The Final

by Sharon K. O’Kelley

 

Part One – Bouncing Barney

 

Barney is in the triangular room ABC. 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.

                                                                                                               

Discussion

 

1. To explore this phenomenon, I decided to look at the specific situation in which Barney (indicated by point D in figure 1) starts at the midpoint of side BC.

 

Figure 1

 

Notice that by constructing the parallel lines as called for in the directions, Barney ends up where he started at point D after touching each side once. Therefore, it can be said that Barney made one circuit around the inside of the triangle.

The Triangle Proportionality Theorem states if a line intersects two sides of a triangle and is parallel to the third side then it divides the two sides into segments of proportional lengths. Therefore, since D is the midpoint of side BC and segment ED is parallel to AC, then E must be the midpoint of AB as well. The same reasoning can be applied to find that F is the midpoint of AC. Segments DE, EF and FD can then be called the midsegments of triangle ABC. According to the Triangle Midsegment Theorem, when a midsegment of a triangle is parallel to the third side of the triangle, its measure is one-half the measure of the third side. Therefore, the measure of DE is one-half the measure of side AC, the measure of DF is one-half the measure of side AB, and EF is one-half the measure of side BC. The perimeter of triangle EFD (or Barney’s circuit) is thus one-half the perimeter of triangle ABC.

 

2.  What if Barney started at a point that marked one-fourth the length of side BC as shown in figure 2?

Figure 2

Following the given directions in constructing parallel lines, Barney returned to point D after touching each wall twice. What else can we discover?

We know that segment BD is one-fourth the measure of segment BC, and by construction, we know that segment OD is parallel to segment AC. Using the Triangle Proportionality Theorem, we can state that since OD cuts off segment BD which is one-fourth of segment BC, then BO is also one-fourth of segment AB. The same reasoning can be applied around the triangle. If AQ and OB are each one-fourth of side AB, then AQ is one-third the length of segment AO. Since parallel lines have been constructed, it can also be concluded that triangles AQS and AOU are similar by the Angle-Angle Theorem of Similarity. (Angle A is congruent to Angle A and angles AQS and AOU are congruent because they are corresponding angles of parallel lines.) Because these two triangles are similar, their corresponding sides are proportional. Therefore, the length of QS must also be one-third the length of side OU. This relationship holds for UV and OD and their counterparts as well. We can also use the Angle-Angle Theorem of Similarity to state that triangles AQS and ABC are similar with the measure of QS being one-fourth the measure of side BC. This relationship holds around the triangle as well. Thus the following can be concluded….

QS +OU = BC.

OD + QV = AC.

UV + DS =  AB.

 Therefore, QS + OU + OD + QV + UV + DS = BC + AC + AB.

Hence, the perimeter of Barney’s circuit in this case equals the perimeter of triangle ABC.

 

3. What if Barney started at a random point on segment BC?

 

From the explorations above, I can deduce that Barney will always make it back to his starting point, no matter where he starts, because of the Triangle Proportionality Theorem. In this situation, the parallel segments are always marking off proportional distances that relate each side to another side; therefore, all the sides are interconnected proportionally through the parallel lines. Thus the distances that are cut off by the parallel lines have to be the same for all three sides which means Barney will eventually end up where he started. Barney will also at most touch a wall two times. Combine the proportionality theorem with the similar triangles created by the parallel lines as demonstrated above and the conclusion can be drawn that the perimeter of Barney’s circuit will always equal the perimeter of the triangular room except in instances where Barney starts at the midpoint of one wall. Consider figure 3.

 

Figure 3

In figure 3, Barney’s starting point at D has been randomly placed. Explorations with Geometer’s Sketchpad reveal that he ends back up at point D after touching each wall twice and that the perimeter of his circuit is equal to the perimeter of triangle ABC.

To see a GSP animation of this situation, go here.

 

4. What if Barney starts on a point outside of triangle ABC? Consider figure 4.

 

Figure 4

 

Notice that when Barney starts outside the triangle at point D, his circuit goes around the outside of the triangle. What is interesting to note is that no matter how far Barney is from a wall when he starts the circuit, the ratios between the sides of the initial triangle to the long sides of the circuit parallel to those sides are constant as demonstrated in the figure above.

 

For Part Two – Ceva’s Theorem, go here.

 

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