[Catalyst 2018]

the emergence of NUMBER
THEORETIC QUESTIONS
from a geometric investigation
ABSTRACT
Taking any regular even-sided polygon, one
can make cross-cuts from each vertex and
connect it to the midpoint of one of the
opposite sides to create a smaller regular
polygon contained within the original. The
ratio between the area of this new polygon
and the original has been shown to have
some interesting values, specifically 1/5 for
squares and 1/13 for hexagons. Building
on this, a general formula that gives the
ratio for an arbitrary regular polygon can
be found. In this paper, we will explore
how to generalize this result even further
by allowing the crosscuts to connect to any
side of the polygon, rather than just to the
opposite side. We then explore the rationality
of this expression, and discover interesting
connections to number theoretic techniques
looking into the relationship between
Chebyshev Polynomials and the minimal
polynomials of cosine.
INTRODUCTION
Recent results have shown that there is an
interesting relationship between the area of a
regular polygon and the area of the polygon
formed through
crosscuts.
Essentially, a
second regular
polygon is
created from
the first by
connecting
each vertex to
the midpoint
of the opposite side. Because even-sided
polygons have two sides that can be opposite
to any vertex, this implies that one side
should be chosen and that direction should
be repeated for every other vertex (See
Figure 1). The area ratio has been calculated
explicitly for squares and hexagons, and were
found to be 1/5 and 1/13, respectively. 1,5,6
In addition, for an odd-sided polygon, the
crosscuts will connect in the center, so that
the ratio will always be 0. After these specific
results, the next question is whether a
generalized formula can be found that agrees
with these, as well as gives ratios for a regular
polygon with an arbitrary number of sides. It
is also important to explore the rationality of
this function, as it is rational ratios that have
motivated this research.
Fig. 1: The crosscut formation of regular polygons when n=3, 4, 6.
INITIAL FORMULAS
The following are some necessary formulas
that will be used later in the paper. The first
is a simplified expression for the ratio of the
areas, which gives R=a S2
/a B2
,where a S
is the
apothem of the smaller polygon, and
a B
is for the original polygon. The
second is an explicit formula that
gives the ratio in terms of only the
number of sides of the polygons: R n
=
1/(1+4cot 2 (θ)) , where θ=π/n. This is
the starting point for this paper, as
this result will be generalized using
similar methods as those used in the
proof of R n
.
GENERALIZED EXPRESSION
We want this new generalization to
allow our crosscut to be connected
to an arbitrary side of the polygon,
instead of requiring that it connect to the side
opposite of the vertex where it started. This
new formula should give the area ratio only in
terms of the number of sides of the polygon,
n, and the number of vertices skipped
before connecting the crosscut, denoted k.
To do this, we first define a coordinate axis
system that is
suitable for our
calculations. This
coordinate axis
can be created
by placing one
vertex and the
center of the
polygon on the
line y=C that
runs parallel to the x-axis. Then, the crosscut
originating from this vertex would create an
angle with this line. Connecting the other
endpoint of the crosscut to the center of the
polygon forms a triangle that has one known
angle, ψ=2πk/n+(1/2)(2π/n)=(π(2k+1))/n, at
the center. (We assume in this proof that
0< ψ≤π, but we will later show that this
formula actually holds for all values of ψ.) In
addition to this known angle, two of the sides
are known, as one, a B
, is the apothem and
another, r, is the radius of the polygon. This
setup is detailed in Figure 2.
As only one angle in this triangle is known, it
is easiest to form two right triangles, so that
known trigonometric formulas more readily
apply. This is accomplished by taking a line
By Jacob Kesten
from E to E’, where is connects to the line
y=Cat a right angle. In Figure 2, this new line
is denoted b. Now using trigonometry, it is
possible to find the slope of the crosscut by
calculating the lengths of b and c. This gives
Fig. 2: A diagram depicting the two triangles used to compute the slope
of the specific crosscut. The first picture shows placement of the triangle
on the polygon and coordinate axes. The second shows how the 2 right
triangles were formed, and the third shows how this can be used to
compute the length of the smaller apothem.
the slope of the crosscut as m=(–a B
sin(ψ))/
(r–a B
cos(ψ))=(–cos(θ)sin(ψ))/(1–cos(θ)cos(ψ)).
Knowing the slope makes it possible to
write an equation for the line containing
the crosscut, as well as the equation of
the line perpendicular to the crosscut that
contains aS. These are given by y 1
=mx+mr+C
and y 2
=–x/m+C, respectively. Finding the
intersection of these two lines allows us to
find a S
2
in terms of m and a B2
. Dividing both
sides by a B2
, we get that a S2
/a B
2
= R n,k
= (m 4 +m 2 )/
(cos 2 θ(1+m 2 ) 2 ) = (1–cos 2 (ψ))/(1–2cosψcos θ
+cos 2 θ), where θ=π/n and ψ=(2k+1)θ. Notice
that this formula depends only on n and k, as
desired.
We now check that this formula holds for
values of π < ψ < 2π. This follows from the
fact that for π < ψ < 2π, ψ=π+ψ’ where ψ‘ < π,
and cosψ=cosψ’ so that the formula will be
the same as the one for the corresponding
complement angle. Essentially, the case
where π < ψ < 2π reduces to a case where
ψ≤π by looking at the angle going in the
opposite direction of the obtuse angle. An
example of this is given in Figure 3 using
different crosscut octagons. Flipping the
octagons on the bottom shows that they will
be identical to the ones in the first row, so
that they will produce identical ratios.
Now that we have found the formula R n,k
,we
can check that it matches with the values
already found for specific polygons. One such
example is when the crosscuts connect to the
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