**Final Assignment:**

**By Nikhat Parveen**

**PART A**:

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. 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.

**fig.1**

**Analysis:**

We can start with a point on the side of a triangle that will represent the Barney's moving point. Since Barney is moving from a point on BC parallel to AC. First We will construct a point on BC and then construct a parallel line to AC. Then we will construct the point of intersection between the parallel line and the side of a triangle. We will construct the other parallel lines that will represent the Barneys path and complete the figure shown as below:

**Barney's Walking Path**

fig.2 Click here for the GSP file

**Tracing Barney's path:**

Barney start at point B on BC and walk parallel to AC (pink path) and touches the wall AB at '1'

Barney turns around walk parallel to BC (green path) and hit the wall AC at '2'.

Barney bounces and walks parallel to AB (blue path) and hit the wall BC at '3'

B' again walks parallel to AC (purple) and touches AB wall again at '4'.

He turns around and walks parallel to BC again (dark blue) and hit the wall AC at '5'

Finally Barney walks parallel to AB and come to his original point where he started from and touches the wall sixth time.

So, in all Bouncing Barney touches wall six times unless he starts at the midpoint of one of the side of the triangle, then he will touch the wall only three times.

But how sure are we that Barney will return to his original point?

Let's try to prove .........

**Proof:**

Consider the parallel lines and the triangles formed by following Barney's path as shown in the figure below.

Notice that DE is parallel to BC, GF | | AC and RH | | AB by the construction laid down by Barney's walking or moving commands.

According to the parallel lines property-
*A line parallel to one side of triangle and
intersecting the other two sides divides the two sides proportionally*.

Since the parallel line DE intersects side AB and side AC in D and E, it divides the two sides AB and AC proportionally ie., DB/AB = EC/AC

Likewise, GF is parallel to AC and intersects the side AB and BC in G and F. Therefore, AG/AB = FC/BC.

Similarly, HR is parallel to AB and intersects side AC in H and side BC in R dividing it proportionally.

By parallel line property we have BR/BC = AH/AC.

Consider the figure below:

*Some properties listed below will be used in
proof:*

*If two parallel lines are cut by a transversal
the corresponding angles are equal.*

*Similar triangles are triangles which have
their corresponding angles equal and their corresponding sides in proportion.*

We know from above that
DR | |
AC and
DB/AB = BR/BC. Also notice that angle
BDR
BAC
*
(corresponding angles)* and
BRD
BCA
*
(corresponding angles)*. Therefore by the
conditions of similarity
Δ DBR **~** Δ ABC.

Considering the above figure we have
DE | |
BC and BD/AB = EC/AC. Angle ADE is
congruent to angle ABC *(corresponding angles)* and angle AED is congruent to angle ACB
*(corresponding angles)*. Therefore, by
the conditions of similarity Δ ADE **~** Δ ABC.

Similarly
EF is parallel to AB
and divides the sides proportionally, EC/AC = CF/BC. Angle FEC is congruent to
angle BAC *(corresponding angles)* and angle
EFC is congruent to angle ABC *(corresponding angles).* Therefore, by
the conditions of similarity Δ EFC **~** Δ ABC.

By the similar hypotheses, Δ AGH **~** Δ
ABC. So, we have three triangles that are similar to the original triangle
ABC. Let's see the complete path of Barney as shown by the figure below.

All the three purple triangles are congruent to the original triangle ABC. Notice that the three quadrilaterals BDYR, FZEC, and XGAH form three parallelograms (why? because, the pair of sides for all these three quadrilaterals are both equal and parallel) with in the triangle ABC.

Since these parallelograms share the sides with
the triangles DBR, EFC and AGH. We have
BD
RY,
BR
DY and
DR
DR (every
segment is congruent to itself) By SSS Δ
BDR Δ YDR. Also, *by the property of
parallelogram, diagonal of a parallelogram divides the parallelogram into two
congruent triangles. * By the similar hypotheses, Δ ZFE **~**
Δ EFC and Δ XGH **~** Δ AGH. Therefore,
all the light blue triangles are congruent to all the purple
triangles that are congruent to the original triangle ABC.

Δ XYZ **~** Δ
ABC by AAA postulate since YXZ
BAC;
XYZ
ABC &
XZY
ACB.

By our analysis of similar triangles and the parallelograms formed by the Barney's walking command, Barney does return to his original point (R) from where he started, if say Barney doesn't return to his origin then he must return to some other point say R', then we want to prove that BR = BR'.

By parallel line property, BR'/BC = AH/AC but AH/AC = AG/AB = FC/BC

By transitivity, we have BR'/BC = FC/BC --> BR' =
FC. But FC =
GH and we have already proved that Δ
DBR **~ ** Δ AGH. Therefore, by CPCT
GH =
BR which implies BR' = BR (by transitivity). Therefore,
this proves that Barney does returns to
his origin point.

**Other conjectures:**

If when Barney's starting point is one of the trisection point i.e. when BR = RF = FC on the side of the triangle from where he starts, then Barney crosses the center of the triangle, centroid and we get a total of nine small triangles. Each angle of these small triangles are congruent to either angle A, B or C.

But the trapezoid on the upper part ie., DGHE has angles that is equal to the sum of A and B, & A and C. For example, DGH = sum of A and C and GHE = sum of A and B.

So from the above findings we can conjecture:

Based on the Barney's bouncing directions his path will always results in congruent triangles with each angle of the triangular room as long as his path is confined to the three pairs of parallel lines.

If Barney starts from another side instead of BC, he will still form the congruent triangles with the original triangle and every thing remains the same as above.

What do you think will happen if Barney's path starts outside the triangular room? Make a guess!

Click here to find out more.