BOUNCING BARNEY

by: Katie Gilbert

A. Bouncing Barney. We discussed this investigation in class. Your challenge now is to prepare a write-up on it, exploring the underlying mathematics ideas and conjectures.

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 begin let's take a look at the different paths Barney can take, starting from different points. First, let's take a look at the path Barney takes when he starts inside the triangle. Let's investigate the following:

Does he come back to his starting point? If yes, how many paths does it take to return to his point of origin?

If Barney does not get to the starting point, how long does it take Barney to "fill up" the triangle with his paths?

From our picture above it looks like Barney will get back to the starting point, and to get back to the starting point he traveled six paths. Let's now investigate a different starting point; we will pick the point that trisects the side as his starting point.

We see here that Barney again returns to his starting point and again his journey consists of six paths .

Now let's investigate when Barney starts at the midpoint of the side, will he return to his starting point and how many paths will he take?

Interesting! It appears as if he again returns to his point of origin, but it only takes him three paths to get back to his starting point.

Now let's take a look at when he starts at one of the vertices of the triangle.....

He again returns to his point of origin, but once again it takes him three paths to return to his starting point.

If you wish to explore this further before we begin to use geometry to solve this problem click here to vary Barney's Starting point and see what his paths look like.

So what we have found from our explorations, is that Barney always returns to his starting point, and that it either takes him three or six paths to get back to his point of origin. If we take a look at our case where Barney starts at one of the vertices, we see that his distanced "bounced"/ traveled is equal to the perimeter of our triangle. We will us GSP to investigate our cases above and compare the perimeter of our triangle to the length of Barney's journey.

Click here to view calculated measurements of Barney's journey compared to our triangle.

We see that if Barney's point of origin is inside our triangle then the length of his journey IS equal to the perimeter of our triangle. We will use the properties of a parrallelogram to prove our findings....

We will first define a paralleogram and point out properties we will use......

In geometry, a parallelogram is a quadrilateral with two sets of parallel sides. The opposite sides of a parallelogram are of equal length, and the opposite angles of a parallelogram are congruent.

Let's take a look at the different parallelograms in our Triangle and then use the fact that opposite sides of a parallelogram are equal.

We see above that length of AE = length of OF, and length of AO = length of EF.

From this picture we see that the length of BG = length of EH, and the length of BE = length of GH.

Lastly, we see that the length of OC = length of DG, and length of OD = length of CG.

So we can conclude that:

We know the perimeter of our triangle is the sum of the length of the sides, so we can write it as:

We also know the length of each side is equal to the sum of its parts:

We will substitute these "sub-lengths" into our original equation and then substitute there equivalent lengths:

We can write Barney's Journey as the sum of his paths:

We then compare the two equations and we see that.......

So when we use the properties of a parallelogram we see when Barney starts inside the triangle (except when his starting point is at the median of the side, which we will discuss in a moment) that the length of the paths he takes to "bounce" back to his starting point (his complete journey) are equal to the perimeter of our triangle. This is obviously the case if Barney starts at one of the vertices.

So let's find out what is happening when Barney starts at the median of the side, still using what we discovered above.....

Using the properties of the medial triangle we know that the perimeter of our medial triangle = 1/2 perimeter of our big triangle, so we will us this to figure out what is created by Barney's path when he starts at the midpoint of the side....

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So we can say if Barney starts at the midpoint of the side, the length of his journey is equal to half the perimeter of our triangle. If you want the length of Barney's journey to equal the perimeter of our triangle, Barney travels around his path twice if he starts at the mipoint of the side.

Lastly, let's explore what happens if Barney starts outside the triangle, does he get back to his point of origin? How many paths does he take?

Well, we see once again he does get back to his starting point, and he takes six paths to get there. But in this case the length of his journey is obviously not equal to the perimeter of our triangle.

When we look at the length of Barney's Journey, we see that it is obviously not equal to the perimeter of the triangle, so we will use directed measurements to investigate further.

Above we have labeled the direction Barney travels inside the triangle as a positively directed measurement, and if he travels in the opposite direction we label that as a negatively directed measurement. If you take a look at the arrows as Barney "bounces" inside the triangle and compare it to the direction he travels outside you can see when we have a positively directed measurement/movement (labeled with +), and when Barney "bounces" in a negatively directed measurement/movement (labeled with -). Let's find the sum of these directed measurements and see what we get....

So we see that when we use directed measurements to measure the length of Barney's Journey outside our triangle, we again get the length of his journey equal to the perimeter of our triangle.

So in summation we have found:

2) If Barney starts inside the triangle:

A) starts at midpoint of side: it will take him three paths to return to his origin and the length of his journey will equal 1/2 perimeter of our triangle; or he can go around twice and the length of his journey will be equal to the perimeter of our triangle.

B) starts at vertices of triangle: his journey consists of 3 paths and his journey is equal to the perimeter of our triangle, in fact his journey is the original triangle.

C) if he starts at any other point inside our triangle: he returns to his starting point using six paths and the length of his journey is equal to the perimeter of our triangle.

3) If Barney starts outside the triangle, it takes him six paths to return to his starting point and the length of his journey using directed measurements is equal, once again, to the perimeter of our triangle.