Bouncing Purple Dinosaur

Presented By

Dana TeCroney

Barney is in the triangular room shown here. He walks from a
point, D, on BC parallel to AC. When he reaches AB, he turns and walks parallel
to BC (the line/wall he just came from). When he reaches AC, he turns and walks
parallel to AB. What I wish to prove is that Barney will eventually return to
his starting point. How many times will Barney reach a wall before returning to
his starting point?

Well, first look at the entire situation. The path of travel is shown below by
the red lines.

By construction, there are a plethora of parallel lines. In the picture below, all the blue
lines are parallel, all the green lines and parallel, as are all the brown
lines. Note that all the lines
that are not parallel act as transversals.

First, I will argue that the triangles formed in the corners are
congruent. Consider one
corner. In the figure below, angle
EBD is congruent to angle JDG (G is along the side BC) since they are
corresponding angles. Angle JDG is
also congruent to angle EJD because they are alternate interior angles, and
hence, angle EBD and angle EJD are congruent.

Angles EDB and angle DEJ are also alternate interior angles and
hence congruent. Now since ÆEDB
and ÆDEJ share side ED, they are congruent by an angle-angle-side argument.

A similar argument can be made for the other two corners of the
triangle.

Another set of situations created by the parallel line is the
formation of parallelograms. In
the figure below, the opposite sides are parallel, and hence the shaded figure
is a parallelogram. A result that
follows is that side DE is congruent to CF.

Why is this important?
It provides the missing link to show the all the triangles in the
corners are congruent. In the
figure below, angles EJD and LFG are congruent because they are corresponding
one another (parallel lines FG and ID cut by transversal EF). Similarly, angles JED and FLG are
congruent (parallel linseed and HG cut by transversal EF), and hence, ÆEJD is
congruent to ÆLFG by a side-angle-angle argument.

The transitive property then gives us the fact that ÆEBD is
congruent to ÆFGC. This implies
that side BD is congruent to side CG.

A similar argument can be made to show that the shaded orange region
below is a parallelogram and hence side CG is congruent to HI. Side HI is also congruent to BD for the
same reasons.

The moment of triumph,

Remember, Barney walks from a point, D, on BC, parallel to AC.
When he reaches AB, he turns and walks parallel to BC (the line/wall he just
came from). When he reaches AC, he turns and walks parallel to AB. Barney will then walk parallel to AC
and reach AB at point H, where he turns and walks to I. From there he turns and walks until he
hits BC. Since he's walking
parallel to AB, his finishing distance will be the same as the distance from H
to I (side HI). In the arguments
above, it was shown that HI is congruent to BD, which was the starting
distance, hence Bouncing Barney will finish in the same place that he started.

If you would like to experiment with Bouncing Barney, PRESS HERE. What do you think would happen if side
BC were extended to be a line, so
D is outside the triangle?
Interestingly enough, the result still holds, to see it, PRESS HERE.