BOUNCING BARNEY

EXAM

SHADRECK S CHITSONGA

BOUNCING BARNEY

QUESTION

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.

Let us start by looking at the GSP diagram shown in figure 1. Pay special attention to the parallel lines and the lengths of the distances.

Figure1

In figure 1, Barney starts at point I on the side BC. He moves on IH parallel to AC to reach H on AB. Then moves HG parallel BC to G on AC. From there he moves GF parallel to AB to reach F on BC. From F he moves FE parallel to AC to reach E on AC. He then moves ED parallel to BC to reach D on AC. Finally he moves DI parallel to AB to reach I on BC.

Looking at the diagram and the distances shown, there are a number of conjectures one can make.

1. The distance that Barney covers from the starting point

back to the same point is equal to the perimeter of the

triangle.(Note: Barney is starting on a given side of the

triangle). CLICK here to see whether the same thing holds when

Barney starts from another point on a different side of the triangle.

Drag point D along the side AB.

2. To complete his journey, Barney must reach each wall of

the triangle twice. In total Barney travels in six segments to return

to the starting point. For an exception see part 3.

3. If Barney starts from the point that is the mid-point of a

side of the triangle, Barney reaches each wall once before

returning to the staring point. He moves parallel to each

side once. (CLICK HERE for the diagram)

For our proofs let us start by looking at a specific case.

We will consider a situation when Barney starts from the point that is the mid-point of a given triangle.

Figure 2

We started with D as the mid-point of side AB, and constructed DF parallel BC, then FE parallel to AB and finally ED parallel to AC. Just by looking at the distances measured we see that F and E are also the midpoints of sides AC and CB respectively. That should not be a surprise because of the intercept theorem.

Let us forget the distances and write a formal proof.

Proof

In the ABC, D is the mid-point of AB (given)

DF is parallel to BC (construction)

Therefore FD bisects AC (mid-segment theorem)

AF = FC

Now D and F are the midpoints, it means that

FD = (1/2) BC, Similarly FE parallel AB, which means that

CE = EB, and FE = (1/2)

Since E is the mid-point of BC, a line drawn through E parallel to AC must pass through the mid-point of side AB. In this case we know that the mid-point of AB is D, therefore this line passes through D.

Also ED = (1/2) AC.

Conclusions from the proof:

1. From the proof we can see that the total

distance Barney covers is equal

to FD + FE + ED=(1/2) BC + ( 1 / 2) AB +(1/2) AC.

=(1/2){BC + AB + AC}

= (1/2){perimeter of ABC}

2. The proof is showing us that if Barney starts

from the mid-point , he ends up at the mid-point.

Proof to show that the distance Barney covers is equal to the perimeter of triangle ABC if Barney starts from any point on the side other than the midpoint

Figure 3

The perimeter of ABC is given by AB + BC + CA. This is equivalent

to a + b +c ( where c = AB , a= BC b= CA )

Because of the number of parallel lines we have drawn in the diagram, it is easy to see that we have a number of parallelograms formed. In the diagram, the lower case letters are standing for the length of each side, for example the length of ID= m, the length of DE=c-2n.

Now let us trace the path Barney takes. We will start from D on AB and follow the path through DI, IF, FE, EH, HG, and GD.

Adding all the distances we see that:

DI+ IF+ FE+EH+HG+GD = m +n +c-3n +n+q+m+a-3m+m+n+q+b-

3q+q.

=3m +3n +3q-3m-3n-3q+a+b+c

= a + b+ c

= perimeter of ABC

______________________________________________________________________________

Proof to show that Barney will return to the starting point

Figure 4

We now want to prove in general that no matter where Barney starts from, he will end up on the same point. We will still consider D as the starting point. Our proof will mainly involve the application of the basic proportionality theorem also known as Thales.

Theorem.

If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio.

Proof (Barney’s Journey)

Given triangle ABC with D a point on side AB.

But AD is parallel to BC (construction)

= (Constant proportionality)

IF is drawn parallel to AB,

= (Constant proportionality)

FE is drawn parallel to BC,

= (Constant proportionality)

HE is drawn parallel to AC,

= (Constant proportionality)

HG is drawn parallel to AB,

= (Constant proportionality)

If one follows the arguments above carefully it is

easy to that = = = = = (i)

We can also see that , ,

Now let us go back and consider the point G, the line from G must be drawn Parallel AC.

Using Thales Theorem, this line must divide sides AB and AC in the same ratio.

Let us say for the moment that this line intersects AB at point D*.

GD* is parallel to AC

and = (constant proportionality)

But in (i) we showed that =

This means that AD = AD* and because A is a common point, we can

conclude that D* and D are the same point.

YES, BARNEY WILL RETURN TO THE ORIGINAL POINT.

What about if Barney starts from if Barney starts from a point exterior to segment BC?

Refer to figure 5

Figure 5

From the GSP diagram in figure 5, we see that when Barney starts from point D outside segment BC, he still comes back to the starting point. He completes his journey in six segments.

Conjecture

If barney starts from a point outside the segment BC, he covers a distance equal in length to the perimeter of the original triangle + 2(perimeter of the little triangles formed at each vertex of ).

The perimeter of each little triangle is the same as the sum of the three short segments DI, EF and GH that are parallel to the sides AB, BC and CA respectively.

Proof

Figure 6

Using notation in figure 6 it is very easy to see why the sides are marked that way. For example Segment EF= Segment BG=n, because EFGH is a parallelogram (by construction).

We can use similar argument for the other sides.

Note a, b, and c are the lengths of the sides of triangle ABC.

Therefore the perimeter of triangle ABC = a+ b + c.

The total distance Barney travels = a + n + m + b + l + n + c + m + l

= a + b + c + 2 ( m + n + l )

= perimeter of ABC + 2(sum of