABC is similar to triangle . Therefore,

In other words, . Similarly, it can be shown that , and .

Figure 14.11

 

 

 to  denote the quarter points along AB, BC, and CA in Figure 14.11.

 

12 December 2006

Ajay Ramful

 

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