Bouncing Barney

by Margaret Morgan

Bouncing Barney Problem:

Barney is walking inside a triangular region. He begins at any point on a side of the triangle and walks on a path parallel to another side until he reaches the third side. Here, he turns and walks on a path parallel to the side he just came from until he reaches another side of the triangle at which point he turns and continues on a path parallel to the side he just came from. He continues this process until he returns to his starting point.

From the above picture, it appears that Barney is able to return to the point at which he started. Let us prove that this is the case. We will begin by naming our points to make them easier to talk about.

By construction, we know the following:

DE || GH || BC

HI || EF || AC

FG || ID || AB

This gives us the following parallelograms and associated facts:

 The length of segment BE = The length of segment DI The length of segment BI = The length of segment ED

Suppose, when Barney left point I, he arrived on segment AC at a point other than D--let's call it D'. Then AHID' would be a parallelogram and it would follow that the length of HI = the length of AD'.

But, we know the length of HI is equal to the length of AD, so D and D' must be the same point.

Proof that the length of HI is equal to the length of AD even when the segment ID is removed:

 The length of HI = the length of GC--see above The length of ED = the length of FC --see above The length of AE = the length of FG The measure of

Next, let us prove that Barney's path is equal in distance to a path that goes from vertex to vertex on the original triangle. Or, more simply stated, the length of Barney's path is the same as the perimeter of the original triangle.

From above we know that the following pieces of the triangle and pieces of Barney's path are of equal length.

 Piece of Triangle Piece of Barney's Path AE GF EB ID BF HG FC DE CD EF AD HI

Since the pieces of Barney's path and the pieces of the triangle are in a one to one correspondence, the path and the perimeter of the triangle are of equal length.

Barney will reach a wall at most 6 times (the sixth time being when he returns to the starting point). When Barney starts at a vertex or the midpoint of a side, Barney will reach a wall three times (the third time will be when he returns to his starting point). Unless you allow him to go around twice.

Special Cases

The first special case I want to consider is if Barney starts at the midpoint of one of the sides of the triangle.

 It appears that in this case, Barney bounces from mid-point to mid-point on the triangle and that the triangle is divided into four congruent triangles.

Proof that if we start at a midpoint, we go to the midpoints of the other two side.

From our parallelograms above, we can deduce that the lengths of the following segments and the measures of the following angles are equal:

 BE = DI = AH EF = DC = AG BF = IC = HG

Which means triangles AHG, EBF and DIC are congruent:

If D is the midpoint of AC, then DC = AD. Thus the congruent triangles give us that AD = EF = AG--so Barney will only touch the midpoint of AC.

Since in this special D and G are the same point, H and E must be the same point (Because H and E are constructed in the same way--line parallel to BC and through point D or G). Similarly, I and F must be the same point.

We know BE = AH, so the point E/H is the midpoint of AB. We also know BF = IC, so the point I/F is the midpoint of BC.

Now, we can show that all four triangles resulting from this special case are congruent. For simplicity sake, lets refer to the midpoints as D, E, F. We know triangles ADE, DFC, and AEF are congruent already, so we need only show triangle DEF is congruent to one of these.

DE = FC, EF = DC, and DF = DF, so triangles DEF and DFC are congruent by SSS.

Because we know Barney's path (twice around triangle DEF) is equal in length to the perimeter of ABC, we can conclude that the perimeter of DEF is half the perimeter of ABC.

The next special case I want to consider is when we choose D, so that segments EF, GH, and ID intersect at a single point.

It appears that this divides the triangle into nine congruent triangles.

AEJD is a parallelogram so AE = DJ, AD = EJ, and ED = ED, so triangle AED is congruent to triangle EDJ.

EDJG is a parallelogram so ED = JG, EJ = DG and JD = JD, so triangle EDJ is congruent to triangle JDG.

JDGF is a parallelogram so JD = FG, DG = JF, and JG = JG, so triangle JDG is congruent to triangle JGF.

JFGC is a parallelogram so JG = FC, GC = JF, and FG = FG, so triangle JGF is congruent to triangle FGC.

GFIJ is a parallelogram so IJ = FG, IF = JG, and JF = JF, so triangle JGF is congruent to triangle IJF.

HIFJ is a parallelogram so HJ = IF, HI = JF, and IJ = IJ, so triangle IJF is congruent to triangle HIJ.

HBIJ is a parallelogram so HJ = BI, BH = JI, and HI = HI, so triangle HIJ is congruent to triangle BHI.

HJDE is a parallelogram so HJ = ED, HE = JD, and EJ = EJ, so triangle EDJ is congruent to triangle HEJ.

Because the nine triangles are congruent, we see that Barney has trisected the sides of the original triangle.

Finally, let's explore what happens when we extend the sides of the triangle and allow Barney to start somewhere out the triangle.

It appears that two of our conjectures still hold. Barney still returns to his starting point and he still runs into a wall six times (the sixth being when he returns to his starting point.

But, our conjecture that the length of Barney's path is the same as the perimeter of the original triangle no longer holds.

By moving Barney's starting point, we see that there is not a constant ratio between the length of Barney's path and the perimeter of the triangle.

But does this mean there is no relationship? It appears that if we add the lengths of the line segments from the starting point to E, from H to F, and from G to I and then subtract the lengths of line segments from E to H, from F to G, and from I back to the starting point, we get the perimeter of the original triangle. Click here to see a GSP sketch that you can manipulate.

Let's see if we can prove that SE + HF + IG - SI - EH - GF = BD + BC + CD.

SBEH is a parallelogram by construction. So, SE = BH = BC + CH.
HCGF is a parallelogram by construction so HC = GF. By substitution, SE = BC + GF OR SE - GF = BC

EHFD is a parallelogram by construction. So, HF = ED = EC + CD.
ECIS is a parallelogram by construction so EC = SI. By substitution, HF = CD + SI OR HF - SI = DC

SDGI is a parallelogram by construction. So, IG = SD = SB + BD.
EHSB is a parallelogram by construction so EH = SB. By substitution, IG = BD + EH OR IG - EH = BD

Thus, SE - GF + HF - SI + IG - EH = BC + DC + BD.