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opposite side of an odd-sided polygon, where<br />
the ratio should be 0. In this case, we have<br />
that k=(n–1)/2, so that we get the following<br />
result: R n,k<br />
= (1–cos 2 π)/(1–2cosπ cosθ+cos 2 θ)<br />
= 0/(1–2cosθ +cos 2 θ) = 0, as expected. In<br />
addition, using n=6 and k=2, we get that<br />
R 6,2<br />
=(1–cos 2 5π/6)/(1–2cos 5π/6 cos π/6 +cos 2<br />
π/6) = (1–(–√3/2) 2 )/(1–2(–√3/2)(–√3/2)+(–√3/2) 2 )<br />
= (1– 3/4)/(1+2(3/4)+3/4) = (1/4)/(1+3/2+3/4)<br />
= (1/4)/(13/4) = 1/13, which is exactly what<br />
Fig. 3: Octagon crosscuts using k=1,2,3,4,5,6. Comparing the<br />
2 lines shows that the octagons using k=1,2,3 are the same as<br />
those using k=4,5,6.<br />
was found in earlier papers. The other known<br />
results can also be checked using a similar<br />
method.<br />
RATIONALITY<br />
Now we try to find which values of n and k<br />
will produce a rational value for R n,k<br />
. We can<br />
easily check different values of n and k and<br />
find that the following are rational: R n,0<br />
, R 4,1<br />
,<br />
R 6,2<br />
, R 6,1<br />
, R 8,2<br />
, R n(odd),(n-1)/2<br />
. However, there are an<br />
infinite number of combinations for n and k,<br />
and there is no straightforward way to prove<br />
that these are or are not the only rational<br />
values of R n,k<br />
. One approach is to look at<br />
different values for the possible parts of the<br />
ratio that could be irrational. For example,<br />
when both cosθ and cosψ are rational, then<br />
R n,k<br />
is rational. This approach works for some<br />
combinations, but a problem arises when<br />
both cos 2 θ and cos 2 ψ are irrational, as there<br />
is no way to tell what will happen with R n,k<br />
.<br />
This is because both the sum and product<br />
of irrational numbers can be rational, thus<br />
even though all components of R n,k<br />
are<br />
irrational, the final output could still be a<br />
rational number. Because this is the majority<br />
of the values that will occur, an approach that<br />
directly applies to this case is needed.<br />
One possible approach is to look at the<br />
polynomial representation for R n,k<br />
and the<br />
minimal polynomials for cosθ. The minimal<br />
polynomial of the value cosθ is the smallest<br />
nonzero, monic polynomial with rational<br />
coefficients that has cosθ as a solution. For<br />
example, the minimal polynomial of √2 is<br />
x 2 –2=0. Note that this polynomial cannot<br />
be reduced, and thus it is the smallest<br />
polynomial with √2 as a solution. All algebraic<br />
numbers have a minimal polynomial, by<br />
definition, and thus for each value of n, cosθ<br />
has a minimal polynomial. 2<br />
First, we use R n,k<br />
to create a polynomial<br />
expression of one variable. Using the multiple<br />
angle formula, we know that cosψ=cos((2k+1)<br />
θ)=cos(sθ)=T s<br />
(cosθ) where T s<br />
(cosθ) is the<br />
s th Chebyshev Polynomial. For example,<br />
cos(4θ)=8cos 4 (θ)–8cos 2 (θ)+1because T 4<br />
(x)=8x 4 –<br />
8x 2 +1. 7 This allows us to express R n,k<br />
as an<br />
expression of θ: R n,k<br />
=(1–T s2<br />
(cosθ))/(1–2T s<br />
(cosθ)<br />
cos θ +cos 2 θ, where s=2k+1. Setting x=cosθ,<br />
and R n,k<br />
=r, where r is a rational value, we get<br />
the polynomial expression P(R n,k<br />
)=1-T s<br />
2(<br />
x) –<br />
r+2rxT s<br />
(x)–rx 2 =0.<br />
It is known that if a certain value solves a<br />
polynomial, then the minimal polynomial<br />
(minpoly) of that value has to be able to divide<br />
the larger polynomial, with a remainder of<br />
0. 4 Thus we want to see what happens when<br />
we use different combinations of n and k,<br />
and see if there is any way to find a pattern<br />
in the remainders of P(R n,k<br />
)/minpoly(cosθ).<br />
In this case, a random collection of values<br />
from within the bounds n≤100 and k1<br />
–1) ∏ ψ2 (x) for n even.<br />
d<br />
d|n, d>2<br />
where ψ n<br />
(x) is the minimal polynomial of<br />
cos(2π/n) multiplied by a constant and d|n<br />
means that d is a divisor of n, so that the<br />
product runs through all the divisors of n. 3<br />
Exploring this relationship in more detail is<br />
an important and rapidly expanding area<br />
of theoretical mathematics, and could be<br />
very helpful in creating a final proof of the<br />
rationality of Rn,k. In our exploration, we<br />
came across a need to understand more<br />
about the relationship between these two<br />
concepts and have seen how a simple<br />
question concerning area ratios can produce<br />
complicated mathematical questions that<br />
are still being explored purely for theory.<br />
This is just one application of the possible<br />
knowledge that can be gained by the abstract<br />
investigation of these two concepts and the<br />
inner workings of their natural connection.<br />
A special thanks to Dr. Zsolt Lengvarszky of<br />
the Louisiana State University, Shreveport,<br />
Mathematics Department for his help and<br />
mentorship in the project.<br />
WORKS CITED<br />
[1] Ash, J.M., et al., Constructing a Quadrilateral Inside<br />
Another One, Mathematical Gazette, 2009, 528, 522–532.<br />
[2] Calcut, J.S., Rationality and the Tangent Function, http://<br />
www2.oberlin.edu/faculty/jcalcut/tanpap.pdf (accessed Feb.<br />
8, <strong>2018</strong>).<br />
[3] Gürtaş, Yusuf Z., Chebyshev Polynomials and the Minimal<br />
Polynomial of Cos(2/n), The American Mathematical Monthly,<br />
2017, 124, 74-78.<br />
[4] Leinster, T., Minimal Polynomial and Jordan Form, http://<br />
www.maths.ed.ac.uk/~tl/minimal.pdf (accessed Feb. 8, <strong>2018</strong>).<br />
[5] Mabry, R., Crosscut Convex Quadrilaterals, Math Mag,<br />
2011, 84, 16–25.<br />
[6] Mabry, R., One-Thirteenth of a Hexagon, n.d. 1-11.<br />
[7] Mason, J.C., Handscom, D.C., Chebyshev Polynomials;<br />
Chapman and Hall: Boca Raton, 2003; sec 1.2.1.<br />
DESIGN BY Kaitlyn Xiong<br />
EDITED BY Olivia Zhang<br />
CATALYST | 37