Bouncing Barney

By Pei-Chun Shih

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.

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

2.    How many times will Barney reach a wall before returning to his starting point?

3.    Explore and discuss for various starting points on line BC, including points exterior to segment BC.

4.    Discuss and prove any mathematical conjectures you find in the situation.

Assume that Barney starts at the point D and ends his journey at point DŐ where D  DŐ as shown above. Since BarneyŐs paths are all parallel to the three sides of triangle ABC, the following are given: DE || HG || AC, HI || EF || BC, and FG || DŐI || AB. Therefore, angle ABC = angle IDŐC because they are corresponding angles of parallel lines AB and DŐI.

Next, letŐs connect the starting point, D, with point I as shown below. We want to prove that triangle HBG is congruent to triangle IDC that are colored with gray and green respectively. As we can see from the triangle ABC on the right below, there are three parallelograms, BEFG, DEFC, and GHIC, since their opposite sides are parallel as mentioned earlier. Therefore, BG = EF = DC and HG = IC by the property of a parallelogram that the opposite sides are equal in length.  Besides, the two angles BGH and DCI are equal since they are corresponding angles of the parallel lines HG and AC. Hence, triangle HBG and IDC are congruent by the Side-Angle-Side Congruence Axiom. So angle HBG = angle ABC = angle IDC.

However, weŐve got the relation that angle ABC = angle IDŐC from the fist section of our proof. Therefore, angle IDŐC must equal angle IDC since both of these two angles are equal angle ABC. Thus, angle DIDŐ = 0 and line segments ID and IDŐ are congruent. We have proved that Barney will eventually return to his starting point, D, since D = DŐ.

Now we know that Barney will eventually return to his starting point, D. Here I am going to use another point of view to verify this conclusion.

There are several parallelograms inside triangle ABC as shown above since BarneyŐs paths are always parallel to the three sides of triangle ABC.  These parallelograms are BEFG, GHIC, and AEDI as marked in yellow, blue, and gray respectively. By the property of a parallelogram, the opposite sides are equal in length. Therefore, BG = EF, BE = FG, CI = GH, CG = HI, AE = ID, and AI = DE. The total path of BarneyŐs journey is the sum of the following segments: DE + EF + FG + GH + HI + ID which equal AI + BG + BE + CI + CG + AE. After rearranging it, we get (AI + CI) + (BG + CG) + (BE + AE) = AC + BC + AB, which equals the sum of the three sides of triangle ABC. Since the total paths of BarneyŐs journey are equal to the perimeter of triangle ABC, we can view BarneyŐs traveling paths as he starts from point D and goes along the perimeter of triangle ABC then back to the original point D in the end. This is another way to verify that Barney will eventually return to the starting point.

Now, letŐs extend our discussions to the situation which the starting point is outside triangle ABC. Suppose that Barney starts his journey at point D where exterior to segment BC and follows the same pattern as he did earlier. Will he return to his starting point as before? Assume that Barney ends his journey at point DŐ where D  DŐ as shown below left. Since BarneyŐs paths are always parallel to the three sides of triangle ABC, the following are given: DE || AC || HG, HI || BC || EF, and FG || AB || DŐI. Therefore, angle ABC = angle IDŐC because they are corresponding angles of parallel lines AB and DŐI.

Next, letŐs connect the starting point, D, with point I as shown above right. We want to prove that triangle HBG is congruent to triangle IDC that are colored with blue and yellow respectively. As we can see from the diagram above right, there are three parallelograms, BEFG, DEFC, and GHIC, since their opposite sides are parallel as mentioned earlier. Therefore, BG = EF = DC and HG = IC by the property of a parallelogram that the opposite sides are equal in length.  Besides, the two angles BGH and DCI are equal since they are corresponding angles of the parallel lines HG and AC. Hence, triangle HBG and IDC are congruent by the Side-Angle-Side Congruence Axiom. So angle HBG = angle ABC = angle IDC.

However, weŐve got the relation that angle ABC = angle IDŐC earlier. Therefore, angle IDŐC must equal angle IDC since both of these two angles are equal angle ABC. Thus, angle DIDŐ = 0 and line segments ID and IDŐ are congruent. Again, as we have proved before, Barney will eventually return to his starting point, D no matter his starting point is inside or outside triangle ABC.

How many times will Barney reach a wall before returning to his starting point? Well, it depends on where his starting point is.

Suppose that his starting point D is always on the line BC. If point D locates outside triangle ABC, then he will always reach a wall five times before returning to his starting point.  If D is any point inside triangle ABC except the midpoint of BC, then he also needs to reach a wall five times before returning to D. However, if point D is the midpoint of BC, then he will reach a wall only two times before returning to D.