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Summer Tuggle

Final Assignment part I

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.

Barney starts at point G and walks to H, I, J,K, L and then back to G following the rules stated in the problem.

Using Geometer’s Sketchpad, you can manipulate the construction to make all the intersections concurrent with one of the vertices of triangle ABC.

It seems that Barney’s green path is the same as the perimeter of the original triangle.  If proven that the green path is equal to the blue path, then we could assume that Barney will always end up where he started. The next construction shows point G in a different spot than in the first two pictures and again it seems that G is the beginning and ending point.

BH + HK + KA + AL + LI + IC + CJ + JG + GB = perimeter of ΔABC

And GH + HI + IJ + JK + KL + LG = Barney’s path

KJ is parallel to AI and JI is parallel to KA and KAIJ is a quadrilateral.  Since the opposite sides are parallel then KAIJ is parallelogram where opposite sides are also equal.

KA=JI and AI= KJ and since AI=AL + LI then KJ=AL + LI.  We can now substitute into our perimeter equation:

BH + HK + KA + AL + LI + IC + CJ + JG + GB = perimeter of ΔABC

BH + HK + JI + KJ + IC + CJ + JG + GB = perimeter of ΔABC

The same logic follows with quadrilateral BKLG.  Since GL is parallel to BA and KL is parallel to BG then BKLG is a parallelogram with opposite sides congruent.  BG = KL and BK = GL.  Since BK = BH + HK then GL = BH + HK and the following substitutions can be made:

BH + HK + JI + KJ + IC + CJ + JG + GB = perimeter of ΔABC

GL + JI + KJ + IC + CJ + JG + KL = perimeter of ΔABC

To complete the substitutions we need to look at quadrilateral GHIC.  HI is parallel to GC and GH is parallel to CI so GHIC is a parallelogram with opposite sides equal. GH = CI and GC = HI and since GC = GJ + JC then HI = GJ + JC and the following substitutions can be made:

GL + JI + KJ + IC + CJ + JG + KL = perimeter of ΔABC

GL + JI + KJ + GH + HI + KL = perimeter of ΔABC

Since GH + HI + IJ + JK + KL + LG = Barney’s path

then Barney’s path now equals the perimeter of ΔABC which means that he will always start and end in the same place.