Barney is in the triangular room shown below. 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.
First, let's look at a complete path that Barney follows:
It seems like Barney comes back to his starting point P and he completes his path in 6 steps. Click here to observe his path for different starting points.
You will see that there is a point on BC and if Barney starts at that point, he will come back to his starting point in 3 steps:
This should be the midpoint of BC:
If Barney starts at the midpoint of BC, a path parallel to AC will take him to the midpoint of AB. Then, a path parallel to BC will take him to the midpoint of AC, The final path parallel to AB will take him back to the starting point, the midpoint of BC.
But why does he eventually return to his starting point for points other than the midpoint?
Let's look at where Barney is after his fifth step:
Looking at the parallelograms we see that
EF = AG
FC = GH
AGH = EFC (both sides parallel to each other)
So, AGH is congruent to EFC.
DB = EF
DP = EC
BDP = FEC
So, EFC is congruent to BDP.
Hence, AGH, EFC and DBP are congruent triangles.
As a result, GH = BP.
Now, we know that GH is parallel to BP and they have equal lengths. Any segment from H to BC which is parallel to GB must intersect BC at P. Assume, it does not. Say, it intersects in point Q:
In this way, we construct a parallelogram BGHQ with opposite sides GH and BQ which have different lengths. This is impossible. Therefore, Barney must come back to his starting point P in his last step.
What happens if Barney's starting point is outside of BC?
It seems like it again takes 6 steps for Barney to come to his starting point. It should not be difficult to prove that he will come back to his starting point looking at the previous proof.
What about the length of Barney's path?
1. If he starts from a point BC other than tthe midpoint, the length of his path is equal to the perimeter of the triangle.
PD + DE + EF + FG + GH + HP = (PD + FG) + (EF + HP) + (GH + DE)
= (EC + AE) + (AG + GB) + (BP + PC)
= AC + AB + BC = P (ABC)
2. If he starts from the midpoint of BC, the length of his path is equal to the half of the perimeter of ABC.
Each step he will take will be the half length of each side.
3. If he starts from a point outside of BC, the length of his path will be the perimeter of the triangle ABC plus twice the perimeter of PBD in the following figure:
Proof should be easy after the previous one.
A final observation about Bouncing Barney:
If you trace the intersection points of his paths (also their extensions) when P moves along BC and outside of BC, the locus will be three median lines of ABC. Click here to observe.