Problem 1: 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.
I began by constructing Barney’s path on Geometer’s Sketchpad. When I manipulated Barney’s starting point on segment BC, I found that Barney always returns from his starting point. He also appears to take one of two paths. Either he returns to his starting point after touching each wall twice (Figure 1), or after touching each wall once. (Figure 2) When he runs into each wall once I will call that 1 circuit, and when he hits each wall twice I will call that 2 circuits.
1. It appears that when Barney only needs 1 circuit to end at his starting point (Figure 2), it is because he started at the midpoint of segment BC (or at one of the vertices of rABC).
I began by assuming that Barney starts his journey at the midpoint of segment BC. I know that when a line is parallel to one side of a triangle and passes through the other two sides, then it divides the two sides into proportional segments. Therefore, since Barney started at the midpoint of segment BC, he must also hit segment AB at its midpoint, E. By similar reasoning, he must also hit segment AC at its midpoint, F. And when he leaves the midpoint of segment AC, he will end at D, the midpoint of segment BC, his starting point. Q.E.D.
It falls from this proof that Barney’s path is the medial triangle of rABC since he hits the midpoint of each side. I also know that when a midsegment (the segment between the midpoints of two sides of a triangle) is parallel to the third side of the triangle, its measure is half the measure of the third side. This means
DE = ½AC, EF = ½BC, and FD = ½AB
DE + EF + FD = ½(AC + BC + AB)
So Barney’s path is half the length of the perimeter of rABC.
2. If Barney starts anywhere other than the midpoint or one of the vertices of side BC of rABC, he needs two circuits to return to his starting point (Figure 1).
Suppose Barney does not meet return to his starting point and instead ends his two circuits at a point J on segment BC. (Figure 3)
Let Barney start his path at a point D on side BC. He then runs into segment AB at a point E. Since when a line is parallel to one side of a triangle and passes through the other two sides it divides the two sides into proportional segments, I know that
By the same reasoning, when Barney runs into segment AC at point F, I know that
So by transitivity,
Therefore BD = BJ. Which is a contradiction. So D and J are the same point, and Barney returns to his starting point in 2 circuits. Q.E.D.
Conjecture: The length of Barney’s path equals the perimeter of rABC, when he starts anywhere other than the midpoint or one of the vertices of side BC of rABC.
LEBD LFGC by corresponding angles
LBDE LGCF by corresponding angles
Therefore, rBED rGFC by Angle, Side, Angle
and m(ED) = m(FC)
LBHG LAEF by corresponding angles
LAEF LHBG by corresponding angles
m(AE) = m(HB)
Therefore, rAEF rHBG by Angle, Side, Angle
and m(HG) = m(AF)
Since m(ED) = m(FC) and m(HG) = m(AF),
m(AF) + m(FC) = m(HG) + m(ED)
m(AC) = m(HG) + m(ED)
m(AB) = m(FG) + m(ID)
m(BC) = m(HI) + m(EF)
m(AB) + m(BC) + m(AC) = m(FG) + m(ID)+ m(HI) + m(EF) + m(HG) + m(ED) Q.E.D.
3. What if Barney starts on a point outside of rABC? (Figure 4)
It appears that he will still return to his starting point.
Since RS || AB and ST || BC, LRST LABC
Since ST || BC and RT || AC, LSTR LBCA
Since RT || AC and RS || AB, LSRT LBAC
So, rABC and rRST are similar. And since D, E, F, G, H, and I are on rRST, Barney will always return to his starting point. Q.E.D.
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