BOUNCING BARNEY
by: Katie Gilbert
A. Bouncing Barney. We discussed
this investigation in class. Your challenge now is to prepare
a write-up on it, exploring the underlying mathematics ideas and
conjectures.
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.
To begin let's take a look at the different
paths Barney can take, starting from different points. First,
let's take a look at the path Barney takes when he starts inside
the triangle. Let's investigate the following:
Does he come back to his starting point?
If yes, how many paths does it take to return to his point of
origin?
If Barney does not get to the starting
point, how long does it take Barney to "fill up" the
triangle with his paths?
From our picture above it looks like
Barney will get back to the starting point, and to get back to
the starting point he traveled six paths. Let's now investigate
a different starting point; we will pick the point that trisects
the side as his starting point.
We see here that Barney again returns
to his starting point and again his journey consists of six paths
.
Now let's investigate when Barney starts
at the midpoint of the side, will he return to his starting point
and how many paths will he take?
Interesting! It appears as if he again
returns to his point of origin, but it only takes him three paths
to get back to his starting point.
Now let's take a look at when he starts
at one of the vertices of the triangle.....
He again returns to his point of origin,
but once again it takes him three paths to return to his starting
point.
If you wish to explore this further before we begin
to use geometry to solve this problem click
here to vary Barney's Starting point and see what his paths
look like.
So what we have found from our explorations,
is that Barney always returns to his starting point, and that
it either takes him three or six paths to get back to his point
of origin. If we take a look at our case where Barney starts at
one of the vertices, we see that his distanced "bounced"/
traveled is equal to the perimeter of our triangle. We will us
GSP to investigate our cases above and compare the perimeter of
our triangle to the length of Barney's journey.
Click here to view calculated
measurements of Barney's journey compared to our triangle.
We see that if Barney's point of origin is
inside our triangle then the length of his journey IS equal to
the perimeter of our triangle. We will use the properties of a
parrallelogram to prove our findings....
We will first define a paralleogram and point
out properties we will use......
In geometry, a parallelogram
is a quadrilateral with two sets of parallel sides. The opposite sides of a parallelogram
are of equal length, and the opposite
angles of a parallelogram are congruent.
Let's take a look at the different parallelograms
in our Triangle and then use the fact that opposite sides of a
parallelogram are equal.
We see above that length of AE = length of OF,
and length of AO = length
of EF.
From this picture we see that the length of BG = length of EH, and the length
of BE = length of GH.
Lastly, we see that
the length of OC = length
of DG, and length of OD = length of CG.
So we can conclude
that:
We know the perimeter
of our triangle is the sum of the length of the sides, so we can
write it as:
We also know the length
of each side is equal to the sum of its parts:
We will substitute
these "sub-lengths" into our original equation and then
substitute there equivalent lengths:
We can write Barney's
Journey as the sum of his paths:
We then compare the
two equations and we see that.......
So when we use the
properties of a parallelogram we see when Barney starts inside
the triangle (except when his starting point is at the median
of the side, which we will discuss in a moment) that the length
of the paths he takes to "bounce" back to his starting
point (his complete journey) are equal to the perimeter of our
triangle. This is obviously the case if Barney starts at one of
the vertices.
So let's find out what
is happening when Barney starts at the median of the side, still
using what we discovered above.....
Using the properties
of the medial triangle we know that the perimeter of our medial
triangle = 1/2 perimeter of our big triangle, so we will us this
to figure out what is created by Barney's path when he starts
at the midpoint of the side....
.
So we can say if Barney
starts at the midpoint of the side, the length of his journey
is equal to half the perimeter of our triangle. If you want the
length of Barney's journey to equal the perimeter of our triangle,
Barney travels around his path twice if he starts at the mipoint
of the side.
Lastly, let's explore what happens
if Barney starts outside the triangle, does he get back to his
point of origin? How many paths does he take?
Well, we see once again he does get back to
his starting point, and he takes six paths to get there. But in
this case the length of his journey is obviously not equal to
the perimeter of our triangle.
When we look at the length of Barney's Journey,
we see that it is obviously not equal to the perimeter of the
triangle, so we will use directed measurements to investigate
further.
Above we have labeled the direction
Barney travels inside the triangle as a positively directed measurement,
and if he travels in the opposite direction we label that as a
negatively directed measurement. If you take a look at the arrows
as Barney "bounces" inside the triangle and compare
it to the direction he travels outside you can see when we have
a positively directed measurement/movement (labeled with +), and
when Barney "bounces" in a negatively directed measurement/movement
(labeled with -). Let's find the sum of these directed measurements
and see what we get....
So we see that when we use directed
measurements to measure the length of Barney's Journey outside
our triangle, we again get the length of his journey equal to
the perimeter of our triangle.
So in summation we have found:
1) Barney will always return to his original
starting point
2) If Barney starts inside the triangle:
A) starts at midpoint of side: it will
take him three paths to return to his origin and the length of
his journey will equal 1/2 perimeter of our triangle; or he can
go around twice and the length of his journey will be equal to
the perimeter of our triangle.
B) starts at vertices of triangle:
his journey consists of 3 paths and his journey is equal to the
perimeter of our triangle, in fact his journey is the original
triangle.
C) if he starts at any other point inside
our triangle: he returns to his starting point using six
paths and the length of his journey is equal to the perimeter
of our triangle.
3) If Barney starts outside the triangle, it
takes him six paths to return to his starting point and the length
of his journey using directed measurements is equal, once again,
to the perimeter of our triangle.
CLICK
HERE if you wish to explore Bouncing
Barney further.
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