Bouncing Barney

by Molly McKee

Prove that Barney will eventually return to his starting point.

Let’s begin by examining what happens when Barney begins at a vertex of ABC.

Perimeter of ABC = length of AB + BC +AC

Barney’s Path = length of AB + BC + AC.

Perimeter of ABC = Barney’s Path

Now let’s examine what happens when Barney begins at the midpoint of BC.

Since S is the midpoint of BC, we know that

A transversal is a line which crosses two parallel lines.

Properties of Transversals

opposite angles are congruent

1=4, 5=8, 2=3, and 6=7

alternate interior angles are congruent

4=5 and 2=7

alternate exterior angles are congruent

1=8 and 3=6

corresponding angles are congruent

1=5, 2=6, 3=7, and 4=8

When AB is a transversal, we can see that angles p1 and A are congruent.

We can also let BC be a transversal because it also crosses the two parallel lines. When BC is a transversal, we can see that angles S and C are congruent.

Since angle B is contained in both triangles, we can say that ABC is similar to p1BS because all of their angles are congruent.

Since the triangles are similar, then we can say

Recall that

Therefore

Perimeter of ABC = length of AB + BC +AC

Barney’s Path = half the length of (AB + BC + AC).

Half of the Perimeter of ABC = Barney’s Path

Barney starts at a quarter-point of BC

Barney starts at an eighth-point of BC

Now let’s investigate Barney’s path if he begins outside of ABC.

Now we can calculate each portion of Barney’s journey as follows:

AB + 1/8 of AB

1/8 of BC

AC + 1/8 of AC

AC + 1/8 of AC

1/8 of AB

BC + 1/8 of BC

1/8 of AC

After all portions are added together, we can see Barney’s journey is 10/8 of the perimeter of ABC.

Perimeter of ABC = length of AB + BC +AC

Barney’s Path = 10/8 of (AB + BC + AC).

10/8 of the Perimeter of ABC = Barney’s Path

Investigate Barney’s journey

you can move his path around to see what happens