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.

Answer:

LetŐs
start from the middle of segment BC.

Barney
will hit two walls before getting back because when we draw parallels from our
starting point T, we make a triangle RST similar to triangle ABC, and 4
congruent triangles are constructed.

Let
Barney start at a point when he is (1/3) of the distance of BC from B, call
this point T.

We
get nine triangles that are all congruent and all similar to triangle
ABC. So perhaps it is 3^{2} and 2^{2} instead of 2*2 and
3*2.

Conjecture:

Based
on the previous two that we have y = n^{2} where n is the number of
segments that BC is broken down into, and y is the number of triangles formed.

So,
let Barney start at a distance that is a (1/4) of BC from point B, call this
point V.

Our
conjecture does not hold.

There
are 7 triangles and 3 trapezoids.

Conjecture:

Perimeter
of ABC = Perimeter of the dashed lines

The
area of the smaller triangles have the sides of the same length => areas are
the same.

Apparently,
the areas of the triangles are equal and they are always equal.

Let
Barney be outside the room at point P. If he starts walking, he will just
keep going on forever or until he gets tired.

So,
in this case Barney will never be inside the room.

CevaŐs
Theorem

Consider
any triangle ABC. Select a point P inside the triangle and draw lines AP, BP,
and CP extended to their intersections with the opposite sides in points D, E,
and F respectively.

1.
Explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various
locations of P.

2.
Conjecture? **Prove it!** (you may need draw some parallel lines to produce some
similar triangles, OR, you made need to consider areas of triangles within the
figure) Also, it probably helps to consider the ratio

3.
In your write-up, after the proof, you might want to discuss how this theorem
could be used for proving concurrency of the medians (if P is the centroid),
the lines of the altitudes (if P is the orthocenter), the bisectors of the
angles (if P is the incenter), or the perpendicular bisectors of the sides (if
P is the circumcenter). Concurrency of other special points?

4.
Explore a generalization of the result (using lines rather than segments to
construct ABC) so that point P can be **outside** the triangle. Show
a working GSP sketch.

Answer:

To
explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various
locations of P, click on the link below:

Clearly,
each product in the different triangles is the same, hereŐs why:

By
CevaŐs Theorem, lines AD, BE, and CF are concurrent if and only if

So
construction gives us:

Because
DBPD
and DCPD
have the same height, we have

Similarly,

It
follows that,

Similarly,

Additionally,

From
the two equations for BD/CD, we get:

BD/CD
= Area(APB)/Area(APC)

If
we had started with the other two sides instead of BC, we would have gotten the
following:

CE/AE
= Area(CPB)/Area(APB)

AF/BF
= Area(APC)/Area(CPB)

Multiplying
these three equations gives us:

Since
the cevian is the line segment joining a vertex of a triangle to any given
point on the opposite side, it is convenient to apply the CevaŐs Theorem for
proving the concurrency.

** **

This
theorem is useful to prove of the concurrency of the medians (if P is the
centroid), the lines of the altitudes (if P is the orthocenter), and the
bisectors of the angles (if P is the incenter) or other special point P.

Besides
showing the concurrency of segments, we also have AF = EA, FB = BD, DC = CE.

Say,
for instance p is outside of the triangle, click below to see a sketch of this:

Seemingly,
our property holds even if p is outside of the triangle.