Bouncing Barney

by

Rui Kang

The problem:

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 this 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.

Overview: In the following investigation, we will explore seven different cases for this same problem,

starting with the easier cases and moving on to the harder ones.

In particular, these cases are:

Case 1: Barney starts at a vertex of triangle ABC
Case 2: Barney starts at the midpoint of a side of triangle ABC
Case 3: Barney starts at any point on a side of triangle ABC, and the point is inside triangle ABC
Case 4: Barney starts at any point on the interior of triangle ABC
Case 5: Barney starts at the centroid of triangle ABC
Case 6: Barney starts at the orthocenter of triangle ABC
Case 7: Barney starts at a point on the extension of a side and outside the triangle

We claim that for each of these cases, Barney will always return to his starting point.

However, the path and distance he has traveled vary from case to case. Next, let's explore each of these cases more in-depth.

Case 1: Suppose Barney starts at vertex B.

This case is fairly straight forward. If Barney starts at B and has to walk parallel to one of the sides of triangle ABC,

he will be walking along either side BC or side AB.

Since it does not matter which one he chooses to walk on first,

let's say, Barney walks on BC first and there is no problem since BC is parallel to itself. Then his path will look like:

Barney will hit wall or bounce three times (including the time he hits B(P)) before he goes back to the point wheret he starts.

And the distance he has traveled in this case is simply |BC| + |AC| + |AB| or the perimeter of triangle ABC.

Case 2: Suppose Barney starts at the midpoint of side BC, let's call it P.

Barney's path in this case is like:

Barney will hit wall or bounce three times (including the time he hits P) before he goes back to the point where he starts.

We claim that the distance he has traveled in this case is half of the perimeter of triangle ABC. Here is one way to prove it:

(Note: in the following proofs, I used "bar" instead of "| |" to represent the lenght of a side of a triangle)

The path that Barney has traveled also creates a pattern of similar or congruent triangles. For example,

Triangles AEF, BEP, PFC are all similar to triangle ABC.

Triangles AEF, BEP, PFC, and PEF are all congruent to each other. This means that triangle PEF is also similar to triangle ABC.

Case 3: Barney starts at any arbitrary point on side BC (except for point B, point C, and the midpoint)

Barney's path in this case is like:

Barney will hit wall or bounce six times (including the time he hits P) before he goes back to the point where he starts.

We claim that the distance he has traveled in this case is the perimeter of triangle ABC. Here is one way to prove it:

The path that Barney has traveled also creates a pattern of similar or congruent triangles. For example,

Triangles AHI, BEP, and GFC are all congruent to each other by the SSS criteria.

In addition, let's call the intersection of HG and PI, K, the intersection of PI and EF, L, and the intersection of HG and EF, M.

Then we can easily prove that triangles AHI, HIK, BEP, EPL, CFG, and MGF are all congruent to each other.

From the above, we also know that triangles AHI, HIK, BEP, EPL, CFG, and MGF are all similar to triangle ABC.

In addition, triangle HIK is similar to triangles PKG and KLM. Similarly, triangle MGF is similar to triangles KML and HME;

triangle ELP is simlar to triangles KLM and ILF.

Finally, all of these triangles, HIK, PKG, KLM, MGF, HME, ELP, ILF are similar to triangles ABC, AHI, BEP, and CFG by transivity.

Any other similar or congruent triangles that you can see for Case 3?

Case 4: Barney starts at any point on the interior of triangle ABC.

The path that Barney travels in this case is like:

When Barney bounces from J and travels parallel to AC toward side BC and before he hits BC at E again,

he passes through point P, i.e., he goes back to the original point where he starts.

Note that Barney still will hit wall or bounce six times before he goes back to the point where he starts,

except that path 7 and 1 in fact can be seen as one path divided into two parts with P lying on the path.

We observe that although the path that Barney travels is different than that for Case 3,

the distance he travels is exactly the same as that in Case 3.

Both distances equal the perimeter of triangle ABC.

In addition, the patterns of similar or congruent triangles created by the paths for both cases also match with each other.

Therefore, we will skip the proof for this case and move on to a more challenge case.

However, you are welcome to prove this case by modeling after my proof for Case 3.

Case 5: Barney starts at the centroid of triangle ABC.

We know that Barney for sure will go back to the centroid eventually because in Case 4,

we selected any arbitrary point on the interior of ABC.

Since a centroid is always inside triangle ABC, we know Barney will travel back to the centroid in the end.

However, the path that Barney has traveled is very different than that in Case 4. In particular, it looks like:

Barney will hit wall or bounce only twice before he goes back to his starting point P

When Barney bounces from F and travels parallel to BC toward side AB and before he hits AB at U,

he passes through point P, i.e., he goes back to the original point where he starts.

If we use GSP's "measure" function, we will find that the distance that Barney has traveled is one third of the perimeter of triangle ABC, i.e.,

,

but can we prove it?

We have noticed so far, this is the shortest distance Barney has traveled.

The path that Barney has traveled under this case also creates patterns of similar or congruent triangles, for example:

triangle CEF is congruent to PEF, both are then similar to triangle ABC,

in addition, triangles CEF, AUF, and BEV are all similar to triangle ABC.

In fact, triangle PEF is also similar to triangle ABC by the AAA criteria, can you prove it?

Furthermore triangle APF is simlar to ALC; APU is similar to ABL; BPV is similar to ABM; BPE is similar to BMC and etc.

Case 6: Barney starts at the orthocenter of triangle ABC.

There are actually two subcases if Barney starts at the orthocenter because orthocenter can lie either inside or outside a triangle.

When all three angles of a triangle are less than 90 degree, the orthocenter lies inside the triangle.

When one of the three angles of a triangle is greater than 90 degree, the orthocenter lies outside the triangle.

Observe that under both cases Barney has eventually traveled back to his starting point.

Subcase 1:

For the case that the orthocenter lies inside triangle ABC, we are not surprised that Barney has eventually travelled back to his starting point.

This is because under Case 4, we have already shown that the conclusion is true for all arbitrary points we select inside a triangle.

Also by case 4, we know that the total distance that Barney has traveled if he starts at the orthocenter,

if inside triangle ABC, is also the perimeter of triangle ABC.

Now let's skip the proof for subcase 1 of Case 6 and move on to subcase 2.

Subcase 2:

We observe that even when the orthocenter lies outside of triangle ABC, Barney still has eventually traveled back to his starting point, i.e., the orthocenter.

Of course, the distance he has traveled must be longer than the perimeter of triangle ABC this time,

but we will save the proof for Case 7 below because the path, the distance, and the pattern of similar or congruent triangles

that are generated from subcase 2 of Case 6 matches those generated from Case 7.

Case 7: Barney starts at a point on the extension of a side (i.e., BC) and outside the triangle.

We observe that under this case, Barney is still able to eventually go back to his starting point, and the path he has travelled looks like:

He will hit wall or bounce a total of six times (including the time he hits P) before he goes back to his starting point.

The distance he has traveled depends on how farther away P is from vertex B in this case

(if we put p to the right of C then, the distance depends on how farther away P is from C instead of B)

An easier way to look at the distance, though, is:

Perimeter of triangle ABC + 2* Perimeter of triangle AHI

= Perimeter of triangle ABC + 2 * Perimeter of triangle PEB

= Perimeter of triangle ABC + 2* Perimeter of CFG.

During this proof, we have also shown that the distance that Barney has travelled can be written as |HE| + |IF| + |PG|,

if you prefer this expression to using perimeters of triangles.

This proof also implies that the path that Barney has traveled in this case creates a series of similar or congruent triangles as in previous cases,

for example,

Triangle CFG, AHI, and BPE are all congruent to each other and simultaneously similar to triangle ABC.

In addition, triangle ABC is also similar to triangles CPI, AEF, and BHG.

In fact, subcase 2 of Case 6 and Case 7 can be generalized to all such cases where Barney starts from a point located outside of triangle ABC.

Summary:

(1) No matter where Barney starts, he will always be able to travel back to his starting point if he follows the rules set up in this problem.

(2) Barney will hit the wall or bounce at most six times and at least two times.

(3) The paths he travels under each of the cases create patterns of similar or congruent triangles.

(4) The distance he travels depends on where he starts and the path he travels.

Under most cases, the distance he travels equals the perimeter of the triangle room if Barney starts at a point located inside the triangle room.

(5) The shortest distance he travels is 1/3 of the perimeter of the triangle room, when he starts at the centroid of the triangle room.

(6) When he starts at the midpoint of any side of the triangle room, he may also travel less than the perimeter of the triangle room.

Under this case, he travels half of the perimeter of the triangle room.

(7) Sometimes, he has to travel more than the perimeter of the triangle room. There are two cases when this happens.

One is that Barney starts on the extension of a side of the triangle room. The point he starts at is in fact outside of the triangle room.

The other case is when he starts at the orthocenter of a triangle room with one of its angles greater than 90 degree.

Under this case, Barney also starts at a point outside of the triangle room.

Under the last two circumstances, the distance that Barney travels cannot be decided for sure,

although there are certain ways to express this distance that makes its calculation easier.

The rule of thumb is that under these last two circumstances,

the distance that Barney has to travel in order to go back to his starting point is longer than the perimeter of the triangle room,

and that the farther away Barney is from the closest vertex of the triangle room when he starts bouncing,

the longer he has to travel in order to go back to his original starting point.

Finally, these last two circumstances can be generalized to all the cases when Barney starts at a point outside of the triangle room.