Bouncing Barney

Barney is in the triangular room 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? Discuss and prove any mathematical conjectures you find in the situation.

First, we should construct Barney's path around his triangular room. After picking a random point on the triangle on line AC, I made the first line of his path from the starting point, P, parallel to BC to AB.

Then, we make our next line of Barney's path from the point on AB, parallel to AC, to BC.

Then, we make our next line of his path from the point on BC, parallel to AB, to AC.

Then he turned and walked parallel to BC until he hit AB.

Next he turned and walked parallel to AC until he ran into wall BC.

Then he turned and paced parallel to AB until he hit AC and realized that he was back where he started.

As we can see, that he ended back at his starting point on his 6th bounce.

But what happens when he starts from a different point, does it still only take 6 bounces for him to return to his starting point?

We can see that it does. Click here so that you can manipulate the starting point on your own.

Now what if the starting point is the midpoint?

We can see that it only takes 3 bounces for Barney to make it back toi the starting point.

What about if Barney starts at one of the vertices, what will happen then?

If he starts at the vertice, then he will return to his starting point in 3 bounces because he is moving from vertice A to vertice B and then to vertice C and back to vertifce A, which is where he started.

Now, what if Barney starts outside the triangle?

He will still return to his starting point at on his 6th bounce. The distance of the path traveled is longer than the length of the sides of the triangular room.

Now that we can assume that Barney will always return to his starting point. But we still need to prove that this is always true. Let's look at our diagram and pick a random starting point.

We can see that Barney's path created many parallelograms and triangles within the triangular room. Let's see what we can tell about our triangular room and the traingles Barney's path created to see if Barney will alwas return to his starting point.

From the path that Barney took, we know that:

and

and

And we also know that if 2 or more parallel lines are cut by a transversal , then their corresponding angles are congruent.

All the parallel lines above were cut by transversals so we have the following congruent angles:

and

and

Therefore we know by the AA theorem that if two angles on one triangle are equal, respectively, to two angles on another triangle, then the triangles are similar. Therefore we know that the following triangles are similar:

So we know that the yellow triangles are similar to each other as well as the triangular room. We can see that the yellow triangles and the green triangles form a parallelogram. We know this because EF and AC are parallel and EK and LF lie on EF and AP and GC lie on AC.

We know that the diagonal of of a parallelogram cuts the parallelogram into two equal triangles. Therefore we know that:

and

and

Then by the transitivity property, we then know that all the triangles are congruent to each other. Therefore the parallelograms are congruent to each other as you can see in the diagram below.

Now that we know that the parallelograms are congruent, we still need to show that Barney always ends at his starting point. Now we can assume that if Barney does not end at his starting point, then after his fifth bounce, then he must hit at another point in AC. Let's just mark that unknown point as R in the diagram below.

So we know from our above work that:

and we can assume that

and then by transitivity property

so

implies

and since we know that

therefore

So the endpoint R must also be P, therefore R = P. So Barney will always end at his starting point.

By Carolyn Amos

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