Barney Come Home!

By Paulo Tan

 

Suppose Barney is in a triangular region ABC.  If Barney starts at point D and travels on a parallel path to line AF, then he will end up on line AB at point E (fig1).  The dashed red line represents BarneyÕs path.  He then travels parallel to line BC and ends up on line AC at point F.  If Barney continues this geometric travel process will he ever come home to point D?

 

Fig1

 

 

 

WeÕll start by analyzing the path that Barney has traveled so far (fig2).  Notice that BarneyÕs path form two parallelograms, BEFG and DEFC.  

 

 

Fig2

 

Triangle EBD appears to be congruent to triangle FGC.  LetÕs try to prove it:

 

 

Since opposite sides and angles of parallelograms are congruent then from parallelogram BEFG:

 

á      Angle EBD is congruent to angle GFE

á      Line segment BE is congruent to line segment GF

 

From parallelogram DEFC:

 

á      Angle FCG is congruent to angle DEF

á      Line segment FC is congruent to line segment ED

 

In addition, since line EF is parallel to line BC, then:

 

á      Angle DEF is congruent to angle EDB

á      Angle GFE is congruent to angle FGC

 

By the transitive property:

 

á      Angle EBD is congruent to angle FGC

á      Angle FCG is congruent to angle EDB

 

It follows that the third angle of each triangle must be congruent: angle BED is congruent to angle GFC.  Thus, by the Side Angle Side Postulate triangle EBD is congruent to triangle FGC.

 

 

The next two paths result in Barney landing on point H on side AB and then on to point I on side AC (fig3).

 

Fig 3

 

 

Once again, notice that there are two new sets of parallelograms that are formed: parallelograms GHAF and HICG (fig4).

 

Fig 4

 

Earlier we proved that triangles FGC and EBD are congruent.  Using a similar proof we could show that triangle FGC is congruent to triangle AHI.  As a result we have that triangles EBD, FGC, and AHI are congruent to one another. 

 

We could think of BarneyÕs path in terms of modular arithmetic. 

 

á      The first, fourth, seventh, tenth, É paths are all parallel to side AC. 

á      The second, fifth, eighth, eleventh, É paths are all parallel to side BC.

á      The third, sixth, ninth, twelfth, É paths are all parallel to side AB. 

 

Figure 5 shows BarneyÕs path in terms of the different equivalence classes (shown in different colors) in modulo three.  Notice that the sixth path takes Barney from point I back home to point D. 

 

By using modular arithmetic we know that the sixth path must be parallel to side AB.  In addition, we have shown that triangles EBD, AHI, and FGC are all congruent.  Therefore, if BarneyÕs sixth path did not intersect at point D, then we would have a contradiction of the proven facts. 

 

Fig 5

 

The result of BarneyÕs parallel path is that congruent triangles will be formed.  Thus, we could conclude that BarneyÕs journey inside the triangular region will take at most six paths before he returns home no matter where he starts.  

 

Click HERE for a GSP file to see BarneyÕs path from different starting points on line segment BC. 

 

The GSP file confirms that BarneyÕs journey will six paths for all of the starting points on side BC except for the endpoints and a particular point on side BC that only seems to only require Barney to take three paths before returning home (fig 6). 

 

Fig 6

 

In figure 6, points D, E, and F appear to be the midpoints of their respective sides.  Lets prove this. 

 

         We know:

¤    Line AB is parallel to line DF

¤    Line BC is parallel to line EF

¤    Line AC is parallel to line DE

 

Thus using the Alternate Interior and Corresponding angle theorem we conclude that:

 

¤     Angle aÕs are congruent to each other (fig 7)

¤    Angle bÕs are congruent to each other

¤    Angle cÕs are congruent to each other

 

                 

        

 

 

 

Fig 7

 

Therefore triangles AEF, EBD, FDC, DFE are all similar to one another.  In addition, AEF, EBD, FDC, DFE are all similar to ABC.  It follows that:

 

á      Line segment BC is twice the length of line segment EF

á      Line segment AB is twice the length of line segment DF

á      Line segment AC is twice the length of line segment DE

 

Thus by the Mid-segment Theorem we have, D, E, and F as the midpoints of their respective sides.

 

Hence, if Barney starts at the midpoint then he only has three paths to take before he comes home.  At any other point, besides the endpoints, Barney must take six paths before coming home. 

 

 

Next, letÕs see what happens when Barney starts his journey outside of triangle ABC (fig8).  Click HERE for a GSP file to see BarneyÕs path from outside of triangle ABC. 

 

Fig 8

 

We see that Barney will still take six paths before returning home. 

 

 

Suppose that Barney was in a regular pentagonal region (fig 9).  If Barney takes his parallel paths as he did with the triangle will he ever return home?

 

Fig 9

 

We could see that in this case we will need to use modulo five.  This will be very useful in reducing the confusion caused by so many more lines that will be involved in this case.  We need to establish the order of lines that Barney will travel parallel to. In a triangle there are only two choices.  In a pentagon there are four.  In this case, Barney will start on side 5.  He will first travel parallel to side 1, then side 3, then side 5, then side 2, and side 4.

 

As expected, we see that after five paths, Barney returns to the starting side.

                                                                                                                                          

Repeating the process, we see that Barney, as when he was in the triangle, returns home after two cycles (fig 10).

 

Fig 10

 

When Barney travelÕs inside the regular pentagon, his path leaves behind some congruent quadrilaterals (fig 11). 

 

Fig 11

 

We see that quadrilaterals DWAX and CYFZ congruent to because:

 

á      Quadrilaterals DWAX and CYFZ are parallelograms since opposite sides are parallel.

 

á      Since we represents a regular polygon, then base angles are congruent.

 

á      Opposite sides of parallelograms are congruent. 

 

As Barney travels around the pentagon other congruent quadrilaterals are formed (fig 12).  Just like we proved quadrilaterals DWAX and CYFZ are congruent we could similarly show that all these ÒcornerÓ quadrilaterals are congruent to one another.  Thus, as with the triangle, Barney must return home, otherwise there will be contradiction of the proven congruencies.

 

Fig 12

 

What if BarneyÕs starting point is the midpoint?  Just as in the triangle BarneyÕs journey is cut to just one cycle (fig 13).

 

Fig 13

 

 

What if the region is not a regular polygon (fig14)?

 

 

 

Fig 14

 

 

 

Fig 14b

 

 

 

Fig 14c

 

 

GSP suggests that Barney will make it home in a polygon (14c).  However, the quadrilaterals that his path makes are not congruent since the polygon is not regular. 

 

Can Barney make it back home in five paths?  Figure 15 shows there exists a starting point where it appears Barney will be able to take five paths to get back home.

 

Fig 15

 

How about a regular heptagon (fig 16)?

 

 

Fig 16

 

 

Notice again we have what appear to be congruent ÒcornerÓ quadrilaterals.  We should be able to prove using similar techniques as we did with the triangle and the regular pentagon.

 

We should expect similar results with the midpoint as the starting point (fig 17).

 

 

Fig 17

 

Thus, it appears that any odd nth-gon will help Barney return home. 

 

It remains to show that Barney could return home in a region with an even number of sides, although it seems unlikely. 

 

 

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