Barney is in the triangular room shown here. He walks from a point 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.
We will prove that Barney will eventually return to his
starting point and show that he will return to this point after bouncing off a
wall 6 times. Finally, we will
discuss various starting points for Barney.
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Let Barney start at an
arbitrary point D (excluding the midpoint) on BC and let him travel parallel to
AC to the point E on line AB. Let
him continue traveling parallel to point F on AC, and on in a similar fashion
to point G (as pictured above).
Now we note that thus far,
Barney’s Bouncing about has formed two parallelograms, BEFG and DEFC.
We note that the triangle EBD
is congruent to the triangle FGC.
This follows from the fact that
angle EBD is congruent to angle GFE, and the line segment BE is congruent to
the line segment GF, since opposite sides and angles of parallelograms are
Similarly, considering DEFC we
have that the angle FCG is congruent to angle DEF, and the line segment FC is
congruent to the line segment ED.
Furthermore, since the line EF
is parallel to the line BC, we have that angle DEF is congruent to angle EDB and
angle GFE is congruent to angle FGC.
And so, angle EBD is congruent
to angle FGC and angle FCG is congruent to EDB, by transitivity.
Hence, by the side angle
postulate, triangle EBD is congruent to triangle FGC.
Letting go of Barney’s leash
and allowing him to continue bouncing about we have the following:
Imagine that: Barney has traced
out another two parallelograms GHAF and HIGC.
Arguing as before we have that
triangle FGC is congruent to triangle AHI and so triangle EBD is congruent to triangle
FGC, which is congruent to triangle AHI.
Now for Barney’s last
departure, we see he must arrive at point D since if not we would have a
contradiction with respect to the congruence relations above.
So Barney bounces home on his
When will Bouncing Barney stop?
We may never know!!!!
But at least we can always find
him! We need only know how many
times he has bounced.
We already know that if Barney
has bounced 0 times he is at point D.
If he has bounced exactly1 time: point E. If he has bounced exactly 2 times: point F. Exactly 3 bounces: point G. Exactly 4 bounces: point H. Exactly 5
bounces: point I. Exactly 6
bounces: point D (Barney is home at last!).
So we have the following
relation for the nth bounce:
Barney will be at point:
D if .
E if .
F if .
G if .
H if .
I if .
We know that Barney will always
return home on his 6th bounce (or equivalently is nth bounce
congruent modulo 6) as long as home is not the midpoint, since our point was
arbitrarily chosen (excluding the midpoint).
So what happens if Barney
starts at the midpoint? Well, we
do not have to wait so long for Barney to come home. That is supposing he bounces at a constant rate.
Consider the following pictured
path of Bouncy Barney:
Now D, E, and F are all
midpoints of BC, AB, and AC respectively since:
Line AB is parallel to line DF
Line BC is parallel to line EF
Line AC is parallel to line DE
So by the Alternate Interior
and Corresponding Angle Theorem we know that each of the three triangles AEF,
EBD, and FDC have corresponding congruent angles.
Therefore, all of the triangles
are similar and so
BC = 2EF
AB = 2DF
Hence, we have that D, E, and F
are midpoints, by the Mid-segment theorem.
And so when Barney starts to
bounce about at one of the midpoints, he will return home on the 3rd
bounce. Again we could express his
return with congruence relations as in the previous case. The only difference is we will now
consider modulo 3.
The line segment joining a vertex of a triangle to any given
point on the opposite side is called a cevian. Hence, if D, F, E are points on the respective sides BC, CA,
AB of triangle ABC, the segments AE, BF, CE are cevians.
Ceva’s Theorem: If three cevians AE, BF, CE, one through
each vertex of a triangle ABC, are concurrent, then
Now, let ABC be any
triangle. 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.
Now, saying that the three line segments are concurrent is
the same thing as saying that all of the lines pass through on point, say P (as
Next, we note that the areas of triangles with equal
altitudes are proportional to the bases of the triangles. Just look at the pretty above and we
can see that
Similarly, we have that, and .
Now just follow your nose, i.e.
multiply these, and we see that
Thus proving Ceva’s Theorem.
The converse also holds!!!!!!
That is, if three cevians AD,
BE, CF satisfy
then they are concurrent.
Well, let us suppose that the first
two cevians meet at P, just as before, and that the third cevian through this
point P is CG.
Now from our previous theorem we
Wait! We supposed that
Therefore, we must have that
and so G coincides with F.
Imagine that. So we have that AD, BE, CF are