Final Project

 

By: Laura Lowe

 

 

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.

 

Figure 1

 

Figure 2

 

 

 

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).

 

Proof:

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). 

 

Proof:

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)

 

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

Similarly,

                                          

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.

 

Proof:

LEBD  LFGC by corresponding angles

LBDE  LGCF by corresponding angles

m(BD)  m(GC)

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)

 

Similarly,

m(AB) = m(FG) + m(ID)

m(BC) = m(HI) + m(EF)

 

Therefore

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) 

 

Figure 4

 

It appears that he will still return to his starting point.

 

Proof:

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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