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[Catalyst 2018]

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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

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