EMAT 6680 :: Clay Kitchings :: Final Write Up :: Bouncing Barney

Problem: Bouncing Barney

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.

Preliminary Drawing: Barney’s starting point is point D. We shall assume he completes his path at a point D’, such that .

Let us assume that , which is equivalent in this figure to assuming . We shall also assume that the common, color-coded paths are parallel (the blue sides will be proven parallel). That is:

HD || FK || AC FE || HG || BK

**Note: Any reference to D’ shall the assumption that unless otherwise noted***

Note #2: It could also be proven with the assumption that D’ – D = 0 and showing a contradiction.

Construct Segment ED. We need to show that quadrilateral FEDB is a parallelogram because we would like to have FB || ED.

We have an enormity of similar triangles, some congruent triangles, and some parallelograms in our figure. We will locate and identify some congruent triangles and parallelograms.

Let us start with some parallelograms we already know exist:

The following are parallelograms because they have parallel, opposite sides by definition of Barney’s path (two of which are shown above):

FECK, FEGP, PGCK, HGCD, HBKG, AGKF, and HPKD

From these parallelograms, we derive the following congruent segments (opposite sides of a parallelogram are congruent):

FE = PG = KC HD = PK = GC HB = GK = AF

We can now show that by AAS:

Why?

· m(BHD) = m(KGC) and m(HBD) = m(GKC) by the corresponding angles property (due to parallel lines and transversals)

· HB = GK by equality statements above

· We have AAS.

We know triangle FAE ~ triangle BAC by AA similarity

Why?

· Both share <A, and m<(AFE) = m<(HBD) by corresponding angles (from the parallel lines and transversal)

· This implies m<(FAE) = m(BHD)

Now we can write: .

Why?

· AF = HB

· m<(AFE) = m<(HBD)

· m<(FAE) = m(BHD)

· Use ASA triangle congruence theorem

Now, by CPCTC, we have FE = BD, and FE || BD by definition of Barney’s path. This implies FEDB is a parallelogram. Why? A quadrilateral is a parallelogram if it has one pair of opposite sides that are both congruent and parallel.

Let us return to the original assumption that D’ is a distinct point from D. By assumption, Barney is traveling a parallel path to AB from point E (to point ED’). This implies AED’B is also a parallelogram.

Now, let us consider our assumption that .

We know that Barney must travel from point E parallel to AB according to the definition of his path traveled. Now, we have already verified that FEDB is a parallelogram. This means that m(FED) = m(FBD), and since FED’B is a parallelogram, m(FED’)=m(FBD) since opposite angles of parallelograms are congruent. This implies:

**m(FED) – m(FBD) = 0 and m(FED’) – m(FBD) = 0

We know from angle sums: m(FED’) = m(FED) + m(DED’)

Now substitute the highlighted (yellow) parts to obtain:

m(FED) + m(DED’) – m(FBD) = m(FED) – m(FBD) + m(DED’) = 0

However, we know from ** that m(FED) – m(FBD) = 0, and we’ve assumed that m(DED’) > 0. Therefore we now have:

0 + m(DED’) = 0

This isn’t possible because m(DED’) > 0, so we have a contradiction. Therefore our original assumption that m(DED’) > 0 was invalid. Thus, D=D’ and Barney ends his journey exactly where he started it!!!!!