For further exploration with this problem, see this gsp file.

## 1. Prove that Barney will eventually return to his original starting point.

Let Barney start at some point D on BC. He will walk parallel to AB until he reaches point E on AC. Then he walks parallel to BC until he reaches point F on AB. Then he walks parallel to AC until he reaches point G on BC.

Stop and observe that 2 parallelograms have been constructed: CEFG and DEFB. Because of the properties of parallelograms, we can see that angle B = angle EDC, angle C = angle EDC, and BE = GF. By AAS, ΔBFG≅ ΔDEC.

Continue on Barney's path: Barney walks from point G parallel to AB until he reaches point H on AC. Then he walks parallel to BC until he reaches point I on AB. Then he will walk from I parallel to AC until he reaches point D on BC (aka his starting point). Similar to above, we now have constructed parallelograms AFGH and CDHI. So, angle BFG = angle A, angle AHI = angle C, and CD = IH. By AAS, ΔHAI≅ ΔGFB. Because ΔBFG≅ ΔDEC≅ ΔHAI, ΔBFG≅ ΔHAI. So AI≅ ED and AI ||ED, so AIED must also be a parallelogram.

Below are the three congruent triangles constructed by Barney's path:

There are 6 parallelograms constructed by Barney's path:

CEFG, EDFB, AFGH, CDHI, AIED, BIHG

Thus, Barney will always end in the same place he started and hit the wall 5 times before returning to his starting point when he starts at an arbitrary point on one of the sides of the triangle.

There are two exceptions:

1. Barney starts on a vertex.

In this case, Barney will travel along the perimeter of the triangle and will therefore only hit the wall twice before returning to his starting point.

2. Barney starts on a midpoint on one of the sides of the triangle.

We will explore this further below.

## 2. What is the distance that Barney travels?

The more obvious case for exploring the distance that Barney travels is when he starts on a vertex. He will clearly travel the distance of the perimeter of the triangle.

For all other cases except when he starts on the midpoint of one of the sides, we must explore how far Barney travels.

By looking at the gsp file, it appears the Barney always travels the same distance as the perimeter of the triangle. To prove that this is true we need to show that:

AB+BC+AC = DE+EF+FG+GH+HI+ID.

We know that AB, BC, and AC can be rewritten:

AB = AF + FB

BC = BD + DC

AC = AE + EC

Call the intersection point between DI and GH point K. Because of properties of parallelograms and what we know about congruent triangles created by Barney's path, we realize that FB = DE, BD= FE, FG= EC, AF = GH, IH = DC, and ID = AE.

Using substitution we now see that:

AB = FG + ID

BC = HI + EF

AC = DE + GH

## AND... AB+BC+AC = DE+EF+FG+GH+HI+ID.

In conclusion, we know that the distance that Barney walks is equal to the perimeter of the triangle when he starts on one of the sides of the triangle.

## 3. What happens when Barney starts at the midpoint of one of the sides of the triangle?

Without loss of generality, lets say he starts at the midpoint of BC. He then walks parallel to AC until he reaches AB at the midpoint. Then we walks parallel to BC until he reaches the midpoint of AC and changes direction to walk parallel to AB until he reaches his starting point.

Barney has now walked the distance of 1/2 of the perimeter of the triangle ABC and has only hit 2 walls before returning to his starting point.

## 4. What happens if Barney starts walking at the centroid?

If he starts at the centroid and travels parallel to AC, when he hits AB, the point D will be 2 times the distance from B as from A. Barney will then travel parallel to BC until he reaches AC at point E. At point E, he will be 2 times as far from C as from A. Then, Barney will travel parallel to AB until he reaches the centroid again.

If he starts from the centroid, Barney will walk 1/3 of the perimeter of the triangle ABC and hit 2 walls before returning to his starting point.

## 5. What happens if Barney starts walking from the orthocenter?

The path that Barney takes if he starts from the orthocenter has a similar shape to the path that he takes when he starts from an arbitrary point (not the midpoint) on one of the sides of the triangle. Therefore, we know from what we proved before that Barney will hit 6 walls before returning to his starting point and will walk the distance of the perimter of the triangle.